Question

select the correct answer.
richard wants to compare the data transfer rate of two internet connections. the difference between the means of samples taken for the two connections is 18 mbps. the standard deviation of the distribution of the difference between the sample means is 6.64. which statement is true if we are testing the null hypothesis at the 95% confidence level?
a. the difference between the two means is significant at the 95% confidence level, so the null hypothesis must be rejected.
b. the difference between the two means is significant at the 95% confidence level, so the null hypothesis must be accepted.
c. the difference between the two means is not significant at the 95% confidence level, so the null hypothesis must be rejected.
d. the difference between the two means is not significant at the 95% confidence level, so the null hypothesis must be accepted.

Explanation:

Step1: Recall 95% confidence level z - value

For a 95% confidence level, the critical z - value (two - tailed test) is approximately \(z = 1.96\). We can use the formula for the z - score of the difference between sample means: \(z=\frac{\bar{x}_1-\bar{x}_2}{SE}\), where \(\bar{x}_1 - \bar{x}_2\) is the difference between the sample means and \(SE\) is the standard deviation of the distribution of the difference between sample means.
Given that \(\bar{x}_1-\bar{x}_2 = 18\) Mbps and \(SE=6.64\).

Step2: Calculate the z - score

Substitute the values into the z - score formula: \(z=\frac{18}{6.64}\approx2.71\).

Step3: Compare z - score with critical value

We compare the calculated z - score (\(z\approx2.71\)) with the critical z - value for 95% confidence level (\(z = 1.96\)). Since \(|2.71|>1.96\), the difference between the two means is significant at the 95% confidence level. When the difference is significant at the 95% confidence level, we reject the null hypothesis.

Answer:

A. The difference between the two means is significant at the 95% confidence level, so the null hypothesis must be rejected.