Question

The head circumference is measured for 25 girls and their younger twin sisters. The mean of the older twin girls was 50.23 cm and the mean of the younger twins was 49.96 cm. The standard deviation of the differences was 1 cm. Is this difference significant at a significance level of 5% (i.e., α = 0.05)?

Answer 1

The difference in head circumference between older and younger twin girls is not significant at the 5% level, as the calculated test statistic (1.35) is lower than the critical value (1.96) for a two-tailed test.

Step-by-step explanation:

To determine if the difference in head circumferences between older and younger twin girls is significant at the 5% significance level, we can conduct a hypothesis test. Calculate the test statistic using the mean difference (μ1 - μ2), which is 50.23 cm - 49.96 cm = 0.27 cm, and divide it by the standard deviation of the differences (1 cm) normalized by the square root of the sample size, √25. The resulting test statistic is (0.27 cm) / (1 cm / √25) = 1.35.

We can now refer to a standard normal distribution table to find the p-value corresponding to the test statistic of 1.35. At a 5% significance level, the critical value for a two-tailed test is approximately 1.96. Since our test statistic is less than the critical value, we fail to reject the null hypothesis and conclude that the difference is not significant.

Answer 2

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