Question

A professor wishes to test a claim that the mean test score of her first class is lower than the mean test score of her second class. A random sample of size 27 with a mean of 65.7 and a standard deviation of 5.4 is selected from her first class. A random sample of size 20 with a mean of 69.5 and a standard deviation of 12.1 is selected from her second class. Assume that test scores in both classes are normally distributed with unknown but unequal standard deviations. Test the claim at a significance level of 0.01.

What are the correct hypotheses? Select the correct symbols in the order they appear in the problem.

H₀: Select an answer μ₂ σ² x̅₂ s² x̅₁ μ₁ p   ? ≠ = < > ≥ ≤  Select an answer x̅₂ x̅₁ 0 s² σ² μ₂ p μ₁

H_a: Select an answer σ² μ₂ s² x̅₁ p x̅₂ μ₁  ? > ≥ ≠ < ≤ =  Select an answer 0 x̅₂ μ₂ σ² p s² x̅₁ μ₁

What is the test statistic? (Report answer accurate to three decimal places.)
Test statistic =

What is the p-value? (Report answer accurate to four decimal places.)
p-value =

The decision to the hypotheses is to... Select an answer reject the null hypothesis fail to reject the null hypothesis accept the alternative hypothesis reject the claim accept the null hypothesis

The final conclusion is: Select an answer There is not enough evidence to reject the claim There is not enough evidence to support the claim There is enough evidence to reject the claim There is enough evidence to support the claim  that the mean test score of her first class is lower than the mean test score of her second class.

Answer 1

X1 = 65.7 X2 = 69.5 Also, the provided sample standard deviations are: Si = 5.4 S2 = 12.1 and the sample sizes are ni = 27 and n2 20. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi ? M2 Ha: M1 M2 This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.01, and the degrees of freedom are df = 24.628. In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal: -2.488, for Hence, it is found that the critical value for this left-tailed test is to a = 0.01 and df = 24.628. The rejection region for this left-tailed test is R = = {t:t< -2.488}. (3) Test Statistics Since it is assumed that the population variances are unequal, the t-statistic is computed as follows: t= X1 - X2 si + să ni n2 65.7 – 69.5 1.311 5.42 27 + 12.12 20 (4) Decision about the null hypothesis Since it is observed that t = -1.311 > tc = -2.488, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.101 and since p = 0.101 > 0.01, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean mi is less than M2, at the 0.01 significance level.

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