Answer:
Step-by-step explanation:
The data for the test are the differences between the head circumferences of the older twin girls and that of their sisters
μd = mean head circumferences of the older twin girls minus the mean head circumferences of the sisters
For the null hypothesis
H0: μd ≥ 0
For the alternative hypothesis
H1: μd < 0
It is a left tailed test
The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 25 - 1 = 24
The formula for determining the test statistic is
t = (xd - μd)/(sd/√n)
Where
xd = difference in sample means
μd = difference in population means
sd = standard deviation of the difference
From the information given,
xd = 50.23 - 49.96 = 0.27
t = (0.27 - 0)/(1/√25)
t = 1.35
We would determine the probability value by using the t test calculator.
p = 0.095
Since alpha, 0.05 < than the p value, 0.095, then we would fail to reject the null hypothesis.
Therefore, at 5% significance level, the difference is not significant
corresponds to the part of the distribution lying within 2 standard deviations of the mean (since 500-2*100=300 and 500+2*100=700). The empirical rule states that approximately 95% of the distribution is expected to fall in this range.