**Educational Website Content: Statistical Analysis Exercise****Chapter 7: Hypothesis Testing for Science Achievement Scores****Background:**An educational publication claims that the mean score for grade 12 students on a science achievement test is more than 145. To test this claim, a random sample of 36 grade 12 test scores has been selected. The scores are as follows:188, 80, 165, 222, 175, 195, 201, 143, 119, 81, 118, 119, 109, 134, 200, 110, 199, 181, 79, 135, 124, 205, 90, 120, 216, 167, 198, 183, 173, 187, 143, 166, 147, 219, 206, 97 The test is conducted at a 1% significance level (α = 0.01).**Tasks:**a) **State the Hypotheses:**- **Null Hypothesis (H₀):** µ ≤ 145 - **Alternative Hypothesis (H₁):** µ > 145 b) **Calculate the p-value:** Use statistical software (StatCrunch) to compute the p-value associated with the test statistic. Include the software printout with your exercise submission.c) **Decision at α = 0.01:** Determine whether to reject or not reject the null hypothesis based on the calculated p-value.d) **Interpretation:** Provide an interpretation of the results in the context of the claim made by the publication.**Note:** This exercise focuses on hypothesis testing and requires critical analysis of statistical outputs to make informed decisions regarding the claim.

Given: An education publication claims that the mean score for grade 12 students on a science achievement test is more than 145. You want to test this claim. You randomly select 36 grade 12 test scores. The results are listed below. At a = 0.1.Solution:: a)Let denote the true mean score is greater than 145. Ususally this claim to be tested (with condition of inequality) forms the alternative hypothesis and the negation of this claim forms the null (with the condition of equality). To test: Vs at say 5% level Since the population standard deviation is unknown, the appropriate statistical test to test the above hypothesis would be a One sample t test, with test statistic given by: with rejection region t > t0.05,35. b. Substituting the values in the test stastistic formula: = 1.42 The exact p-value of the test can easily be obtained using the excel function: We get p-value = 0.082 c. Since p-value = 0.082 > 0.04 is not significant at 1% level, we may not support the claim d. 90% CI for population mean can be computed using the formula: For 35 degrees of freedom at 10% level significance, the critical value is obtained as: We get tcritical = 1.690 Substituting the values: = (143.02, 167.76)