Question

Do men take less time than women to get out of bed in the morning? The 60 men observed averaged 6.8 minutes to get out of bed after the alarm rang. Their standard deviation was 5.8. The 44 women observed averaged 2 minutes and their standard deviation was 2.6 minutes. What can be concluded at the αα = 0.10 level of significance? Assuming Unequal Variances

1. For this study, we should use Select an answer z-test for the difference between two population proportions t-test for the difference between two independent population means t-test for a population mean t-test for the difference between two dependent population means z-test for a population proportion
2. The null and alternative hypotheses would be:   
3.   

H0:H0:  Select an answer μ1 p1  Select an answer ≥ ≠ ≤ =  Select an answer μ2 p2  (please enter a decimal)   

H1:H1:  Select an answer p1 μ1  Select an answer > < ≠ =  Select an answer p2 μ2  (Please enter a decimal)

3. The test statistic ? t z
4. The test statistics is  (please show your answer to 3 decimal places.)
5. The p-value is  (Please show your answer to 4 decimal places.)
6. The p-value is ? > ≤  αα
7. Based on this, we should Select an answer accept fail to reject reject  the null hypothesis.
8. Thus, the conclusion is that ... Select an answer At the 10% significance level, the data provides sufficient evidence to conclude that the population mean time for men to get out of bed in the morning is less than the population mean time for women to get out of bed in the morning. At the 10% significance level, the data provides insufficient evidence to conclude that the population mean time for men to get out of bed in the morning is less than the population mean time for women to get out of bed in the morning.

Answer 1

The provided sample means are shown below: X1 = 6.8 X2 = 2 Also, the provided sample standard deviations are: $1 = 5.8 S2 = 2.6 and the sample sizes are ni - 60 and n2 = 44. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = M2 Ha: M1 < u2 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 = 0.10, and the degrees of freedom are df = 86.82. In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal: Hence, it is found that the critical value for this left-tailed test is te = -1.291, for a = 0.10 and df = 86.82. The rejection region for this left-tailed test is R = {t:t< -1.291}. (3) Test Statistics Since it is assumed that the population variances are unequal, the t-statistic is computed as follows: tt X1 - X2 si + ni n2 6.8 – 2 = 5.679 5.82 60 + 2.62 44 (4) Decision about the null hypothesis Since it is observed that t = 5.679 > tc = -1.291, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 1, and since p=1 > 0.10, 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 0.10 significance level.

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