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

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

Step-by-step explanation:

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

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

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

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.

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step-by-step explanation:

first you take the 6 dollars and add 10 to it then add the 50 to that then multiply by 8 with xy and your answer will be

hope this