Question

A sample of 25 observations is selected from a normal population where the sample standard deviation is 4.85. The sample mean is 16.80.  

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 95% confidence interval for the population mean. (Round the t-value to 3 decimal places. Round the final answers to 3 decimal places.)

The 95% 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

a)

The standard error of the mean is =s/sqrt(n)
= 4.85/sqrt(25)
= 0.97

b)

sample mean, xbar = 16.8
sample standard deviation, s = 4.85
sample size, n = 25
degrees of freedom, df = n - 1 = 24

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.064

ME = tc * s/sqrt(n)
ME = 2.064 * 4.85/sqrt(25)
ME = 2.002

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (16.8 - 2.064 * 4.85/sqrt(25) , 16.8 + 2.064 * 4.85/sqrt(25))
CI = (14.798 , 18.802)

The 95% confidence interval for the population mean is between 14.798 and 18.802

c)

Decrease