Question

Let's say that one of the items the university measured in the study was reaction time. That is a typical measurement that is taken when judging concussions as well as looki the beginning of the study they recorded a baseline average reaction time for the group of 41 football players at .239 seconds. At the end of their study they retested the players and the average reaction time was 233 with a standard deviation of .021. Use this data to create a 95% confidence interval for μ and explain if the reaction time at the conclusion of the study showed a significant decrease in reaction time or not.

Answer 1

The 95% confidence interval is given by: (0.226, 0.240)

On this case we can't conclude that we have a significant reduction on the reaction time since the upper bounf of the interval is higher than the value of 0.239.

Explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X =0.233` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s=0.021 represent the sample standard deviation

n=41 represent the sample size

2) Calculate the confidence interval

Since the sample size is large enough n>30 but we don't know the population deviation. The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

First we need to calculate the degrees of freedom given by:

`df=n-1=41-1=40`

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,40)".And we see that
`t_(\alpha/2)=2.02`

Now we have everything in order to replace into formula (1):

`0.233-2.02(0.021)/(√(41))=0.226`

`0.233+2.02(0.021)/(√(41))=0.240`

The 95% confidence interval is given by: (0.226, 0.240)

On this case we can't conclude that we have a significant reduction on the reaction time since the upper bounf of the interval is higher than the value of 0.239.