Question

A researcher randomly selects 6 fathers who have adult sons and records the​ fathers' and​ sons'...

A researcher randomly selects 6 fathers who have adult sons and records the​ fathers' and​ sons' heights to obtain the data shown in the table below. Test the claim that sons are taller than their fathers at the

alpha equals 0.10α=0.10

level of significance. The normal probability plot and boxplot indicate that the differences are approximately normally distributed with no outliers so the use of a paired​ t-test is reasonable.

Observation

1

2

3

4

5

6

Height of father​ (in inches)

73.5

69.9

65.9

70.8

67.4

68.1

Height of son​ (in inches)

71.4

71.7

70.6

75.0

71.7

69.4

Find the critical​ value(s).

Find the test statistic.

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Answer #1

Given :-

Sample 1 = Height of son

Sample 2 = Height of father

The following table is obtained:

Sample 1 Sample 2 Difference = Sample 1 - Sample 2
71.4 73.5 -2.1
71.7 69.9 1.8
70.6 65.9 4.7
75.0 70.8 4.2
71.7 67.4 4.3
69.4 68.1 1.3
Average 71.633 69.267 2.367
St. Dev. 1.868 2.715 2.606
n 6 6 6

From the sample data, it is found that the corresponding sample means are:

From the sample data, it is found that the corresponding sample means are: X1 71.63:3 X2 69.267 Also, the provided sample standard deviations are 81 1.868 82 2.715 and the sample size is n 6. For the score differences we have D- 2.367 SD- 2.606 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ha: D>0 This corresponds to a right-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 0.10, and the degrees of freedom are df -5 Hence, it is found that the critical value for this right-tailed test is t 1.476, for a - 0.10 and df - 5 The rejection region for this right-tailed test is R-t :t> 1.476. (3) _Test Statistics The t-statistic is computed as shown in the following formula: 2.367 2.225 SD/Vn 2.606/V6(4) _Decision about the null hypothesis Since it is observed that 2.225 〉 tc 1.476, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p 0.0383, and since p 0.0383 < 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 is greater than μ2, at the 0.10 significance level.

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