Question

A sample of 28 observations is selected from a normal population where the sample standard deviation is 4.40. The sample mean is 16.40.  

a. Determine the standard error of the mean. (Round the final answer to 2 decimal places.)

The standard error of the mean is            .

b. Determine the 98% confidence interval for the population mean. (Round the t-value to 3 decimal places. Round the final answers to 3 decimal places.)

The 98% confidence interval for the population mean is between  and  .

c. If you wanted a narrower interval, would you increase or decrease the confidence level?

(Click to select)  Increase  Decrease

Answer 1

Here, sample size , n = 28

sample mean , = 16.40 and smaple standard deviation , s = 4.40

a) The standard error of the mean = (s/n)

= 4.40 / (28)

= 0.83

b)Degrees of freedom = n -1 = 27

From the t critical values table we find the t value for df = 27 and 98% confidence level.

t = 2.473

The confidence interval is = t * s/n

= 16.40 2.056

= (14.344 , 18.456)

The 98% confidence interval for the population mean is between 14.344 and 18.456.

c)We know that, the width of the confidence decreases with decrease in confidence level. Therefore, if we want a narrower confidence interval, we would decrease the confidence interval.