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- 25. The State of California claims the population average of the amount of ice cream each Californian eats in the month of September is 6.85 pints with population standard deviation of 1.35 pints. An SRS of 500 Californians resulted in a sample average of 6.75 pints eaten per person in the month of September . At alpha=0.05, is there evidence to support the State of California's claim that Californians eat an average of 6.85 pints of ice cream in the month of September? Write a conclusion using the context of the problem.23. The State of California claims the population average of the amount of ice cream each Californian eats in the month of September is 6.85 pints with population standard deviation of 1.35 pints. An SRS of 500 Californians resulted in a sample average of 6.75 pints eaten per person in the month of September At alpha = 0.05 , is there evidence to support the State of California's claim that Californians eat an average of 6.85 pints of ice cream in the month of September? Find the p-value ?2. Consider a study where students are measured on whether they had an internship during their time at WKU (Y/N) and whether they had a job at graduation (Y/N). If we wanted to test whether having an internship was associated with having a job at graduation (i.e., internship holders were more likely to have jobs), why would the chi-square test be inappropriate for this hypothesis? How should we analyze our data?
- 24. The State of California claims the population average of the amount of ice cream each Californian eats in the month of September is 6.85 pints with population standard deviation of 1.35 pints. An SRS of 500 Californians resulted in a sample average of 6.75 pints eaten per person in the month of September. At alpha = 0.05 , is there evidence to support the State of California's claim that Californíans eat an average of 6.85 pints of ice cream in the month of September? True or False : Since the p-value is greater than alpha we fail to reject the null hypothesis. True FalseTo evaluate the effect of a treatment, one sample of n=8 is obtained from a population with population mean: u=40 and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M=35. 1. If the sample variance is s^2=32, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha=.05? (show work please, and state both experimental and statistical hypotheses, and state your decision regarding the H0. (all 4 steps of hypothesis test) 2a. If the sample variance is s^2=72, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha=.05? (please show your work and state your decision regarding the H0.) 2b. Calculate Effect size by using r^2 (percentage of variance explained) and state the meaning of effect size.The desired percentage of SiO2 in a certain type of aluminous cement is 5.5. To test whether the true average percentage is 5.5 for a particular production facility, 16 independently obtained samples are analyzed. Suppose that the percentage of SiO2 in a sample is normally distributed with ? = 0.32 and that x = 5.21. (Use ? = 0.05.) (a) Does this indicate conclusively that the true average percentage differs from 5.5?State the appropriate null and alternative hypotheses. H0: ? = 5.5Ha: ? ≠ 5.5H0: ? = 5.5Ha: ? ≥ 5.5 H0: ? = 5.5Ha: ? < 5.5H0: ? = 5.5Ha: ? > 5.5 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = State the conclusion in the problem context. Do not reject the null hypothesis. There is sufficient evidence to conclude that the true average percentage differs from the desired percentage.Reject the null hypothesis. There is sufficient evidence…
- Even a very small effect can be significant if the sample is large enough. Suppose, for example, that a researcher obtains a correlation (computed from the raw data) of r = 0.60 for a sample of n = 10 participants. a. Is this sample sufficient to conclude that a significant correlation exists in the population? Use a two-tailed test with α = .05. In your response, be sure to specify the critical value for r. b. If the sample had n = 25 participants, is the correlation significant? Again, use a two-tailed test with α = .05. In your response, be sure to specify the critical value for r.A researcher obtains t=2.25 for a repeated measures study using a sample of n=10 participants. Based on this t value, what is the correct decision for a two- tail test at .05 and .01 alpha level?1. The sample mean weights for two varieties of lettuce grown for 16 days in a controlled environment are 3.259 and 1.413 and the corresponding sample standard deviations are .400 and .220. If the sample sizes for the two varieties are 9 and 6 respectively, what would be the pair of hypotheses to test if the two varieties of lettuce have the same average weight? (Given: weight of each variety of lettuce is normally distributed). A. H0: μ1 ≠ μ2 vs H1: μ1 = μ2 B. H0: μ1 = μ2 vs H1: μ1 ≠ μ2 C. H0: μ1 > μ2 vs H1: μ1 ≤ μ2 D. H0: μ1 ≤ μ2 vs H1: μ1 > μ2 2. At 5% level, what are the critical values for testing equality of mean weights in problem 1? A. 2.18 B. -2.18 and 2.18 C. -1.78 D.-1.78 and 1.78 3.What is the best decision using critical value approach in problem 1? A. The computed test statistic falls in the critical region and we do not reject the null hypothesis. B. The computed test statistic does not fall in the critical…
- 1. The sample mean weights for two varieties of lettuce grown for 16 days in a controlled environment are 3.259 and 1.413 and the corresponding sample standard deviations are .400 and .220. If the sample sizes for the two varieties are 9 and 6 respectively, what would be the pair of hypotheses to test if the two varieties of lettuce have the same average weight? (Given: weight of each variety of lettuce is normally distributed). A. H0: μ1 ≠ μ2 vs H1: μ1 = μ2 B. H0: μ1 = μ2 vs H1: μ1 ≠ μ2 C. H0: μ1 > μ2 vs H1: μ1 ≤ μ2 D. H0: μ1 ≤ μ2 vs H1: μ1 > μ2 2.What would be the degree of freedom for the test statistic in problem 1? A. 6 B. 9 C. 12.7 D. 14 3. What would be the computed test statistic in problem 1? A. 2.93 B. 3.57 C. 8.44 D. 11.481. The sample mean weights for two varieties of lettuce grown for 16 days in a controlled environment are 3.259 and 1.413 and the corresponding sample standard deviations are .400 and .220. If the sample sizes for the two varieties are 9 and 6 respectively, what would be the pair of hypotheses to test if the two varieties of lettuce have the same average weight? (Given: weight of each variety of lettuce is normally distributed). A. H0: μ1 ≠ μ2 vs H1: μ1 = μ2 B. H0: μ1 = μ2 vs H1: μ1 ≠ μ2 C. H0: μ1 > μ2 vs H1: μ1 ≤ μ2 D. H0: μ1 ≤ μ2 vs H1: μ1 > μ2 2. What is the best decision using critical value approach in problem 1? A. The computed test statistic falls in the critical region and we do not reject the null hypothesis. B. The computed test statistic does not fall in the critical region and we do not reject the null hypothesis. C. The computed test statistic falls in the critical region and we reject the null hypothesis. D.The computed…1. The sample mean weights for two varieties of lettuce grown for 16 days in a controlled environment are 3.259 and 1.413 and the corresponding sample standard deviations are .400 and .220. If the sample sizes for the two varieties are 9 and 6 respectively, what would be the pair of hypotheses to test if the two varieties of lettuce have the same average weight? (Given: weight of each variety of lettuce is normally distributed). A. H0: μ1 ≠ μ2 vs H1: μ1 = μ2 B. H0: μ1 = μ2 vs H1: μ1 ≠ μ2 C.H0: μ1 > μ2 vs H1: μ1 ≤ μ2 D. H0: μ1 ≤ μ2 vs H1: μ1 > μ2 2. What is the best decision using critical value approach in problem 1? A. The computed test statistic falls in the critical region and we do not reject the null hypothesis. B. The computed test statistic does not fall in the critical region and we do not reject the null hypothesis. C. The computed test statistic falls in the critical region and we reject the null hypothesis. D. The computed test statistic does not fall…