Consider a paired t-test . The setting is that two normal random variables (X) a

Question

Consider a paired t-test. The setting is that two normal random variables (X) and (Y) are sampled. We pair each of the samples from the first variable with each of the samples from the second variable, and we check the "null hypothesis" that the two elements in each pair have the same mean, i.e. X=Y, against the alternative hypothesis that XY. Now create a new random variable (D) by taking the difference between the two elements in each pair. The Null Hypothesis is defined as D=0, and the alternative hypothesis D0. Note that if the two variables are normally distributed, their difference (D) is also normally distributed. We then calculate the confidence interval for the mean of D .

Select an option: yes, no, or can't tell, for each question:

1) Suppose that for a given data set of D, a 95% confidence interval is calculated to be (-3.45,1.78). If you were to perform a hypothesis test at the 5% significance level to test the null hypothesis, would you reject it?

2) Suppose that for a given data set of D, a 99% confidence interval is calculated to be (-10.77,-2.35). If you were to perform a hypothesis test at the 1% significance level to test the null hypothesis, would you reject it?

3) Suppose that for a given data set of D, a 97% confidence interval is calculated to be (25.6,41.1). If you were to perform a hypothesis test at the 3% significance level to test the null hypothesis, would you reject it?

Explanation / Answer

1) Here in the 95% confidence interval, we see that 0 lies in side the interval and therefore at 5% level of significance, we cannot reject the null hypothesis here.

2) Here in the given 99% confidence interval (-10.77,-2.35), we see that the whole interval lies below 0 and therefore, we can reject the null hypothesis here because 0 does not lie in the confidence interval. Therefore Reject the null hypothesis here.

3) Similar to the previous part, here we see that the whole confidence interval (25.6,41.1) lies above 0 and therefore 0 does not lie inside this interval and therefore Reject the null hypothesis here.

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