Question

a)A sample of 25 patients in a doctor's office showed that they had to wait for an average of 38 minutes before they could see the doctor. The sample standard deviation is 12 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.

b)A sample of 30 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 99% confidence, compute the margin of error.

Answer 1

Solution :

1) Given that,

Point estimate = sample mean = = 38

sample standard deviation = s = 12

sample size = n = 25

Degrees of freedom = df = n - 1 = 25 - = 24

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.05

t,df = t0.05,24 = 1.711

Margin of error = E = t/,df * (s /n)

= 1.711 * (12 / 25)

Margin of error = E = 4.11

The 95% upper confidence interval estimate of the population mean is,

+ E  

= 38 + 4.11 = 42.11

upper bound = 42.11

Given that,

Point estimate = sample mean = = 35

sample standard deviation = s = 10

sample size = n = 30

Degrees of freedom = df = n - 1 = 30 - 1 = 29

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t0.005,29 = 2.756

Margin of error = E = t/2,df * (s /n)

= 2.756 * (10 / 30)

Margin of error = E = 5.03