In testing for differences between the means of two independent populations the null hypothesis is: O A. H iH,-H2=2 O B. Ho:H1 -H2<0 OC. Ho:H1-H2>0 O D. H H,-H2=0
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- Test the claim below about the mean of the differences for a population of paired data at the level of significance a. Assume the samples are random and dependent, and the populations are normally distributed. Claim: H, 20; a = 0.05. Sample statistics: d= -2.2, s, = 1.5, n= 14 Identify the null hypothesis by writing its complement. O A Ho: Ha0 H: Ha =0 O B. Hg: Hg s0 OC. Ho: Ha =0 O D. Ho: Ha 0 H: Hg s0 The test statistic is t=. (Round to two decimal places as needed.)The table below gives beverage preferences for random samples of teens and adults. We are asked to test for independence between age (i.e., adult and teen) and drink preferences. This problem is an example of a Teens Adults Coffee 50 200 E!! Tea 100 150 Soft 200 200 Drink Other 50 50 Select one: O a. normally distributed variable O b. test for independence O c. binomial distributed variable O d. multinomial populationTest the claim that the proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.2 significance level. The null and alternative hypothesis would be: Но : им — ИF Ho: Им — Иr Но: им — Ar Ho: Pм — Pr Ho:Pм — Pr Ho:рм — Pғ µf Ho:PM PF Ho:PM = PF Ho:PM H1: µM > µF H1:µM + HF H1:µM PF The test is: left-tailed right-tailed two-tailed Based on a sample of 20 men, 25% owned cats Based on a sample of 80 women, 35% owned cats The test statistic is: (to 2 decimals) (Hint: difference in proportions, not z) The positive critical value is: (to 2 decimals) (Hint: the proportion, not z) Based on this we: O Fail to reject the null hypothesis O Reject the null hypothesis
- Suppose we have a case-control study that completes and yields a study OR = 4.5. If we were told the true OR in the population was actually 2.25 (we normally wouldn’t know this, but suppose hypothetically someone was able to know this and told us), this again means our study OR is biased. Is our study OR biased towards or away from the null? A)Away from the null. B)Towards the null.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 hpCollege students and STDs: A recent report estimated that 25% of all college students in the United States have a sexually transmitted disease (STD). Due to the demographics of the community, the director of the campus health center believes that the proportion of students who have a STD is higher at his college. He tests H0: p = 0.25 versus Ha: p > 0.25. The campus health center staff select a random sample of 150 students and determine that 43 have been diagnosed with a STD. Conduct a hypothesis test to address the director’s hypothesis. Use a 5% significance level to make your decision. Use the applet to determine the P‑value. Click here to open the applet. Which of the following is an appropriate conclusion based on the results? The data provides strong evidence that the proportion of students at his college who have a STD is more than 25%. The data suggests that the proportion of students at his college who have a STD is 29%. Of the students surveyed at his college, 29%…
- 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, 4.3541Test 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 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
- 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…2. A researcher had rats engage in a lever-press task for either a small, medium, or larger reward. The number of lever presses was the dependent variable. The data were as follows: Reward Size Small Medium Large 1 4 6 4 3 5 1 4 4 2 5 T = 13 SS = 2.75 T= 6 T= 20 SS = 9 SS = 2 EX2 = 165 a. Do the data indicate that reward size had a significant impact on lever-pressing? Use the hypothesis testing steps & p = .05. Compute any necessary post-hoc tests. b. Compute eta-squared to measure the effect size.▾ Part 1 of 4 A pharmaceutical company claims their new diabetes medication results in less variance in a patient's glucose level than if the patient were on no medication at all. An endocrinologist wishes to test this claim. She divides participants randomly into two groups. Group A consists of 20 diabetics who received the medication; group B consists of 26 diabetics who received a placebo. After two weeks, the blood sugar level of each patient in each group was measured with the following results (in mg/dL): Group A: 77.8, 229.4, 199.9, 110.1, 180.2, 116.1, 139.7, 171.1, 37.4, 158.1, 88.4, 195.5, 246.1, 142.4, 178.1, 105.5, 179.6, 146.1, 78.8, 123.7 Group B: 124.5, 130.1, 136, 162.8, 113.4, 72.8, 142.6, 50.3, 179.8, 197, 230, 194.3, 171, 109.3, 114.4, 107.5, 114.7, 195.3, 127.7, 126.4, 85.6, 166.7, 182.3, 113.5, 216.7, 162.8 Perform a hypothesis test using a 6% level of significance to test the pharmaceutical company's claim. Step 1: State the null and alternative hypotheses. Ho:…