(a) State the null hypothesis H and the alternate hypothesis H₁. H₁:0 H₁:0 (b) Determine the type of test statistic to use.
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- Test the claim that the proportion of people who own cats is smaller than 30% at the 0.025 significance level. The null and alternative hypothesis would be: Ho: p=0.3 Ho: p = 0.3 Ho: p= 0.3 H₁: p 0.3 The p-value is: Ho: p = 0.3 Ho: p = 0.3 H₁: p0.3 H₁:p> 0.3 (to 2 decimals) (to 2 decimals)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: Ho: HN S HD Ho:pN 2 PD Ho:PN = PD Ho:UN = HD Ho: PN HD H:PN PD H1: µN Next Question hpTest the claim that the mean GPA of night students is larger than 2.4 at the 0.025 significance level. The null and alternative hypothesis would be: Ho: µ = — 2.4 Но:р 2.4 Н: + 2.4 H:p> 0.6 H:д> 2.4 Hi:p #0.6 Hi:p < 0.6 Н:р < 2.4 The test is: left-tailed two-tailed right-tailed Based on a sample of 65 people, the sample mean GPA was 2.41 with a standard deviation of 0.03 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
- 7 10Consider the following hypothesis test. Ho: H₁ - H₂ SO Ha: M₁ - H₂ > O The following results are for two independent samples taken from the two populations. Sample 1 Sample 2 n₁ = 40 7₂ = 50 X₁ = 25.4 %1 = 5.7 (a) What is the value of the test statistic? (Round your answer to two decimal places.) = 22.8 x₂ = %₂ = 6 (b) What is the p-value? (Round your answer to four decimal places.) (c) with a = 0.05, what is your hypothesis testing conclusion? Reject Ho. There is insufficient evidence to conclude that μ₁ - H₂ > 0. Reject Ho. There is sufficient evidence to conclude that μ₁ - μ₂ > 0. Do not reject Ho. There is insufficient evidence to conclude that μ₁ - μ₂ > 0. Do not Reject Ho. There is sufficient evidence to conclude that μ₁ −μ₂ > 0. -Solve the problem. When performing a hypothesis test for the ratio of two population variances, the upper critical F value is denoted FR. The lower critical F value, FL, can be found as follows: interchange the degrees of freedom, and then take the reciprocal of the resulting F value found in table A-5. FR can be denoted Fa/2 and FL can be denoted F1-a/2 - Find the critical values FL and FR for a two-tailed hypothesis test based on the following values: n₁ = 4, n₂ = 8, a = 0.05 0.1703, 5.8898 0.1112, 5.0453 0.0684. 5.8898 O 0.1211, 4.3541
- Test the claim that the proportion of men who own cats is larger than 20% at the .05 significance level. The null and alternative hypothesis would be: 0.2 Но:р — 0.2 Но: и 0.2 Ho:H Ho:H H:р + 0.2 Н]:р 0.2 Hi:р > 0.2 Hi:р#0.2 0.2 Но:р || The test is: left-tailed two-tailed right-tailed Based on a sample of 25 people, 25% owned cats 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 mean GPA of night students is larger than 2 at the 0.005 significance level. The null and alternative hypothesis would be: Но:р 2 Нo:д 0.5 Но:и — 2 Нi:р> 0.5 H:n 2 Hi:p#0.5 Hі:р<0.5 Hі:р +2 The test is: two-tailed left-tailed right-tailed Based on a sample of 65 people, the sample mean GPA was 2.02 with a standard deviation of 0.08 The p-value is: (to 2 decimals) Based on this we: Fail to reject the null hypothesis O Reject the null hypothesisIs there a statistically significant difference between males and females in terms of newspaper readership? The proportion of each sex that says they read a newspaper daily is reported below. Results are from a nationally representative random sample.Use α = .01 and the five-step model to conduct the hypothesis test. Write a sentence or two interpreting your results. Males Ps1 = .56 N1 =786 Females Ps2 = .59 N2 =965
- 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…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)Test the claim that the proportion of people who own cats is larger than 20% at the 0.005 significance level. The null and alternative hypothesis would be: H2:p = 0.2 Họ:p 2 0.2 H9:p 0.2 H9:p = 0.2 Hg: u 0.2 H1: u 0.2 The test is: left-tailed two-tailed right-tailed Based on a sample of 300 people, 24% 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