In estimating the mean score on a fitness exam, we use an original sample of size n = 30 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 10,000 bootstrap samples for the distribution:

A. 66 to 74 8.

B. 67 to 73

C. 67.5 to 72.

Respuesta :

Answer:

C. 67.5 to 72.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

The width of the interval is determined by it's margin of error, which is given by the following formula:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

So, as n increases, the margin of error decreases, and the interval gets smaller.

Using 10,000 bootstrap samples for the distribution:

We increase the sample size, which means that the interval gets smaller.

We had 67 to 73, since it got smaller, it will be from a value higher than 67 to a value lower than 73.

So the correct answer is:

C. 67.5 to 72.

The degree of uncertainty or certainty is the confidence interval. From the following confidence intervals, C is the most likely result after the change which is 67.5 to 72.

What is a confidence interval?

A confidence interval depicts the likelihood that a parameter will fall between two values near the mean.

The degree of uncertainty or certainty in a sampling process is measured by confidence intervals. They are often built with confidence levels of 95 percent or 99 percent.

The confidence interval is given as ;

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n} }[/tex]

z score is denoted by z

The margin of error is helpful to find the width of the interval which is determined by the following formula;

[tex]M= z\sqrt{\frac{\pi(1-\pi)}{n} }[/tex]

The value of n is inversly propotional to the margin of error. If the value of n increases, the margin of error decreases, and also the interval gets decreases.

Using 10,000 bootstrap samples for the distribution:

As the valuer of sample size increases the interval gets smaller values.

We had 67 to 73 given in the problem it will be from a value greater than 67 to a value lower than 73.

From all the observations we come to the conclusion that  C is the most likely result after the change which is 67.5 to 72.

Hence C is the most appropriate answer

To learn more about the confidence interval refer  the link;

https://brainly.com/question/2396419