Question

Do students perform worse when they take a test alone than when they take an test in a classroom setting? Eight students were given two tests of equal difficulty. They took one test in a solitary room and they took the other in a room filled with other students. The results are shown below.

test Scores

Alone	70	79	74	80	76	86	67	85
Classroom	74	79	90	82	81	97	70	84

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Assume a Normal distribution. What can be concluded at the αα = 0.10 level of significance?

For this study, we should use Select an answer t-test for the difference between two independent population means z-test for the difference between two population proportions t-test for a population mean z-test for a population proportion t-test for the difference between two dependent population means

1. The null and alternative hypotheses would be:   
2.   

H0:H0:  Select an answer μ1 p1 μd  Select an answer = ≤ ≠ ≥  Select an answer p2 μ2 0  (please enter a decimal)   

H1:H1:  Select an answer μ1 p1 μd  Select an answer = > < ≠  Select an answer 0 p2 μ2  (Please enter a decimal)

2. The test statistic ? z t  =  (please show your answer to 3 decimal places.)
3. The p-value =  (Please show your answer to 4 decimal places.)
4. The p-value is ? ≤ >  αα
5. Based on this, we should Select an answer accept fail to reject reject  the null hypothesis.
6. Thus, the conclusion is that ... Select an answer At the 10% significance level, the data provides insufficient evidence to conclude that the population mean test score taking the test alone is less than the population mean test score taking the exam in a classroom setting. At the 10% significance level, the data provides sufficient evidence to conclude that the population mean test score taking the test alone is less than the population mean test score taking the test in a classroom setting.

Answer 1

Sample 1 Sample 2 Difference = Sample 1 - Sample 2 70 74 -4 79 79 0 74 90 -16 80 82 -2 76 81 -5 86 97 -11 67 70 -3 85 84 1 Average 77.125 82.125 -5 St. Dev. 6.728 8.543 5.757 n 8 8 8 From the sample data, it is found that the corresponding sample means are: Ãj = 77.125 X2 = 82.125 Also, the provided sample standard deviations are: Si = 6.728 S2 8.543 and the sample size is n = 8. For the score differences we have ✓ = -5 SD = 5.757

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: UD = 0 Ha: HD < 0 This corresponds to a left-tailed test, for which a t-test for two paired samples be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.10, and the degrees of freedom are df = 7. Hence, it is found that the critical value for this left-tailed test is te = -1.415, for 0.10 and df 7. a = The rejection region for this left-tailed test is R = {t:t< -1.415}. (3) Test Statistics The t-statistic is computed as shown in the following formula: ū t= -5 5.757/18 -2.457 SD/m (4) Decision about the null hypothesis Since it is observed that t = -2.457 <tc = -1.415, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0218, and since p = : 0.0218 < 0.10, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean mi is less than M2, at the 0.10 significance level.

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