7.4 Find the p-value, Part II. An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at a = 0.01. (a) n = 26, T = 2.485 (b) n = 18, T = 0.5
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- Q1Pick the Test (Z-test, Single Sample T-test, Independent Sample T-test, Paired Sample Ttest, ANOVA): 17. A large university wants to compare the mathematical abilities of its male and female students. A researcher selects 100 men and 100 women at random from each of the four classes, and administers a math test. The men average 500 on the test, with a standard deviation of 120. The women average 450 on the test, with a standard deviation of 110. What test would be appropriate to determine whether the difference between men and women is real or due to chance variation? Test: ______________Test the claim that the mean GPA of night students is significantly different than 2.9 at the 0.2 significance level.The null and alternative hypothesis would be: H0:μ=2.9H0:μ=2.9H1:μ≠2.9 Based on a sample of 70 people, the sample mean GPA was 2.89 with a standard deviation of 0.05The test statistic is: (to 2 decimals)The p-value is: (to 2 decimals) test is two tailed
- Q10 = In a one-tailed hypothesis test with a sample size of 300, the test statistic is determined to be Z = 1.53 Assuming population standard deviation o is known , The p-value for this test is: Q11 = In a two-tailed hypothesis test with a sample size of 300, the test statistic is determined to be Z = 2.21 . Assuming population standard deviation is known, The p-value for this test isA random sample of size 29 is selected from a normal population with mean = and variance = 8. Determine the probability that s? (the sample variance) is between 11.81 and 14.57. 12A sample of n=4 scores has ΣX=8 and ΣX2=40. What is the value of the sample variance, s2?
- Q1: A researcher was interested in the effect of exercise on stress levels. For the general population, the distribution of the Stress Battery Scale scores is normal with the mean of µ = 25 and the standard deviation of σ = 5. A sample of n = 100 participants was asked to exercise for three weeks and then reported their stress levels. The sample mean was ?X = 23. Conduct a hypothesis test and determine the effect of exercise. Use the two-tailed test α = .01 1.Type the null and research hypotheses in complete sentences. 2.Cut off points 3.Standard error and z-score -show your computation process 4.Your conclusion thoroughly.14Researchers conducted a study to determine whether magnets are effective in treating back pain. The results are shown in the table for the treatment (with magnets) group and the sham (or placebo) group. The results are a measure of reduction in back pain. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Treatment Sham μ μ1 μ2 n 13 13 x 0.48 0.41 s 0.76 1.42 a. Use a 0.05 significance level to test the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment. What are the null and alternative hypotheses? A. H0: μ1=μ2 H1: μ1>μ2 B. H0: μ1<μ2 H1: μ1≥μ2 C. H0: μ1=μ2 H1: μ1≠μ2 D. H0: μ1≠μ2 H1: μ1<μ2 The test statistic, t, is nothing. (Round…
- n=10 b = 0.50 sb = 0.02 α=0.05 find the confidence interval for β.A simple random sample of 95 basketball players is taken from different teams and their mean weight is found to be 226 pounds and the standard deviation of the population is 20 pounds. Using 0.05 level of significance, test the claim that the average weight for all basketball players is 231 pounds. Complete the following steps: Step # 1: Write the null hypothesis Ho [ Select ] and the alternative hypothesis H1 [ Select ] and identify the claim [Select] Step # 2: Indicate what type of test has to be used [Select ] and find the critical value(s). ( Select ] Step # 3: Find the test value. [ Select ] Step # 4: Make the decision about the null hypothesis. [ Select ] Step # 5: Summarize the result. (Select JUse the data to calculate the sample variance, s. (Round your answer to five decimal places.) n = 7: 1.6, 3.5, 1.6, 2.1, 3.1, 2.9, 3.0 %3D Construct a 95% confidence interval for the population variance, of. (Round your answers to two decimal places.) to Test Ho: = 0.8 versus H: of + 0.8 using a = 0.05. State the test statistic. (Round your answer to two decimal places.) x2 State the rejection region. (If the test is one-tailed, enter NONE for the unused region. Round your answers to two decimal places.) x2 > x² State the conclusion. H, is rejected. There is sufficient evidence to indicate that the population variance is different from 0.8. H, is rejected. There is insufficient evidence to indicate that the population variance is different from 0.8. H, is not rejected. There is sufficient evidence to indicate that the population variance is different from 0.8. O Ho is not rejected. There is insufficient evidence to indicate that the population variance is different from 0.8.