Question

a) A sample of 25 patients in a doctor's office showed that they had to wait an average of 35 minutes before they could see the doctor. The sample standard deviation is 10 minutes. Assume the population of waiting times is normally distributed.

At 95% confidence, compute the margin of error.

b) A sample of 25 patients in a doctor's office showed that they had to wait for an average of 32 minutes before they could see the doctor. The sample standard deviation is 8 minutes. Assume the population of waiting times is normally distributed.

At 95% confidence, compute the upper bound of interval estimate for the average waiting time of all the patients who visit this doctor.

Answer 1

a) Let X denote the time of waiting.

Here random sample of size (n) = 25.

Sample mean = x̄ = 35 minutes.

Sample standard deviation = s = 10 minutes.

As population standard deviation is unknown we will use t score to find the margin of error.

Confidence level = 95% = 0.95.

α = (1 - 0.95) / 2 = 0.025.

Degrees of feedom = n-1 = 24.

From table, we get, critical t value = t = 2.064.

b) Let X denote the time of waiting.

Here random sample of size (n) = 25.

Sample mean = x̄ = 32 minutes.

Sample standard deviation = s = 8 minutes.

As population standard deviation is unknown we will use t score to find the margin of error.

Confidence level = 95% = 0.95.

α = (1 - 0.95) / 2 = 0.025.

Degrees of feedom = n-1 = 24.

From table, we get, critical t value = t = 2.064.

Upper bound of interval estimate for the average waiting time of all the patients who visit this doctor = x̄ + E = (32+3.3024) minutes = 35.3024 minutes. (Ans)