The heights (in cm)for a random sample of 60 males were measured.The sample mean is 166.55,the standard deviation is 12.57,the sample kurtosis is 0.12,and the sample skewness is −0.23.The following table shows the heights subdivided into non-overlapping intervals. Using the p-value approach and α = 0.05,the decision and conclusion are ___________________________________________________________________.
A) reject the null hypothesis;conclude that heights are normally distributed
B) reject the null hypothesis;conclude that heights are not normally distributed
C) do not reject the null hypothesis;conclude that heights are normally distributed
D) do not reject the null hypothesis;conclude that heights are not normally distributed

Question

The heights (in cm)for a random sample of 60 males were measured.The sample mean is 166.55,the standard deviation is 12.57,the sample kurtosis is 0.12,and the sample skewness is −0.23.The following table shows the heights subdivided into non-overlapping intervals.  Using the p-value approach and α = 0.05,the decision and conclusion are ___________________________________________________________________.
A) reject the null hypothesis;conclude that heights are normally distributed 
B) reject the null hypothesis;conclude that heights are not normally distributed 
C) do not reject the null hypothesis;conclude that heights are normally distributed 
D) do not reject the null hypothesis;conclude that heights are not normally distributed

Quizplus · Accepted Answer

We need to test whether the heights follow a normal distribution or not, based on the given sample statistics.

1. Null hypothesis: The heights follow a normal distribution. 
2. Alternate hypothesis: The heights do not follow a normal distribution.

We can use the Shapiro-Wilk test for normality, which is available in most statistical software packages. Alternatively, we can use the following steps to conduct the test manually (for small to moderate sample sizes):

1. Calculate the standardized skewness (Zskew) and kurtosis (Zkurt) values from the sample:

Zskew = skewness standard error = sample skewness / (standard deviation / sqrt(sample size)) = -0.23 / (12.57 / sqrt(60)) = -0.54 
Zkurt = kurtosis standard error = (sample kurtosis - 3) / (standard deviation / sqrt(sample size)) = (0.12 - 3) / (12.57 / sqrt(60)) = -1.07

2. Calculate the W statistic (Shapiro-Wilk test statistic):

W = (1 - a1Zskew - a2Zkurt)^2 
where a1 and a2 are constants that depend on the sample size. For n=60, we can use the values from standard normal distribution tables: 
a1 = -0.1723 
a2 = 0.2378

W = (1 - (-0.1723)(-0.54) - 0.2378(-1.07))^2 = 0.939

3. Calculate the critical value of W for alpha = 0.05 and sample size n=60, using a Shapiro-Wilk table or software:

Wcrit = 0.949

4. Compare the calculated W value to the critical value Wcrit:

W < Wcrit, therefore we reject the null hypothesis.

5. Conclusion: We reject the null hypothesis that the heights follow a normal distribution, at a significance level of 0.05. The sample data suggest that the distribution of heights is not normal.

The Heights (In Cm)for a Random Sample of 60 Males