Determine the P-value for a test of significance with t= 1.990, n = 95 and Haμ 100. ‡ Type your answer...
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- Let ?1 denote true average tread life for a premium brand of P205/65R15 radial tire, and let ?2 denote the true average tread life for an economy brand of the same size. Test H0: ?1 − ?2 = 5000 versus Ha: ?1 − ?2 > 5000 at level 0.01, using the following data: m = 45, x = 42,100, s1 = 2400, n = 45, y = 36,200, and s2 = 1800.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. Fail to reject H0. The data does not suggest that the difference in average tread life exceeds 5000.Reject H0. The data suggests that the difference in average tread life exceeds 5000. Fail to reject H0. The data suggests that the difference in average tread life exceeds 5000.Reject H0. The data does not suggest that the difference in average tread life exceeds 5000.Let ?1 denote true average tread life for a premium brand of P205/65R15 radial tire, and let ?2 denote the true average tread life for an economy brand of the same size. Test H0: ?1 − ?2 = 5000 versus Ha: ?1 − ?2 > 5000 at level 0.01, using the following data: m = 45, x = 42,100, s1 = 2400, n = 45, y = 36,200, and s2 = 1800. 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. Fail to reject H0. The data does not suggest that the difference in average tread life exceeds 5000.Reject H0. The data suggests that the difference in average tread life exceeds 5000. Fail to reject H0. The data suggests that the difference in average tread life exceeds 5000.Reject H0. The data does not suggest that the difference in average tread life exceeds 5000.A researcher obtained a result of t(9)=1.22, p>.05 for their study using a one sample t-test. Are the results significant using an alpha level of .05 and how many participants were in the group?
- Use the following summary to answer the items that follow: Students in the sleep treatment group have higher exam scores than those in the general student population, z = 2.78, p < .01 What was the alpha value used for the test? What was the decision for this test?J 1 Consider the hypothesis statement to the right using a=0.01 and the data to the right from two independent samples. A) calcuate the appropriate test statistic and interpret the results b) calculate the p value and interpret the resultsChoose the letter of the correct answer. 1. In one tailed test, in which critical values below will the computed z of 2.312 falls in the non rejection region? A. 1.383 B. 1.533 C. 2.228 D. 2.354
- In a test of H0:μ = 100 against Ha:μ>100, a sample of size 80 producers z = 0.8 for the value of the test statistic. The P-value of the test is thus equal to?In a test of H0:μ = 100 against H1:μ>100, a sample of size 80 producers z= 0.8 for the value of the test statistic. The P-value of the test is equal to?A researcher conducts a hypothesis test using a sample of n = 20 with M = 34 and s 2 = 36 from an unknown population. What is the df value for the t statistic?
- A researcher conducts a hypothesis test using a sample of n=20 with M =34 and s^2=36 from an unknown population.What is the df value for the t statistic?A treatment is administered to a sample selected from a population with a mean of u = 34 and a standard deviation of o = 8. After treatment, the effect size is measured by computing Cohen’s d, and a value of d = 0.00 is obtained. Based on this information, what is the mean for the treated sample?Explain what Cohen's d and r2 measure when calculated for a t test