Question

A random group of students was surveyed to determine how many months it has been since they visited the dentist. The sample of 32 students gave a mean average of 19 months with a standard deviation of 4.4 months.

a) Find a 95% confidence interval for the population mean number of months since students have seen the dentist. Write answer in a full sentence.

b) Find a 99% confidence interval for the same value. Write answer in a full sentence.

c) Determine the minimum sample size needed to determine the a 95% confidence interval to within 1 month of the population mean.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 19

sample standard deviation = s = 4.4

sample size = n = 32

Degrees of freedom = df = n - 1 = 31

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,31 = 2.040

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

= 2.040* (4.4 / 32)

= 1.587

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

- E < < + E

19 - 1.587 < < 19 + 1.587

17.413 < < 20.587

There is 95% confident that the population mean number of months since students have seen the dentist. is between 17.413 to 20.587.

b)

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,31 = 2.744

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

= 2.744* (4.4 / 32)

= 2.134

The 99% confidence interval estimate of the population mean is,

- E < < + E

19 - 2.134 < < 19 + 2.134

16.866 < < 21.134

There is 99% confident that the population mean number of months since students have seen the dentist. is between 16.866 to 21.134.

c)

Margin of error = E = 1

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

sample size = n = (Z_/2* / E) ²

n = (1.96 *4.4 / 1)²

n = 74.37

n = 75

Sample size = 75