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

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