{"id":128100,"date":"2023-02-14T00:00:00","date_gmt":"2023-02-14T00:00:00","guid":{"rendered":"https:\/\/nursingstudybay.com\/2023\/02\/14\/e270-hw-6-numerical-analysis-homework-help\/"},"modified":"2023-02-14T00:00:00","modified_gmt":"2023-02-14T00:00:00","slug":"e270-hw-6-numerical-analysis-homework-help","status":"publish","type":"post","link":"https:\/\/www.studyproessays.com\/help\/e270-hw-6-numerical-analysis-homework-help\/","title":{"rendered":"E270 hw 6 | Numerical analysis homework help"},"content":{"rendered":"

<\/p>\n

\n\nNAME:<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n\nEmail:<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n\nEnter your LETTER answers HERE<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n\n\u00a0<\/td>\n\u2193<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n\n1<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\nNote:<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n\n2<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\nWhen done, using your last name and firstname, save THIS FILE as,\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n\n3<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\ne270Lastname Firstname HW6<\/td>\n(No space between E270 and Last name)<\/td>\n<\/tr>\n\n4<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\nand e-mail it to<\/td>\n[email\u00a0protected]<\/span><\/span><\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n\n5<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n\n6<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\nExample:<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n\n7<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\ne270Smith Adam HW6<\/td>\nYES<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n\n8<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\ne270 Smith Adam HW6<\/td>\nNO<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n\n9<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\"\"<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
10<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
11<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
12<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
13<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
14<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\nPAY ATTENTION!<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
15<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
16<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
17<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
18<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
19<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
20<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
1<\/td>\nWhich of the following statements about Type I and Type II errors is correct<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
a<\/td>\nType I:\u00a0 Reject a true alternative hypothesis.\u00a0 Type II: Do not reject a false alternative.<\/td>\n\u00a0<\/td>\n<\/tr>\n
b<\/td>\nType I:\u00a0 Do not reject a false null hypothesis.\u00a0 Type II: Reject a true null hypothesis.<\/td>\n\u00a0<\/td>\n<\/tr>\n
c<\/td>\nType I:\u00a0 Reject a false null hypothesis.\u00a0 Type II: Reject a true null hypothesis.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
d<\/td>\nType I:\u00a0 Reject a true null hypothesis.\u00a0 Type II: Do not reject a false null hypothesis.\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
2<\/td>\nYou are reading a report that contains a hypothesis test you are interested in. The writer of the report writes that the p-value for the test you are interested in is 0.061, but does not tell you the value of the test statistic. From this information you can:<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
a<\/td>\nNot reject the hypothesis at a Probability of Type I error = 0.05, and not reject at a Probability of Type I error = 0.10<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
b<\/td>\nReject the hypothesis at a Probability of Type I error = .05, and reject at a Probability of Type I error = 0.10<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
c<\/td>\nNot reject the hypothesis at a Probability of Type I error = .05, but reject the hypothesis at a Probability of Type I error = 0.10<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
d<\/td>\nReject the hypothesis at a Probability of Type I error = .05, but not reject at a Probability of Type I error = 0.10\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
3<\/td>\nThe random sample below is obtained to test the following hypothesis about the population mean.<\/td>\n<\/tr>\n
\u00a0<\/td>\nH\u2080:\u00a0 \u03bc\u00a0 \u2264\u00a0<\/td>\n120<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nH\u2081:\u00a0 \u03bc\u00a0 ><\/td>\n120<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n152<\/td>\n203<\/td>\n67<\/td>\n177<\/td>\n220<\/td>\n101<\/td>\n23<\/td>\n214<\/td>\n134<\/td>\n<\/tr>\n
\u00a0<\/td>\n49<\/td>\n53<\/td>\n177<\/td>\n23<\/td>\n128<\/td>\n181<\/td>\n10<\/td>\n103<\/td>\n214<\/td>\n<\/tr>\n
\u00a0<\/td>\n198<\/td>\n99<\/td>\n126<\/td>\n183<\/td>\n70<\/td>\n16<\/td>\n148<\/td>\n118<\/td>\n69<\/td>\n<\/tr>\n
\u00a0<\/td>\n182<\/td>\n166<\/td>\n199<\/td>\n59<\/td>\n172<\/td>\n40<\/td>\n177<\/td>\n28<\/td>\n42<\/td>\n<\/tr>\n
\u00a0<\/td>\n126<\/td>\n104<\/td>\n157<\/td>\n123<\/td>\n199<\/td>\n76<\/td>\n106<\/td>\n162<\/td>\n135<\/td>\n<\/tr>\n
\u00a0<\/td>\n174<\/td>\n55<\/td>\n64<\/td>\n126<\/td>\n176<\/td>\n62<\/td>\n13<\/td>\n59<\/td>\n154<\/td>\n<\/tr>\n
\u00a0<\/td>\n14<\/td>\n196<\/td>\n164<\/td>\n186<\/td>\n71<\/td>\n150<\/td>\n186<\/td>\n90<\/td>\n140<\/td>\n<\/tr>\n
\u00a0<\/td>\n177<\/td>\n189<\/td>\n209<\/td>\n50<\/td>\n26<\/td>\n233<\/td>\n16<\/td>\n28<\/td>\n135<\/td>\n<\/tr>\n
\u00a0<\/td>\n169<\/td>\n171<\/td>\n198<\/td>\n116<\/td>\n115<\/td>\n236<\/td>\n176<\/td>\n80<\/td>\n130<\/td>\n<\/tr>\n
\u00a0<\/td>\n59<\/td>\n227<\/td>\n212<\/td>\n167<\/td>\n35<\/td>\n61<\/td>\n136<\/td>\n72<\/td>\n123<\/td>\n<\/tr>\n
\u00a0<\/td>\n220<\/td>\n100<\/td>\n135<\/td>\n171<\/td>\n70<\/td>\n58<\/td>\n92<\/td>\n28<\/td>\n141<\/td>\n<\/tr>\n
\u00a0<\/td>\n52<\/td>\n27<\/td>\n181<\/td>\n138<\/td>\n231<\/td>\n80<\/td>\n115<\/td>\n153<\/td>\n187<\/td>\n<\/tr>\n
\u00a0<\/td>\n235<\/td>\n212<\/td>\n235<\/td>\n167<\/td>\n136<\/td>\n16<\/td>\n73<\/td>\n166<\/td>\n156<\/td>\n<\/tr>\n
\u00a0<\/td>\n209<\/td>\n128<\/td>\n166<\/td>\n66<\/td>\n234<\/td>\n76<\/td>\n207<\/td>\n154<\/td>\n188<\/td>\n<\/tr>\n
\u00a0<\/td>\n210<\/td>\n202<\/td>\n198<\/td>\n14<\/td>\n192<\/td>\n10<\/td>\n11<\/td>\n136<\/td>\n170<\/td>\n<\/tr>\n
\u00a0<\/td>\n214<\/td>\n231<\/td>\n28<\/td>\n94<\/td>\n125<\/td>\n214<\/td>\n31<\/td>\n64<\/td>\n72<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nThe level of significance of the test is \u03b1 = 0.05.\u00a0\u00a0 Compute the relevant test statistic.<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nThis is a(n) _______ (two-tail, upper-tail, lower-tail) test.\u00a0 The test statistic is TS = _______.<\/td>\n\u00a0<\/td>\n<\/tr>\n
a<\/td>\nUpper tail test.\u00a0\u00a0<\/td>\nTS =<\/td>\n1.34<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nDo not reject H\u2080:\u00a0 \u03bc\u00a0 \u2264 120.\u00a0 Conclude that the population mean is not greater than 120.<\/td>\n\u00a0<\/td>\n<\/tr>\n
b<\/td>\nUpper tail test.\u00a0\u00a0<\/td>\nTS =<\/td>\n1.88<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nReject H\u2080:\u00a0 \u03bc\u00a0 \u2264 120.\u00a0 Conclude that the population mean is greater than 120.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
c<\/td>\nUpper tail test.\u00a0\u00a0<\/td>\nTS =<\/td>\n1.88<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nReject H\u2080:\u00a0 \u03bc\u00a0 \u2264 120.\u00a0 Conclude that the population mean is greater than 120.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
d<\/td>\nLower tail test.<\/td>\nTS =<\/td>\n1.88<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nDo not reject H\u2080:\u00a0 \u03bc\u00a0 \u2264 120.\u00a0 Conclude that the population mean is less than 120.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
4<\/td>\nConsider the following hypothesis test.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nH\u2080:\u00a0 \u03bc\u00a0 \u2265<\/td>\n15<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nH\u2081:\u00a0 \u03bc\u00a0 <<\/td>\n15<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nA random sample of n = 15 yielded the following observations<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n8<\/td>\n7<\/td>\n11<\/td>\n11<\/td>\n8<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n13<\/td>\n8<\/td>\n12<\/td>\n13<\/td>\n5<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n21<\/td>\n21<\/td>\n19<\/td>\n15<\/td>\n18<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nUse \u03b1 =<\/span><\/td>\n0.05<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nTS = ______<\/td>\nCV = ______<\/td>\nState the decision rule.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
a<\/td>\n-1.743<\/td>\n-1.761<\/td>\nDo not reject H\u2080.\u00a0 Conclude the mean is not less than 15.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
b<\/td>\n-1.74<\/td>\n-1.64<\/td>\nReject H\u2080.\u00a0 Conclude the mean is less than 15.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
c<\/td>\n1.847<\/td>\n2.145<\/td>\nDo not reject H\u2080.\u00a0 Conclude the mean is not less than 15.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
d<\/td>\n1.847<\/td>\n1.761<\/td>\nReject H\u2080.\u00a0 Conclude the mean is less than 15.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
5<\/td>\nIn a recent study, a major fast food restaurant had a mean service time of 164 seconds.\u00a0 The company embarks on a quality improvement effort to reduce the service time and has developed improvements to the service process.\u00a0 The new process will be tested in a sample of stores.\u00a0 The new process will be adopted in all of its stores, if it resulted in decreased service time.\u00a0 To perform the hypothesis test in the previous question, the sample of 54 stores yields the following data (seconds).<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n157<\/td>\n115<\/td>\n115<\/td>\n115<\/td>\n134<\/td>\n174<\/td>\n128<\/td>\n136<\/td>\n161<\/td>\n<\/tr>\n
\u00a0<\/td>\n127<\/td>\n139<\/td>\n125<\/td>\n145<\/td>\n199<\/td>\n161<\/td>\n182<\/td>\n144<\/td>\n199<\/td>\n<\/tr>\n
\u00a0<\/td>\n156<\/td>\n117<\/td>\n129<\/td>\n193<\/td>\n173<\/td>\n146<\/td>\n128<\/td>\n166<\/td>\n185<\/td>\n<\/tr>\n
\u00a0<\/td>\n147<\/td>\n136<\/td>\n180<\/td>\n184<\/td>\n116<\/td>\n172<\/td>\n116<\/td>\n193<\/td>\n183<\/td>\n<\/tr>\n
\u00a0<\/td>\n184<\/td>\n160<\/td>\n120<\/td>\n161<\/td>\n161<\/td>\n122<\/td>\n191<\/td>\n170<\/td>\n124<\/td>\n<\/tr>\n
\u00a0<\/td>\n130<\/td>\n191<\/td>\n170<\/td>\n190<\/td>\n194<\/td>\n139<\/td>\n114<\/td>\n195<\/td>\n183<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nUse \u03b1 =<\/span><\/td>\n0.05<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n|TS| = ______<\/td>\n|CV| = ______<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
a<\/td>\n2.335<\/td>\n1.674<\/td>\nDo not reject H\u2080.\u00a0 The mean is not less than 164 seconds.\u00a0 Do not adopt the new process.<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
b<\/td>\n2.335<\/td>\n1.674<\/td>\nReject H\u2080.\u00a0 The mean is less than 164 seconds.\u00a0 Adopt the new process.<\/td>\n<\/tr>\n
c<\/td>\n1.674<\/td>\n1.349<\/td>\nReject H\u2080.\u00a0 The mean is less than 164 seconds.\u00a0 Adopt the new process.<\/td>\n<\/tr>\n
d<\/td>\n1.349<\/td>\n1.674<\/td>\nDo not reject H\u2080.\u00a0 The mean is not less than 164 seconds.\u00a0 Do not adopt the new process.<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
6<\/td>\nAccording to Kelley Blue Book, the mean price for one-to three-year-old used cars nationwide is $23,400.\u00a0 to compare the average price of similar used cars in central indiana, a random sample of 120 such cars were selected.\u00a0 The sample mean was $21,824 with a standard deviation of 7,309.\u00a0 Does the sample provide significant evidence that the mean price of one-to-three-year old used cars is different from the national mean price?<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nUse \u03b1 =<\/span><\/td>\n0.05<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
a<\/td>\np-value =<\/td>\n0.044<\/td>\nReject H\u2080.\u00a0 Conclude that the dealership’s price is different from the national mean price.<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
b<\/td>\np-value =<\/td>\n0.124<\/td>\nDo not reject H\u2080.\u00a0 Conclude that the dealership’s price is not different from the national mean price.<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
c<\/td>\np-value =<\/td>\n0.0091<\/td>\nDo not reject H\u2080.\u00a0 Conclude that the dealership’s price is not different from the national mean price.<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
d<\/td>\np-value =<\/td>\n0.0182<\/td>\nReject H\u2080.\u00a0 Conclude that the dealership’s price is different from the national mean price.<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
7<\/td>\nThe 2009 mean annual salary of business degree graduates in accounting was $47,900.\u00a0 In a follow-up study in June 2011, a sample of n = 120 graduating accounting majors yielded a sample mean of $49,500 and standard deviation of $8,200.\u00a0 Does the 2011 study provide a significant proof that the mean salary in 2011 is higher than in 2009?\u00a0 Perform this test of hypothesis at a 5% level of significance.<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
a<\/td>\np-value =<\/td>\n0.0524<\/td>\nDo not reject H\u2080.\u00a0 Conclude that the mean annual salary in 2011 is no greater than in 2009.<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
b<\/td>\np-value =<\/td>\n0.0524<\/td>\nReject H\u2080.\u00a0 Conclude that the mean annual salary in 2011 is greater than in 2009.<\/td>\n<\/tr>\n
c<\/td>\np-value =<\/td>\n0.0162<\/td>\nReject H\u2080.\u00a0 Conclude that the mean annual salary in 2011 is greater than in 2009.<\/td>\n<\/tr>\n
d<\/td>\np-value =<\/td>\n0.0162<\/td>\nDo not reject H\u2080.\u00a0 Conclude that the mean annual salary in 2011 is no greater than in 2009.<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
8<\/td>\nA production line operates with a mean filling weight of 16 ounces per container.\u00a0 Overfilling or under filling is a serious problem, and the production line should be shut down if either occurs.\u00a0 A quality control inspector samples 20 items every 2 hours and at that time makes the decision of whether to shut the line down for adjustment.\u00a0 On sample provides the following data:<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n15.8<\/td>\n16.1<\/td>\n16.2<\/td>\n16.1<\/td>\n16.1<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n16.6<\/td>\n16.3<\/td>\n16.3<\/td>\n15.9<\/td>\n16.1<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n16.2<\/td>\n15.8<\/td>\n16<\/td>\n16.3<\/td>\n15.9<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n16<\/td>\n15.9<\/td>\n16.1<\/td>\n15.9<\/td>\n16.2<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u03b1 =<\/td>\n0.05<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nDecision Rule:<\/span>\u00a0 Reject H\u2080 if <\/span>TS > <\/span>CV<\/span><\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nTS = ______<\/td>\nCV = ______<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
a<\/td>\n2.000<\/td>\n2.093<\/td>\nDo not reject H\u2080.\u00a0 Do\u00a0 not shut the line down for adjustment.<\/td>\n\u00a0<\/td>\n<\/tr>\n
b<\/td>\n2.000<\/td>\n1.96<\/td>\nReject H\u2080.\u00a0 Shut the line down for adjustment.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
c<\/td>\n1.786<\/td>\n1.729<\/td>\nReject H\u2080.\u00a0 Shut the line down for adjustment.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
d<\/td>\n2.04<\/td>\n1.64<\/td>\nDo not reject H\u2080.\u00a0 Shut the line down for adjustment.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
9<\/td>\nThe mean cholesterol level in women ages 21-40 in the United States is 190 mg\/dl.\u00a0 A study is conducted to determine the cholesterol levels among recent female Asian immigrants.\u00a0 The following is the cholesterol level of a random sample of 108 recent female Asian immigrants.<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n239<\/td>\n105<\/td>\n251<\/td>\n216<\/td>\n220<\/td>\n120<\/td>\n218<\/td>\n195<\/td>\n129<\/td>\n<\/tr>\n
\u00a0<\/td>\n125<\/td>\n196<\/td>\n193<\/td>\n108<\/td>\n178<\/td>\n187<\/td>\n111<\/td>\n176<\/td>\n178<\/td>\n<\/tr>\n
\u00a0<\/td>\n141<\/td>\n190<\/td>\n214<\/td>\n180<\/td>\n172<\/td>\n204<\/td>\n118<\/td>\n108<\/td>\n124<\/td>\n<\/tr>\n
\u00a0<\/td>\n238<\/td>\n248<\/td>\n253<\/td>\n208<\/td>\n135<\/td>\n146<\/td>\n122<\/td>\n209<\/td>\n254<\/td>\n<\/tr>\n
\u00a0<\/td>\n209<\/td>\n232<\/td>\n238<\/td>\n251<\/td>\n110<\/td>\n224<\/td>\n249<\/td>\n219<\/td>\n219<\/td>\n<\/tr>\n
\u00a0<\/td>\n124<\/td>\n226<\/td>\n252<\/td>\n189<\/td>\n212<\/td>\n163<\/td>\n205<\/td>\n202<\/td>\n190<\/td>\n<\/tr>\n
\u00a0<\/td>\n195<\/td>\n116<\/td>\n125<\/td>\n250<\/td>\n244<\/td>\n140<\/td>\n237<\/td>\n192<\/td>\n191<\/td>\n<\/tr>\n
\u00a0<\/td>\n224<\/td>\n105<\/td>\n201<\/td>\n194<\/td>\n136<\/td>\n245<\/td>\n118<\/td>\n150<\/td>\n165<\/td>\n<\/tr>\n
\u00a0<\/td>\n132<\/td>\n171<\/td>\n245<\/td>\n166<\/td>\n218<\/td>\n159<\/td>\n130<\/td>\n255<\/td>\n131<\/td>\n<\/tr>\n
\u00a0<\/td>\n185<\/td>\n210<\/td>\n223<\/td>\n153<\/td>\n167<\/td>\n174<\/td>\n239<\/td>\n200<\/td>\n107<\/td>\n<\/tr>\n
\u00a0<\/td>\n235<\/td>\n123<\/td>\n224<\/td>\n221<\/td>\n106<\/td>\n212<\/td>\n212<\/td>\n130<\/td>\n154<\/td>\n<\/tr>\n
\u00a0<\/td>\n200<\/td>\n140<\/td>\n170<\/td>\n202<\/td>\n247<\/td>\n112<\/td>\n153<\/td>\n150<\/td>\n205<\/td>\n<\/tr>\n
\u00a0<\/td>\nDoes the sample provide significant evidence that mean cholesterol level of recent female Asian immigrants is lower than the mean cholesterol level among all females in the United States?\u00a0 State the null and alternative hypotheses.\u00a0 Compute the test statistic and the p-value.\u00a0 State the decision rule.<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nRound x\u0305 to two decimal points and the standard error to three decimal points.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\np-value = ______<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
a<\/td>\n0.069<\/td>\nThe evidence is significant at \u03b1 = 0.05, but not significant at \u03b1 = 0.01.<\/td>\n\u00a0<\/td>\n<\/tr>\n
b<\/td>\n0.069<\/td>\nThe evidence is significant at \u03b1 = 0.10, but not significant at \u03b1 = 0.05.<\/td>\n\u00a0<\/td>\n<\/tr>\n
c<\/td>\n0.034<\/td>\nThe evidence is significant at \u03b1 = 0.05, but not significant at \u03b1 = 0.01.<\/td>\n\u00a0<\/td>\n<\/tr>\n
d<\/td>\n0.034<\/td>\nThe evidence is significant at \u03b1 = 0.01, but not significant at \u03b1 = 0.05.<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
10<\/td>\nWe want to test the hypothesis that mothers with low socio-economic status (SES) deliver babies whose birth weights are lower than “normal”.\u00a0 To test this hypothesis, a list is obtained of birth weights from 100 consecutive, full-term, live-born deliveries from the maternity ward of a hospital in a low-SES area.\u00a0 The mean birth weight is x\u0305 = 115 oz. with a standard deviation s = 24 oz.\u00a0 Nationwide, the mean birth weight in the United States is 120 oz.\u00a0 At \u03b1 = 0.05, does this sample provide significant evidence that the mean birth weight of babies born to mother with low SES is lower than “normal”?<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
a<\/td>\np-value =<\/td>\n0.0188<\/td>\nReject H\u2080 at the 5 percent level of significance.\u00a0 Conclude that the mean birth weight of babies born to low-SES mothers is lower than “normal”.<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
b<\/td>\np-value =<\/td>\n0.0188<\/td>\nReject H\u2080 at the 1 percent level of significance.\u00a0 Conclude that the mean birth weight of babies born to low-SES mothers is lower than “normal”.<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
c<\/td>\np-value =<\/td>\n0.0785<\/td>\nDo not reject H\u2080 at the 5 percent level of significance.\u00a0 Conclude that the mean birth weight of babies born to low-SES mothers is no lower than “normal”.<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
d<\/td>\np-value =<\/td>\n0.0785<\/td>\nDo not reject H\u2080 at the 10 percent level of significance.\u00a0 Conclude that the mean birth weight of babies born to low-SES mothers is no lower than “normal”.<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
11<\/td>\nAt Western University the historical mean scholarship examination score of entering students has been 900.\u00a0 Each year a sample of applications is taken to see whether the examination scores are at the same level as in previous years.\u00a0 The null hypothesis tested is H\u2080: \u03bc = 900.\u00a0 A sample of n = 81 students in this year\u2019s class provided a sample mean score of x\u0305 = 935 and a standard deviation of s = 180.\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nFirst build a 95% confidence interval for the population mean score.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
a<\/td>\nThe confidence interval captures \u00b5\u2080 = 900.\u00a0 Do not reject H\u2080.\u00a0 Conclude that the current mean score is not different from the historical mean score.<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
b<\/td>\nCompared to the MOE, x\u0305 \u2212 \u00b5\u2080 is within the margin of error.\u00a0 Conclude that the current mean score is not different from the historical mean score.<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
c<\/td>\nCompared to the MOE, x\u0305 \u2212 \u00b5\u2080 is outside the margin of error.\u00a0 Conclude that the current mean score is different from the historical mean score.<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
d<\/td>\nBoth a and b are correct.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
12<\/td>\nConsider the following hypothesis test.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\nH\u2080:\u00a0 \u03c0\u00a0 \u2264<\/td>\n0.5<\/td>\nH\u2081:\u00a0 \u03c0\u00a0 ><\/td>\n0.5<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nA sample of n = 200 provided a sample proportion of p\u0305 = 0.57.\u00a0 At \u03b1 = 0.05, what is your conclusion?<\/td>\n<\/tr>\n
\u00a0<\/td>\nTS = ______<\/td>\nCV = ______<\/td>\nState the decision rule.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
a<\/td>\n1.98<\/td>\n1.64<\/td>\nConclude the population proportion is no greater than 0.50.<\/td>\n\u00a0<\/td>\n<\/tr>\n
b<\/td>\n1.98<\/td>\n1.64<\/td>\nConclude the population proportion is greater than 0.50.<\/td>\n\u00a0<\/td>\n<\/tr>\n
c<\/td>\n2.98<\/td>\n1.96<\/td>\nConclude the population proportion is no greater than 0.50.<\/td>\n\u00a0<\/td>\n<\/tr>\n
d<\/td>\n2.98<\/td>\n1.96<\/td>\nConclude the population proportion is greater than 0.50.<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
13<\/td>\nIn the previous question, the prob value for the test is:<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
a<\/td>\n0.0239<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
b<\/td>\n0.0427<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
c<\/td>\n0.0618<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
d<\/td>\n0.0808<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
Next THREE questions are based on the following<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
Consider the following hypothesis test.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\nH\u2080:\u00a0 \u03c0 \u2265 0.75<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\nH\u2081:\u00a0 \u03c0 < 0.75<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
Compute the test statistic and the p-value for the following three cases.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
14<\/td>\nn =<\/td>\n200<\/td>\np\u0305 =<\/td>\n0.70<\/td>\n\u03b1 =<\/td>\n0.05<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
a<\/td>\np-value =<\/td>\n0.0258<\/td>\nConclude that the population proportion is not less than 0.75.<\/td>\n\u00a0<\/td>\n<\/tr>\n
b<\/td>\np-value =<\/td>\n0.0258<\/td>\nConclude that the population proportion is less than 0.75.<\/td>\n\u00a0<\/td>\n<\/tr>\n
c<\/td>\np-value =<\/td>\n0.0516<\/td>\nConclude that the population proportion is not less than 0.75.<\/td>\n\u00a0<\/td>\n<\/tr>\n
d<\/td>\np-value =<\/td>\n0.0516<\/td>\nConclude that the population proportion is less than 0.75.<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
15<\/td>\nn =<\/td>\n200<\/td>\np\u0305 =<\/td>\n0.70<\/td>\n\u03b1 =<\/td>\n0.10<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
a<\/td>\np-value =<\/td>\n0.0516<\/td>\nConclude that the population proportion is less than 0.75.<\/td>\n\u00a0<\/td>\n<\/tr>\n
b<\/td>\np-value =<\/td>\n0.0516<\/td>\nConclude that the population proportion is not less than 0.75.<\/td>\n\u00a0<\/td>\n<\/tr>\n
c<\/td>\np-value =<\/td>\n0.0258<\/td>\nConclude that the population proportion is less than 0.75.<\/td>\n\u00a0<\/td>\n<\/tr>\n
d<\/td>\np-value =<\/td>\n0.0258<\/td>\nConclude that the population proportion is not less than 0.75.<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
16<\/td>\nn =<\/td>\n900<\/td>\np\u0305 =<\/td>\n0.72<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
a<\/td>\np-value =<\/td>\n0.0188<\/td>\nReject H\u2080 at \u03b1 = 0.10, but do not reject at \u03b1 = 0.05.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
b<\/td>\np-value =<\/td>\n0.0188<\/td>\nReject H\u2080 at \u03b1 = 0.05, but do not reject at \u03b1 = 0.01.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
c<\/td>\np-value =<\/td>\n0.0672<\/td>\nReject H\u2080 at \u03b1 = 0.10, but do not reject at \u03b1 = 0.05.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
d<\/td>\np-value =<\/td>\n0.0672<\/td>\nReject H\u2080 at \u03b1 = 0.05, but do not reject at \u03b1 = 0.10.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
17<\/td>\nThe Center for Workforce Development found that 40% of Internet users received more than 15 e-mail messages per day in 2008.\u00a0 In 2012, a similar study on the use of e-mail was repeated.\u00a0 The purpose of the study was see whether use of e-mail has increased.\u00a0 Formulate the null and alternative hypotheses to determine whether an increase has occurred in the proportion of Internet users receiving more than 10 e-mail messages per day.<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nTo test the hypothesis at a 5% level of significance, a sample of 420 Internet users found 189 receiving more than 10 e-mail messages per day.\u00a0 Compute the test statistic and the p-value.<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\np-value = ______.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
a<\/td>\n0.0183<\/td>\nReject H\u2080 at \u03b1 = 0.10.\u00a0 Conclude the population proportion is greater than 0.40.<\/td>\n\u00a0<\/td>\n<\/tr>\n
b<\/td>\n0.0183<\/td>\nReject H\u2080 at \u03b1 = 0.05.\u00a0 Conclude the population proportion is greater than 0.40.<\/td>\n\u00a0<\/td>\n<\/tr>\n
c<\/td>\n0.0544<\/td>\nDo not reject H\u2080 at \u03b1 = 0.05.\u00a0 Conclude the population proportion is not greater than 0.40.<\/td>\n<\/tr>\n
d<\/td>\nBoth a and b are correct.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
18<\/td>\nWe want to test the hypothesis that at least<\/span> 75% of drivers on a freeway violate the speed limit.\u00a0 In a random sample of n = 900 vehicles, 657 violated the speed limit.\u00a0 Compute the sample proportion.<\/span><\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\nState the null and alternative hypotheses and the decision rule.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
a<\/td>\nReject H\u2080 at \u03b1 = 10% and conclude less than 75% of drivers violate the speed limit.\u00a0 But, do not reject H\u2080 at \u03b1 = 5% and conclude 75% or more of drivers violate the speed limit.<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
b<\/td>\nReject H\u2080 at \u03b1 = 5%.\u00a0 Conclude less than 75% of drivers violate the speed limit.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
c<\/td>\nReject H\u2080 at \u03b1 = 5%.\u00a0 Conclude more than 75% of drivers violate the speed limit.<\/td>\n\u00a0<\/td>\n<\/tr>\n
d<\/td>\nReject H\u2080 at \u03b1 = 1%.\u00a0 Conclude less than 75% of drivers violate the speed limit.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
19<\/td>\nAt least 20% of all workers are believed to be willing to work fewer hours for less pay to obtain more time for personal and leisure activities.\u00a0 A recent poll consisting of 600 respondents found 17% willing to work fewer hours for less pay to obtain more personal and leisure time.\u00a0 At 5% level of significance, does the sample result support the claim that at least 20% of all workers are willing to work fewer hours for less pay to obtain more time for personal and leisure activities? Round the proportion to two decimal point.<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n<\/tr>\n
a<\/td>\n1.64<\/td>\n1.84<\/td>\nDo not reject H\u2080.\u00a0 Conclude that no less than 20% are willing to work fewer hours for less pay to obtain more time for personal and leisure activities.<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
b<\/td>\n1.84<\/td>\n1.96<\/td>\nDo not reject H\u2080.\u00a0 Conclude that no less than 20% are willing to work fewer hours for less pay to obtain more time for personal and leisure activities.<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
c<\/td>\n1.84<\/td>\n1.64<\/td>\nReject H\u2080.\u00a0 Conclude that less than 20% are willing to work fewer hours for less pay to obtain more time for personal and leisure activities.<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
d<\/td>\n1.56<\/td>\n1.64<\/td>\nDo not reject H\u2080.\u00a0 Conclude that no less than 20% are willing to work fewer hours for less pay to obtain more time for personal and leisure activities.<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
20<\/td>\nIn the previous question, the p-value is _______.<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
a<\/td>\n0.0594<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
b<\/td>\n0.0329<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
c<\/td>\n0.0233<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n
d<\/td>\n0.0158<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n\u00a0<\/td>\n<\/tr>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

NAME: \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Email: \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Enter your LETTER answers HERE \u00a0 \u00a0\u2026<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[10601,10596,10882,479,537,172,478,10597,10599,10595],"tags":[10524,10522,10890,10561,10883,10889,10515,10521,10696,10518,10888,10887,10886,10885,10884,10525],"class_list":["post-128100","post","type-post","status-publish","format-standard","hentry","category-assignment-help-by-uks-no-1-writing-service","category-assignment-writers-canada-university-cost","category-free-ai-english-assignment-writers-china","category-nursing-course-hero","category-nursing-essay-pro","category-nursing-study-bay","category-nursing-study-pool","category-professional-assignment-writers-usa","category-uae-1-cheap-assignment-writing-service","category-write-a-word-essay","tag-apa-citation-format-assignment","tag-assignment-writers-australia-college-student","tag-best-nursing-case-study-writing-services","tag-buy-essay-usa","tag-case-study-nursing-writing-services-for-care-plans","tag-free-nursing-case-study-writing-services","tag-help-write-a-page-assignment","tag-i-need-someone-to-do-my-assignment-within-hours","tag-need-assistance-completing-the-assignment","tag-need-to-write-an-essay","tag-nursing-case-studies-with-answers","tag-nursing-case-study-paper-example","tag-nursing-case-study-template-examples-of-case-studies-nursing","tag-pharmacology-case-study","tag-soap-note-case-study","tag-write-pages"],"_links":{"self":[{"href":"https:\/\/www.studyproessays.com\/help\/wp-json\/wp\/v2\/posts\/128100","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.studyproessays.com\/help\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.studyproessays.com\/help\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.studyproessays.com\/help\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.studyproessays.com\/help\/wp-json\/wp\/v2\/comments?post=128100"}],"version-history":[{"count":0,"href":"https:\/\/www.studyproessays.com\/help\/wp-json\/wp\/v2\/posts\/128100\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.studyproessays.com\/help\/wp-json\/wp\/v2\/media?parent=128100"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.studyproessays.com\/help\/wp-json\/wp\/v2\/categories?post=128100"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.studyproessays.com\/help\/wp-json\/wp\/v2\/tags?post=128100"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}