the mean GPA of night students is smalle ean GPA of day students at the 0.10 significance level. The null and alternative hypothesis would be: µD Ho:µN HA: HN < HD HẠ:µN > µD HẠ:µN + HD Ho:HN µD Ho:HN = HD The test is: left-tailed right-tailed two-tailed The sample consisted of 30 night students, with a sample mean GPA of 3.2 and a standard deviation of 0.5, and 40 day students, with a sample mean GPA of 3.4 and a standard deviation of 0.3.

the mean GPA of night students is smalle ean GPA of day students at the 0.10 significance level. The null and alternative hypothesis would be: µD Ho:µN HA: HN < HD HẠ:µN > µD HẠ:µN + HD Ho:HN µD Ho:HN = HD The test is: left-tailed right-tailed two-tailed The sample consisted of 30 night students, with a sample mean GPA of 3.2 and a standard deviation of 0.5, and 40 day students, with a sample mean GPA of 3.4 and a standard deviation of 0.3.

Name: Test the claim that the mean GPA of night students is smaller than th
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: Test the claim that the mean GPA of night students is smaller than the mean GPA of day students at the0.10 significance level. The null and alternative hypothesis would be:Ho: HN = µD Họ:µN = µD Họ:µN = µDHA: HN µD HA:µN + µDThe test is

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 54 men and 70 women to participate in the study. Each subject was required to step up and down a 6-inch platform. The pulse of each subject was then recorded. The following results were obtained. Two sample T for Men vs Women N Mean StDev SE Mean Men Women 98% CI for mu Men - mu Women (- 12.20, - 1.00) T-Test mu Men = mu Women (vs H2 O C. Ho: H1 = H2; Ha: H1 #H2 (b) Identify the P-value and state the researcher's conclusion if the level of significance was a = 0.01. What is the P-value? P-value =
With 0.05 significance level, H1: p > 0.5, and the value of test statistics is z = 1.25. Then the P-value is equal to 0.8944 -0.1056 0.1056 None
Test the claim that the proportion of men who own cats is smaller than the proportion of women who own cats at the .05 significance level.The null and alternative hypothesis would be: H0:μM=μFH0:μM=μFH1:μM≠μFH1:μM≠μF H0:pM=pFH0:pM=pFH1:pM>pFH1:pM>pF H0:pM=pFH0:pM=pFH1:pM<pFH1:pM<pF H0:pM=pFH0:pM=pFH1:pM≠pFH1:pM≠pF H0:μM=μFH0:μM=μFH1:μM<μFH1:μM<μF H0:μM=μFH0:μM=μFH1:μM>μFH1:μM>μF Incorrect The test is: left-tailed two-tailed right-tailed Incorrect Based on a sample of 20 men, 35% owned catsBased on a sample of 80 women, 40% owned catsThe test statistic is: Incorrect (to 2 decimals)The critical value is: Incorrect (to 2 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesis Incorrect Submit QuestionQuestion 6
Test the claim that the proportion of men who own cats is significantly different than 20% at the 0.02 significance level. The null and alternative hypothesis would be: Ho:µ = 0.2 Ho:H = 0.2 Ho:p= 0.2 Ho:µ = 0.2 Ho:p= 0.2 Ho:p= 0.2 H1: µ > 0.2 H1:470.2 H1:p7 0.2 H1:µ 0.2 The test is: left-tailed right-tailed two-tailed Based on a sample of 4 people, 25% owned cats The test statistic is: (to 2 decimals) The positive critical value is: (to 2 decimals) Based on this we: O Reject the null hypothesis O Fail to reject the null hypothesis Check Answer
Test the claim that the mean GPA of night students is significantly different than 2.5 at the 0.1 significance level. The null and alternative hypotheses would be: Ho p=0.625 Hop=2.5 Ho: p=0.625 Ho: : H₁: p > 0.625 The test is: right-tailed left-tailed two-tailed о :p = 0.625 Hoμ = 2.5 H₁ μ = 2.5 o H₁:μ> 2.5 H₁: p < 0.625 H₁: p0.625 H₁: μ2.5 H₁: μ< 2.5 Based on a sample of 35 people, the sample mean GPA was 2.47 with a standard deviation of 0.06 The p-value is: -1.645 The positive significance level is: .10 Based on this we: OTT Reject the null hypothesis Fail to reject the null hypothesis X (to 3 decimals) o(to 2 decimals)
You wish to test the following claim (Ha) at a significance level of a 0.01. Ho:H1 = 42 Ha:H1 > H2 You believe both populations are normally distributed, but you do not know the standard deviations for either. However, you also have no reason to believe the variances of the two populations are not equal. You obtain a sample of size n1 = 25 with a mean of M1 = 52.5 and a standard deviation of SD, = 9.1 from the first population. You obtain a sample of size n2 = 18 with a mean of M2 = 48.7 and a standard deviation of SD2 = 5.2 from the second population. 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 conservative under-estimate for the degrees of freedom as mentioned in the textbook. (Report answer accurate to four decimal places.) p-value = The p-value is... O less than (or equal to) a O greater than a This test statistic leads to a decision to... O reject the…
Find the value of the standard score, z, and determine whether to reject the null hypothesis at a 0.01 significance level. Is the alternative hypothesis supported? Ho p= 18.5 meters, H p18.5 meters, n= 144, x = 19.1 meters, o = 1.3 meters %3D The value of the standard score is (Round to two decimal places as needed.)
Test the claim that the mean GPA of night students is significantly different than 2.4 at the 0.2 significance level.The null and alternative hypotheses would be: H0:p=0.6H0:p=0.6H1:p≠0.6H1:p≠0.6 H0:μ=2.4H0:μ=2.4H1:μ≠2.4H1:μ≠2.4 H0:μ=2.4H0:μ=2.4H1:μ>2.4H1:μ>2.4 H0:p=0.6H0:p=0.6H1:p<0.6H1:p<0.6 H0:μ=2.4H0:μ=2.4H1:μ<2.4H1:μ<2.4 H0:p=0.6H0:p=0.6H1:p>0.6H1:p>0.6 The test is: right-tailed left-tailed two-tailed Based on a sample of 31 people, the sample mean GPA was 2.36 with a standard deviation of 0.16The p-value is: (to 3 decimals)The positive significance level is: (to 2 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesis
Use the pulse rates in beats per minute (bpm) of a random sample of adult females listed in the data set available below to test the claim that the mean is less than 79 bpm. Use a 0.05 significance level. LOADING... Click the icon to view the pulse rate data. Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses? A. H0: μ=79 bpm H1: μ≠79 bpm B. H0: μ>79 bpm H1: μ<79 bpm C. H0: μ=79 bpm H1: μ<79 bpm D. H0: μ=79 bpm H1: μ>79 bpm Determine the test statistic. nothing (Round to two decimal places as needed.) Determine the P-value. nothing (Round to three decimal places as needed.) State the final conclusion that addresses the original claim. ▼ Reject Fail to reject H0. There is ▼ not sufficient sufficient evidence to conclude that the mean of the population of pulse rates for adult females is ▼ greater than not less than…
You wish to test the following claim (Ha) at a significance level of a = 0.001. H.:µ1 = µ2 Ha:µ1 # µ2 You believe both populations are normally distributed, but you do not know the standard deviations for either. However, you also have no reason to believe the variances of the two populations are not equal (So use the "pool" option). You obtain the following two samples of data. Sample #1 Sample #2 86.4 60.5 63.3 93.5 94.6 61.1 99.2 55.9 85.8 107 88.9 68.8 88.3 61.9 78.8 59.2 103.8 99.7 74.3 72.5 76.5 62 70.5 80.7 80.2 59.2 64.3 107 89.3 113.2 48.5 95.2 72.1 76.1 115.5 75.1 97.6 88.4 96.4 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 P-value reported from the "2-sample t- test" from the technology you are using. (Report answer accurate to four decimal places.) p-value = The p-value is... O less than (or equal to) a greater than a This test statistic leads…
Test the claim that the mean GPA of night students is significantly different than 2.3 at the 0.05 significance level.The null and alternative hypothesis would be: H0:μ=2.3H0:μ=2.3H1:μ>2.3H1:μ>2.3 H0:p=0.575H0:p=0.575H1:p>0.575H1:p>0.575 H0:μ=2.3H0:μ=2.3H1:μ<2.3H1:μ<2.3 H0:p=0.575H0:p=0.575H1:p≠0.575H1:p≠0.575 H0:μ=2.3H0:μ=2.3H1:μ≠2.3H1:μ≠2.3 H0:p=0.575H0:p=0.575H1:p<0.575H1:p<0.575 The test is: two-tailed left-tailed right-tailed Based on a sample of 80 people, the sample mean GPA was 2.27 with a standard deviation of 0.02The test statistic is: (to 2 decimals)The positive critical value is: (to 2 decimals)Based on this we: Fail to reject the null hypothesis Reject the null hypothesis
You work at Happy Joe's family restaurant and want to see if customer meal satisfaction and gender are related to one another. You take a sample of customers and ask them if they were satisfied with their meal and note their gender. What is the appropriate conclusion at the 0.05 level of significance? Unhappy Indifferent Happy Male 23 24 21 Cross Tabulation of Satisfaction vs. Gender Female Pearson Chi-square= 9.879 DF = 4 P-value = 0.0425 26 16 16 Unknown 11 28 14