When results are not statistically significant, why do you conclude that it is plausible that the two variables are independent rather than concluding that the variables are different independent?
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Q: Test the claim that the mean GPA of night students is larger than 2.1 at the 0.005 significance…
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Q: The null and alternative hypothesis would be: Но:р — 0.55 Но: р — 2.2 Но:р — 0.55 Но: и — Ho:P 0.55…
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Q: Test the claim that the mean GPA of night students is significantly different than 2.4 at the 0.02…
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A: Introduction: Denote p as the true proportion of people who owned cats.
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Q: Test the claim that the mean GPA of night students is significantly different than 2.2 at the 0.01…
A: The claim is the mean GPA of night students is significantly different than 2.2.
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- A researcher uncovers that there is a significant interaction between the factor of marital status (i.e., married or non-married) and participant sex (i.e., male or female) regarding well-being. A researcher decides to compare the difference in well-being between married men and women. What would be the null hypothesis for this comparison? a. µMale ≠ µFemale for married individuals b. µMale ≠ µFemale for non-married individuals c. µMale = µFemale for non-married individuals d. µMale = µFemale for married individualsDoes college major depend on gender? Researchers ask a group of college students about their majors. They also group the students by their gender. The data are recorded in the contingency table below, and a chi-square Test of Independence at the 5% significance level is performed. Arts Humanities Sciences Row Total Men 11 5 19 35 Women 11 22 9 42 Column Total 22 27 28 77 (a) The null and alternative hypotheses are: H0: The two variables are independent, so gender does not affect college major. Ha: The two variables are dependent, so gender does affect college major. (b) Compute the test statistic, rounded to one decimal place. (Use expected frequencies that are also rounded to one decimal place.)Test the claim that the proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.1 significance level. The null and alternative hypothesis would be: = Ho: PM = μF Ho: PM =PF Ho: M = μF Ho: PM = PF Ho: M H₁: μM ‡µF H₁: PM>PF H₁₂ : µM > µF H₁: PM/PF H₁: µM < µF H₁: PMTest 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 hypothesisTest the claim that the proportion of men who own cats is larger than 60% at the .10 significance level. The null and alternative hypothesis would be: Но: и 3D 0.6 Но:р%3D 0.6 Но:и %—D 0.6 Но:р — 0.6 Но:р — 0.6 Но: д — 0.6 Нi:р > 0.6 Hі:р 0.6 Hi:p + 0.6 Hi:д < 0.6 The test is: left-tailed two-tailed right-tailed Based on a sample of 80 people, 62% owned cats The test statistic is: (to 2 decimals) The critical value is: (to 2 decimals) Based on this we: O Reject the null hypothesis O Fail to reject the null hypothesisIn reading the results of a multiple regression analysis that contained 4 predictor variables, the researcher noticed a column labeled Beta. Two of the Beta’s were positive and two were negative. He concluded that a.) Beta’s that were positive were statistically significant b.) Beta’s that were positive had more of an effect c.) Beta’s that were positive were associated with increases in the criterion variable d.) Beta’s that were positive did not affect the criterion because they were “controlled for…”Looking at youth depression scores from a general sample of random children in the US, we want to determine if our sample is significantly different than the national average of scores of children with depression in the US. The national average of scores on a measure of depression is 5 points. We have no reason to assume it will be higher or lower than the US national average. The sample: 11 9 12 3 3 4 13 1. What is the null and alternative hypotheses? Do not specify any directionality. 2. What did you discover (using 95% Cl)? Provide an interpretation.A hypothesis test on a regression coefficient fails to reject the null hypothesis at the 5% significance level. This means:A sociologist wishes to study whether the rank attained by officers in the New York City Police Department (NYPD) is related to gender. Data have been collected for both male and female officers in NYPD. The summary data and test for independence results from StatCrunch are given below. At a= 0.05 level of significance, draw a conclusion about whether these data indicate that the rank of the officer is independent of gender.(a) Is this an observational study or experiment?(b) What are the explanatory and response variables, respectively?Explanatory variable:Response variable:(c) Write down the null and alternative hypotheses of the test for independence.Contingency table results:Rows: RANKColumns: GENDER male female total officer 21900 4281 26181 detective 4058 806 4864 sergeant 3898 415 4313 lieutenant 1333 89 1422 captain 359 12 371 higher ranks 218 10 228 total 31766 5613 37379Test 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 hypothesisThree 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…Test the claim that the mean GPA of night students is significantly different than 2 at the 0.01 significance level. The null and alternative hypothesis would be: Но: р > 0.5 Нo: 2 H1:p > 0.5 H1:µ # 2 H1:p+ 0.5 H1:µ 2 The test is: right-tailed left-tailed two-tailed Based on a sample of 55 people, the sample mean GPA was 2.05 with a standard deviation of 0.03 The p-value is: (to 2 decimals)