Question

In estimating the mean score on a fitness exam, we use an original sample of size n 3 and a bootstrap distribution containing 5000 bootstrap samples to obtain a 95% confidence interval of 67 to 73. A change in this process is described below. If all else stays the same, which of the following confidence intervals (A, B, or C) is the most likely result after the change: Using an original sample of size n 16
O A. 66 to 74
O B. 67 to 73
c. 67.5 to 72.5

Answer 1

Answer:

C) 67.5 to 72.5

Step-by-step explanation:

The value to calculate the confidence interval is given as:

$(x-z_{\frac{\alpha}{2}} \times \frac{\sigma}{\sqrt{n}},x-z_{\frac{\alpha}{2}} \times \frac{\sigma}{\sqrt{n}})$

Margin of error is given by:

$z_{\frac{\alpha}{2} } \times \frac{\sigma}{\sqrt{n}}$

Using a sample of size n = 3, with 95% confidence the confidence interval came out to be 67 to 73.

We need to identify the new confidence interval if the sample size is increased to n = 16

If you observe the formula, you will see that the sample size(n) is in the denominator. This means, if value of n will be increased the value of fraction will be decreased, which will result in an overall smaller value of margin of error. Since, margin of error is smaller, the values of confidence interval will be closer to the mean.

From here, we can conclude that if sample size is increased, the confidence interval will get narrower. From the given options only option C contains a narrow confidence interval. Therefore, correct answer will be option C: 67.5 to 72.6