Question

In a random sample of 144 games of the 2014 NFL regular season, mean passing yards per game were 222 with a standard deviation of 36 yards per game. Recalling that the standard error of a sample mean is the standard deviation divided by the square root of the sample size, what is the 95% confidence interval for the mean passing yards per game

Answer 1

The 95% confidence interval for the mean passing yards per game is between 216.12 yards and 227.88 yards.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.025 = 0.975`, so
`z = 1.96`

Now, find M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 1.96*(36)/(√(144)) = 5.88`

The lower end of the interval is the mean subtracted by M. So it is 222 - 5.88 = 216.12 yards

The upper end of the interval is the mean added to M. So it is 222 + 5.88 = 227.88 yards

The 95% confidence interval for the mean passing yards per game is between 216.12 yards and 227.88 yards.