Heights of men aged 25 to 34 have a standard deviation of 2.9. Use a 0.05 significance level to test the claim that the heights of women aged 25 to 34 have a different standard deviation. The heights (in inces) of 16 randomly selected women aged 25 to 34 are listed below. 62.13 65.09 64.18 66.72 61.15 67.50 64.65 63.09
63.80 64.21 60.17 68.28 66.49 62.10 65.73 64.72
Select the correct conclusion based on the null hypothesis and final conclusion

Based on the given data, there is no evidence to support the claim that the standard deviation of the heights of women aged 25 to 34 have a different standard deviation from men aged 25 to 34

Step-by-step explanation:

The standard deviation for the heights of men aged 25 to 34, σ₀ = 2.9

The population variance, σ₀ = 2.9² = 8.41

The significance level of the test = 0.05

The number of randomly selected women in the sample, n = 16

The given heights of the 16 randomly selected women are listed as follows;

62.13, 65.09, 64.18, 66.72, 61.15, 67.50, 64.65, 63.09, 63.80, 64.21, 60.17, 68.28, 66.49, 62.10, 65.73, 64.72

By using the Mean (Average) and standard deviation of a sample formula from Microsoft Excel, we have;

The mean height of the women, = 63.37563

The standard deviation of the variance, s₂ = 2.2 78511

The null hypothesis, H₀; σ = 2.9

The alternative hypothesis, H₀; σ ≠ 2.9

The chi-square test is given as follows;

T = (N - 1)(s/σ₀)²

Therefore, we have;

T = (16 - 1)(2.278511/2.9)² ≈ 9.2597

The critical value for the lower tail = 6.908

The critical value for the upper tail = 28.845

For the critical region, we have;

Reject H₀ if T <6.908 or T > 28.845

Therefore, given that the critical T is larger than 6.908 and less than 28.845, we fail to reject the null hypothesis, and there is not enough statistical evidence to prove that there is a difference in the standard deviation of the heights of men and women aged 25 to 34

there is no graph so i cant sorry

step-by-step explanation: