Test the claim that the mean GPA of night students is smaller than the mean GPA of day students atthe 0.005 significance level.The null and alternative hypothesis would be:O Ho: PN = PDH₁: PN #PDOHO NZ HDΗ:μ. < μεOHO: UNH₁: UNDDHo: PNPDH₁: PN > PDO Ho: PNZ PDH₁: PN PDOHO: UN ≤ HDH₁: UN > D The test is:O left-tailedOright-tailedO two-tailedThe sample consisted of 50 night students, with a sample mean GPA of 2.47 and a standarddeviation of 0.07, and 50 day students, with a sample mean GPA of 2.51 and a standard deviation of0.02.The test statistic is:The p-value is:Based on this we:O Fail to reject the null hypothesisO Reject the null hypothesis(to 2 decimals)(to 2 decimals)

Answered: Test the claim that the mean GPA of…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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a. For this study, we should use t-test for a population mean b. The null and alternative hypotheses would be: Ho: uv (Please enter a decimal) c. The test statistic z v (please show your answer to 3 decimal places.) %3D d. The p-value = (Please show your answer to 4 decimal places.) %3D e. The p-value is> va f. Based on this, we should fail to reject vthe null hypothesis. g. Thus, the final conclusion is that ...
Test the claim that the mean GPA of night students is smaller than the mean GPA of day students at the 0.01 significance level.The null and alternative hypothesis would be: H0:μN=μDH0:μN=μDH1:μN≠μDH1:μN≠μD H0:pN≥pDH0:pN≥pDH1:pN<pDH1:pN<pD H0:μN≤μDH0:μN≤μDH1:μN>μDH1:μN>μD H0:pN=pDH0:pN=pDH1:pN≠pDH1:pN≠pD H0:μN≥μDH0:μN≥μDH1:μN<μDH1:μN<μD H0:pN≤pDH0:pN≤pDH1:pN>pDH1:pN>pD The test is: left-tailed two-tailed right-tailed The sample consisted of 35 night students, with a sample mean GPA of 2.23 and a standard deviation of 0.05, and 35 day students, with a sample mean GPA of 2.24 and a standard deviation of 0.07.
Test the claim that the mean GPA of night students is smaller than 2.1 at the 0.025 significance level.The null and alternative hypothesis would be: H0:μ=2.1H0:μ=2.1H1:μ≠2.1H1:μ≠2.1 H0:p≤0.525H0:p≤0.525H1:p>0.525H1:p>0.525 H0:μ≤2.1H0:μ≤2.1H1:μ>2.1H1:μ>2.1 H0:p≥0.525H0:p≥0.525H1:p<0.525H1:p<0.525 H0:μ≥2.1H0:μ≥2.1H1:μ<2.1H1:μ<2.1 H0:p=0.525H0:p=0.525H1:p≠0.525H1:p≠0.525 The test is: right-tailed two-tailed left-tailed Based on a sample of 40 people, the sample mean GPA was 2.06 with a standard deviation of 0.04The test statistic is: (to 2 decimals)The p-value is: (to 2 decimals)Based on this we: Fail to reject the null hypothesis Reject the null hypothesis
Test the claim that the mean GPA of night students is smaller than the mean GPA of day students at the 0.005 significance level.The null and alternative hypothesis would be: H0:pN=pDH1:pN≠pD H0:pN≥pD H1:pN<pD H0:μN≥μD H1:μN<μD H0:μN≤μD H1:μN>μD H0:pN≤pDH1:pN>pD H0:μN=μDH1:μN≠μD The test is: two-tailed left-tailed right-tailed The sample consisted of 70 night students, with a sample mean GPA of 2.16 and a standard deviation of 0.07, and 70 day students, with a sample mean GPA of 2.18 and a standard deviation of 0.06. The test statistic is: (to 2 decimals)The p-value is: (to 2 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesis
You wish to test the following claim (HaHa) at a significance level of α=0.01α=0.01. Ho:μ1=μ2Ho:μ1=μ2 Ha:μ1<μ2Ha:μ1<μ2You obtain the following two samples of data. Sample #1 Sample #2 52 60.1 61.7 56.9 52 59.7 45 49.9 71.9 31.1 47.1 56.1 54.9 45 85.1 67 66.1 51.2 87.2 50.7 73.1 43.3 65.2 58.5 50.3 71.3 60.5 48 62.2 78.6 50.7 56.5 37.8 45.6 41.5 58.5 54.9 76.8 62.6 29.5 71.3 83.5 71.3 64.3 63 74.5 76.8 50.7 78.6 59.3 31.1 54.9 61.2 50.3 75.1 72.3 68.5 85.7 63.3 50.3 79.1 73.5 65.2 58.4 62.3 57 52.9 72.3 88.2 57.4 75.8 68.5 84.1 45 60 79.7 58.1 56.6 77.1 60.9 68.5 67.3 62 45 66.4 79.7 53.5 52.9 79.7 63.5 67.1 74.8 53.5 66.1 68.7 64.2 72.3 55.2 75.5 What is the test statistic for this sample? (Report answer accurate to three decimal places.)test statistic = What is the p-value for this sample? For this calculation, use the degrees of freedom reported from the technology you are using. (Report answer…
Test the claim that the mean GPA of night students is larger than 2.3 at the .025 significance level. The null and alternative hypothesis would be: Но : р — 2.3 Но:н — 2.3 Но:р %3D 0.575 Но: д —— 2.3 Но:р %3D 0.575 Но:р 3 0.575 H:р + 2.3 Hi:p 2.3 Н:р#0.575 H:р> 0.575 The test is: left-tailed right-tailed two-tailed Based on a sample of 65 people, the sample mean GPA was 2.34 with a standard deviation of 0.06 The test statistic is: (to 2 decimals) The critical value is: (to 2 decimals) Based on this we: O Fail to reject the null hypothesis O Reject the null hypothesis
Test the claim that the mean GPA of night students is smaller than the mean GPA of day students at the 0.01 significance level.The null and alternative hypothesis would be: H0:μN=μDH0:μN=μDH1:μN≠μDH1:μN≠μD H0:μN≤μDH0:μN≤μDH1:μN>μDH1:μN>μD H0:pN=pDH0:pN=pDH1:pN≠pDH1:pN≠pD H0:pN≥pDH0:pN≥pDH1:pN<pDH1:pN<pD H0:μN≥μDH0:μN≥μDH1:μN<μDH1:μN<μD H0:pN≤pDH0:pN≤pDH1:pN>pDH1:pN>pD The test is: left-tailed right-tailed two-tailed The sample consisted of 45 night students, with a sample mean GPA of 3.23 and a standard deviation of 0.05, and 45 day students, with a sample mean GPA of 3.26 and a standard deviation of 0.08.The test statistic is: (to 2 decimals)The p-value is: (to 2 decimals)Based on this we:
Listed below are the lead concentrations in ug/g measured in different traditional medicines. Use a 0.05 significance level to test the claim that the mean lead concentration for all such medicines is less than 13 ug/g. Assume that the lead concentrations in traditional medicines are normally distributed. 12 11 12 3.5 4.5 18 18 3 15 5 0 What are the null and alternative hypotheses? Ο Α. H : μ= 13 μα/g Η: με 13 μg/g O C. Ho μ= 13 μgig Η: μ> 13 μg/g O B. Ho: µ> 13 µg/g Η: με 13 μαα O D. Ho: u= 13 µg/g Η : μ# 13 μg/9
Test the claim that the mean GPA of night students is larger than 2.2 at the 0.10 significance level.The null and alternative hypothesis would be: H0:p≤0.55H0:p≤0.55HA:p>0.55HA:p>0.55 H0:μ=2.2H0:μ=2.2HA:μ≠2.2HA:μ≠2.2 H0:μ≥2.2H0:μ≥2.2HA:μ<2.2HA:μ<2.2 H0:p=0.55H0:p=0.55HA:p≠0.55HA:p≠0.55 H0:p≥0.55H0:p≥0.55HA:p<0.55HA:p<0.55 H0:μ≤2.2H0:μ≤2.2HA:μ>2.2HA:μ>2.2 The test is: left-tailed two-tailed right-tailed Based on a sample of 35 people, the sample mean GPA was 2.25 with a standard deviation of 0.03The p-value is: (to 2 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesis
Test the claim that the proportion of men who own cats is smaller than the proportion of women who own cats at the .10 significance level. The null and alternative hypothesis would be: Но: дм — Мр Но: рм — МF Но: рм — Pr Но: рм — рF Но: им — МЕ Но: рм — РF PF Ho:HM = µf Ho:PM = Pf H1: µM + HF H1:µM PF H1:µM > µF H1:PM < PF The test is: left-tailed right-tailed two-tailed Based on a sample of 60 men, 35% owned cats Based on a sample of 40 women, 55% owned cats The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: Fail to reject the null hypothesis Reject the null hypothesis
Test the claim that the mean GPA of night students is smaller than the mean GPA of day students at the 0.025 significance level.The null and alternative hypothesis would be: H0:μN=μDH0:μN=μDH1:μN≠μDH1:μN≠μD H0:pN≤pDH0:pN≤pDH1:pN>pDH1:pN>pD H0:μN≤μDH0:μN≤μDH1:μN>μDH1:μN>μD H0:pN≥pDH0:pN≥pDH1:pN<pDH1:pN<pD H0:pN=pDH0:pN=pDH1:pN≠pDH1:pN≠pD H0:μN≥μDH0:μN≥μDH1:μN<μDH1:μN<μD The test is: left-tailed two-tailed right-tailed The sample consisted of 35 night students, with a sample mean GPA of 3.34 and a standard deviation of 0.06, and 35 day students, with a sample mean GPA of 3.36 and a standard deviation of 0.03.The test statistic is: _____ (to 2 decimals)The p-value is: ____ (to 2 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesis
Test the claim that the proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.02 significance level. The null and alternative hypothesis would be: Но: HF Ho: UM = µF Ho:PM — pF Ho:рм PF Ho: им — MF Ho:; PF Нi: рм > иF Hi: рм # pF Hi:рм рF Hi: рм < pF Hi:рм # pғ The test is: right-tailed two-tailed left-tailed Based on a sample of 80 men, 25% owned cats Based on a sample of 40 women, 50% owned cats The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: O Reject the null hypothesis O Fail to reject the null hypothesis

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