Respuesta :
Answer:
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.
