Question

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.

Answer 1

Answer:

$(75-71) - 2.14* 3 \sqrt{\frac{1}{8} +\frac{1}{8}}=0.79$

$(75-71) + 2.14* 3 \sqrt{\frac{1}{8} +\frac{1}{8}}=7.21$

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".

$\bar X$ represent the sample mean for the sample  

$\mu$ 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:

$df = n_1 +n_2 -2 = 8+8-2= 14$

The confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} S_p \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

$S_p =\sqrt{9}=3$

The confidence is 0.95, the value for the significance is $apha=1-0.95=0.05$ and $\alpha/2 =0.025$ and the critical value for this case would be: $t_{\alpha/2} =2.14$

And replacing we got:

$(75-71) - 2.14* 3 \sqrt{\frac{1}{8} +\frac{1}{8}}=0.79$

$(75-71) + 2.14* 3 \sqrt{\frac{1}{8} +\frac{1}{8}}=7.21$