Question

Assume that the differences are normally distributed. Complete parts ​(a) through ​(d) below. Observation 1 2 3 4 5 6 7 8 Upper X Subscript i 40.9 50.0 44.6 42.9 47.2 48.7 51.4 50.2 Upper Y Subscript i 45.6 48.3 46.4 47.8 50.1 50.2 53.2 50.6 ​(a) Determine d Subscript i Baseline equals Upper X Subscript i Baseline minus Upper Y Subscript i for each pair of data. Observation 1 2 3 4 5 6 7 8 di nothing nothing nothing nothing nothing nothing nothing nothing ​(Type integers or​ decimals.) ​(b) Compute d overbar and s Subscript d. d overbarequals nothing ​(Round to three decimal places as​ needed.) s Subscript dequals nothing ​(Round to three decimal places as​ needed.) ​(c) Test if mu Subscript dless than0 at the alphaequals0.05 level of significance. What are the correct null and alternative​ hypotheses? A. Upper H 0​: mu Subscript dless than0 Upper H 1​: mu Subscript dequals0 B. Upper H 0​: mu Subscript dless than0 Upper H 1​: mu Subscript dgreater than0 C. Upper H 0​: mu Subscript dgreater than0 Upper H 1​: mu Subscript dless than0 D. Upper H 0​: mu Subscript dequals0 Upper H 1​: mu Subscript dless than0 ​P-valueequals nothing ​(Round to three decimal places as​ needed.) Choose the correct conclusion below. A. Reject the null hypothesis. There is sufficient evidence that mu Subscript dless than0 at the alphaequals0.05 level of significance. B. Reject the null hypothesis. There is insufficient evidence that mu Subscript dless than0 at the alphaequals0.05 level of significance. C. Do not reject the null hypothesis. There is sufficient evidence that mu Subscript dless than0 at the alphaequals0.05 level of significance. D. Do not reject the null hypothesis. There is insufficient evidence that mu Subscript dless than0 at the alphaequals0.05 level of significance. ​(d) Compute a​ 95% confidence interval about the population mean difference mu Subscript d. The lower bound is nothing. The upper bound is nothing. ​(Round to two decimal places as​ needed.) Click to select your answer(s).

Answer 1

The following table is obtained:

	Sample 1	Sample 2	Difference = Sample 1 - Sample 2
	40.9	45.6	-4.7
	50	48.3	1.7
	44.6	46.4	-1.8
	42.9	47.8	-4.9
	47.2	50.1	-2.9
	48.7	50.2	-1.5
	51.4	53.2	-1.8
	50.2	50.6	-0.4
Average	46.988	49.025	-2.038
St. Dev.	3.804	2.481	2.174
n	8	8	8

For the score differences we have

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μD​ = 0

Ha: μD​ < 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 α=0.05, and the degrees of freedom are df=7.

Hence, it is found that the critical value for this left-tailed test is tc​=−1.895, for α=0.05 and df=7.

The rejection region for this left-tailed test is R={t:t<−1.895}.

(3) Test Statistics

The t-statistic is computed as shown in the following formula:

(4) Decision about the null hypothesis

Since it is observed that t=−2.651<tc​=−1.895, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0164, and since p=0.0164<0.05, 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 μ1​ is less than μ2​, at the 0.05 significance level.

Confidence Interval

The 95% confidence interval is −3.855<μD​<−0.22.

Graphically

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