The following MINITAB output presents the results of a hypothesis test for a populationproportion p.Test and CI for One Proportion: XTest of p = 0.4 vs p < 0.4Variable X N Sample p95% Upper Bound Z-Value P-Valueх73 240 0.3041670.353013-3.030.001Is this a one-tailed or two-tailed test?What is the null hypothesis?Can H, be rejected at the 2% level? How can you tell?a.b.C.d.Someone asks you whether the null hypothesis Hg: p 2 0.45 versus H1: p < 0.45 can berejected at the 2% level. Can you answer without doing any calculations? How?Use the output and an appropriate table to compute the P-value for the test of Ho: p s0.25 versus H1: p > 0.25.Use the output and an appropriate table to compute a 90% confidence interval for p.e.f.

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Answered: The following MINITAB output presents…

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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If the p-value for a chi-square goodness of fit test is 0.06, and significance level is alpha = 0.10, then: If the p-value for a chi-square goodness of fit test is 0.06, and significance level is alpha = 0.10, then: we do not reject the null hypothesis because the p-value is greater than 0.05. we do not reject the null hypothesis. we reject the null hypothesis. the chi-square goodness of fit statistic value is 0.06. we have insufficient information to make a decision.
Suppose a p-value of .003 is found from a hypothesis test. At the alpha=.05 level of significance, should the null be rejected? Reject the Null Do not not reject the null
Which of the following relates to the acceptance of the null hypothesis when it is actually false?1. Type I error 2. Type II error 3. Power 4. Statistical significance
(a) State the alternative hypothesis.(b) State the type of the test required and its rejection area at 0.05 significance level.(c) Calculate the test statistic, Z for the population proportion.
The p-value for a hypothesis test turns out to be 0.00694. At a 9% level of significance, what is the proper decision? Reject H0H0 Fail to reject H0
I randomly sample 219 students and ask them to provide their USF GPA. I wish to determine if the average GPA of all USF students ex mu + 3.25 ceeds 3.25. Unfortunately, I generated the printout shown here: One-Sample T Test Null Hypothesis: H - 3.25 Alternative Hyp: u- 3.25 Variable Mean SE DF GPA 3.3121 0.0340 1.82 218 0.0694 Cases Included 219 What assumption is needed for this analysis to be valid? None. The Central Limit Theorem applies and no other assumptions are needed. The GPAS of all USF students need to be approximately normally distributed The mean GPA of all USF students need to be approximately normally distributed
State the Result: A hypothesis test was conducted at the alpha = 0.01 level of significance. The test resulted in a p-value of 0.044.
Under what circumstances is a t statistic used instead of a z-score for a hypothesis test? Justin wants to know whether a commonly prescribed drug does improve the attention span of students with attention deficit disorder (ADD). He knows that the mean attention span for students with ADD who are not taking the drug is 2.3 minutes long. His sample of 12 students taking the drug yielded a mean of 4.6 minutes. Justin can find no information regarding σx , so he calculated s2x =1.96. Determine the critical region using a one-tailed test with alpha = .05. Conduct the hypothesis test (Do the math and compare the t-critical and t-obtained values). State your conclusions in terms of H0 (Should you reject the H0 or fail to reject/accept the H0). Based on your analysis, is there a relationship between the drug and attention span?
A hypothesis test on a regression coefficient fails to reject the null hypothesis at the 5% significance level. This means:
Test the claim that the mean GPA of night students is larger than 2.5 at the 0.005 significance level.The null and alternative hypothesis would be: H0:p=0.625H0:p=0.625H1:p≠0.625H1:p≠0.625 H0:μ=2.5H0:μ=2.5H1:μ≠2.5H1:μ≠2.5 H0:p≤0.625H0:p≤0.625H1:p>0.625H1:p>0.625 H0:μ≤2.5H0:μ≤2.5H1:μ>2.5H1:μ>2.5 H0:p≥0.625H0:p≥0.625H1:p<0.625H1:p<0.625 H0:μ≥2.5H0:μ≥2.5H1:μ<2.5H1:μ<2.5 The test is: right-tailed two-tailed left-tailed Based on a sample of 75 people, the sample mean GPA was 2.51 with a standard deviation of 0.05The 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
Three students, an athlete, a fraternity member, and an honors student, record the number of hours they slept each night for 20 nights. O JMP Applet imp ? Oneway Analysis of Sleep Hours By Student Oneway Analysis of Sleep Hours By Student 10 Oneway Anova 14 Summary of Fit 12 Rsquare 0.024506 10- Adj Rsquare Root Mean Square Error Mean of Response Observations (or Sum Wgts) -0.00072 1.99517 7.7 60 Analysis of Variance Sum of Mean F Prob > Source DF Squares Square Ratio F 2 Athiete Frat Honors Student 2 5.70000 2.85000 0.7180 0.4931 Student Error 57 226.90000 3.98070 C. Total 59 232.60000 Oneway Anova Means for Oneway Anova Std Lower Upper 95% Summary of Fit Level Number Mean Error 95% Athlete 20 8.10000 0.44813 7.2086 8.9934 Rsquare 0.024506 Frat 20 7.65000 0.44813 6.7588 8.5434 Adj Rsquare Root Mean Square Error Mean of Response Observations (or Sum Wgts) -0.00972 Honors 20 7.35000 0.44813 6.4586 8.2434 1.99517 Std Error uses a pooled estimate of error variance 7.7 60 Analysis of…
Using a sample that has n=50 pair of data, we get a correlation coefficient r=-0.269 for the following null and alternative hypotheses determine and compute the test statistic value: HO: p=0 H1: pe 0 O a t=7.51 O b. z=7.51 O o.z=-0. 16 Odt=-1.08

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