An independent-measures study with n = 8 in each treatment produces M = 75 for the first treatment and M = 71 for the second treatment with a pooled variance of 9. Construct a 95% confidence interval for the population mean difference.

Step-by-step explanation:

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".

represent the sample mean for the sample  

population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

For this case we need to find the degrees of freedom like this:

The confidence interval is given by:

The confidence is 0.95, the value for the significance is and and the critical value for this case would be:

And replacing we got:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

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step-by-step explanation:

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