Of 369 randomly selected medical students, 23 said that they planned to work in a rural community. Construct a 95% confidence interval for the population proportion of all medical students who plan to work in a rural community. Then construct a 99% confidence interval for the population proportion.

Is the 95% confidence interval narrower than the 99% confidence intervals and why?

Step-by-step explanation:

Total number of medical students selected, n is 369

23 said that they planned to work in a rural community. So the probability of success, p = 23/369 = 0.062

Probability of failure, q = 1 - p = 1 - 0.062 = 0.938

Mean, u = np = 369*0.062 = 23

Standard deviation, s = √npq = √369*0.062*0.938 = 4.6

1)

We want to determine a 95% confidence interval for the number of medical students who plan to work in a rural community

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z * standard deviation/√n

It becomes

23 ± 1.96 * 4.6/√369

= 23 ± 1.96 * 0.24

= 23 ± 0.47

The lower end of the confidence interval is 23 - 0.47 = 22.53

The upper end of the confidence interval is 23 + 0.47 = 23.47

2)

We want to determine a 99% confidence interval for the number of medical students who plan to work in a rural community. For a confidence level of 99%, the corresponding z value is 2.58. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z * standard deviation/√n

It becomes

23 ± 2.58 * 4.6/√369

= 23 ± 2.58 * 0.24

= 23 ± 0.62

The lower end of the confidence interval is 23 - 0.62 = 22.38

The upper end of the confidence interval is 23 + 0.62 = 23.62

The 95% confidence interval is narrower than the 99% confidence intervals. This is because as the degree of confidence increases, the margin of error increases. So the interval is wider than that of a lower confidence interval.