# Listed below are the lengths (in minutes) of randomly selected music CDs. Construct a 98%...

## Question:

Listed below are the lengths (in minutes) of randomly selected music CDs. Construct a 98% confidence interval for the mean length of all such CDs.

 56.82 62.44 60.17 40.44 44.51 60.9 68.44 51.4 57.9 64.98 56.88 55.92 47.01 51.55 41.28 50.84 70.27 39.85 58.8 45.08 43.77 48.9 72.52 38.41 48.16 33.96 45.9 29.19 47.38 52.05 68.3 55.48 60.69 64.71 50.96 52.51 54.61 49.75 52.53 58.53

CI: _____ < {eq}\mu {/eq} < _____

## Confidence Interval:

In this question, we will use the t distribution to construct the 98% confidence interval for the mean length of all such CDs. We are using the t distribution to construct the 98% confidence interval because the sample size is small and the population standard deviation is not given.

Given that,

 56.82 62.44 60.17 40.44 44.51 60.9 68.44 51.4 57.9 64.98 56.88 55.92 47.01 51.55 41.28 50.84 70.27 39.85 58.8 45.08 43.77 48.9 72.52 38.41 48.16 33.96 45.9 29.19 47.38 52.05 68.3 55.48 60.69 64.71 50.96 52.51 54.61 49.75 52.53 58.53

Sample size, {eq}n = 40 {/eq}

Degree of freedom, {eq}n-1 = 40 - 1 = 39 {/eq}

Sample mean is defined as:

{eq}\bar{x} = \frac{\sum{x_i}}{n} {/eq}

Excel function for the sample mean:

=AVERAGE(56.82,62.44,60.17,40.44,44.51,60.9,68.44,51.4,57.9,64.98,56.88,55.92,

47.01,51.55,41.28,50.84,70.27,39.85,58.8,45.08,43.77,48.9,72.52,38.41,

48.16,33.96,45.9,29.19,47.38,52.05,68.3,55.48,60.69,64.71,50.96,52.51,54.61,49.75,52.53,58.53)

Sample mean, {eq}\bar{x} =52.84 {/eq}

Sample standard deviation is defined as:

{eq}s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n-1}} {/eq}

Excel function for the sample standard deviation:

=STDEV(56.82,62.44,60.17,40.44,44.51,60.9,68.44,51.4,57.9,64.98,56.88,55.92,

47.01,51.55,41.28,50.84,70.27,39.85,58.8,45.08,43.77,48.9,72.52,38.41,48.16,

33.96,45.9,29.19,47.38,52.05,68.3,55.48,60.69,64.71,50.96,52.51,54.61,49.75,52.53,58.53)

{eq}s =9.93 {/eq}

The 98% confidence interval for the population mean is defined as:

{eq}\bar{x} \pm t_{0.02/2}\times \frac{s}{\sqrt{n}} {/eq}

Excel function for the confidence coefficient:

=TINV(0.02,39)

Now,

{eq}52.84 \pm 2.426\times \frac{9.93}{\sqrt{40}}\\ 49.03 < \mu < 56.65 {/eq}

A 98% confidence interval for the mean length of all such CDs = (49.03 , 56.65).