Question

A sample of 22 observations is selected from a normal population. The sample standard deviation is 23.00, and the sample mean is 51.

a. Determine the standard error of the mean. (Round the final answer to 4 decimal places.)

The standard error of the mean is   

b. Determine the 80% confidence interval for the population mean. (Round the final answers to 2 decimal places.)

   

The 80% confidence interval for the population mean is between ？ and ？

   

c. f you wanted a wider interval, would you increase or decrease the confidence level?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 51

sample standard deviation = s = 23.00

sample size = n = 22

Degrees of freedom = df = n - 1 = 22 - 1 = 21

a) SE = s / n = 23.00 / 22 = 4.9036

b) At 80% confidence level

= 1 - 80%

=1 - 0.80 =0.20

/2 = 0.10

t/2,df = t_0.10,21 = 1.323

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

= 1.323 * (23.00 / 22)

Margin of error = E = 6.49

The 80% confidence interval estimate of the population mean is,

  ± E  

= 51 ± 6.49

= ( 44.51, 57.49 )

The 80% confidence interval for the population mean is between 44.51 and 57.49.

c) If you wanted a wider interval,the confidence interval is increases.