Question

The manufacturer of hardness testing equipment uses stool-ball inderters to penetrate metal that is being tested. However, the manufacturer thinks it would be better to use a diamond indenter so that all types of metal can be tested. Because of differences between the two types of indenters, it is suspected that the two methods will produce different hardness readings. The metal specimens to be tested are large enough so that two Indentions can be made. Therefore, the manufacturer ses both indenters on each specimen and compares the hardness readings Construct a 95% confidence interval to judge whether the two indeters resut in different measurements Note: A normal probably plot and boxplot of the data indicate that the differences are approximately normally distributed with no outiers Click the icon to view the datatable Construct a 98% confidence interval to judge whether the two indenters result in different monsurements, where the differences are computed as 'diamond minus tool ball. The lower bounds The upper bounds (Round to the nearest tenth as needed) State the appropriate conclusion. Choose the correct answer below. O There is insufficient evidence to conclude that the two indenters produce different hardness readings O There is sufficient evidence to conclude that the two indenters produce different hardness readings

Answer 1

Sample 1 Sample 2 Difference Sample 1 - Sample 2 51 53 -2 57 56 1 61 61 0 70 74 -4 68 69 -1 54 55 -1 65 68 -3 51 51 0 53 56 -3 Average 58.889 60.333 -1.444 St. Dev. 7.373 8.124 1.667 n 9 9 9 From the sample data, it is found that the corresponding sample means are: X1 = 58.889 X2 = 60.333 Also, the provided sample standard deviations are: 81 = 7.373 82 = 8.124 and the sample size is n = 9. For the score differences we have D = -1.444 SD = 1.667

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: up = 0 Ha: up 0 This corresponds to a two-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 = 0.05, and the degrees of freedom are df = 8. Hence, it is found that the critical value for this two-tailed testis te= 2.306, for a = 0.05 and df = 8. The rejection region for this two-tailed test is R = {t: 1t| > 2.306}. (3) Test Statistics The t-statistic is computed as shown in the following formula: ta D sp/vn -1.444 1.667/9 -2.6 (4) Decision about the null hypothesis Since it is observed that t = 2.6 >te = 2.306, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0316, and since p= 0.0316 < 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 mi is different than u2, at the 0.05 significance level. Confidence Interval The 95% confidence intervalis -2.726 < up < -0.163.

Lower bound = -2. 726

Upper bound = -0.163

Conclusion :

2nd option is correct.

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