Question

Based on the information from Consumer Reports, a random sample of 86 thirty-gram servings of Duncan Hines Chocolate Chip Cookies had a sample means of 132 calories with a sample standard deviation of 27 calories. A random sample of 75 Pepperidge Farm Chocolate Chip Cookies has a sample mean of 124 calories with sample standard deviation of 33 calories. Build an 80% confidence interval for the true mean difference in calories for these two cookie brands.

Answer 1

Answer:

$(132-124) -1.287 \sqrt{\frac{27^2}{86} +\frac{33^2}{75}}= 1.828$

$(132-124) +1.287 \sqrt{\frac{27^2}{86} +\frac{33^2}{75}}= 14.172$

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 expresseda % 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".

For this case we have the following info given:

$\bar X_1 = 132$ mean for Duncan Hines Chocolate Chip Cookies

$n_1 =86$ represent the sample size for Duncan Hines Chocolate Chip Cookies

$s_1 = 27$ represent the deviation for Duncan Hines Chocolate Chip Cookies

$\bar X_2 = 124$ mean for Pepperidge Farm Chocolate Chip Cookies

$n_2 =75$ represent the sample size for Pepperidge Farm Chocolate Chip Cookies

$s_2 = 33$ represent the deviation for Pepperidge Farm Chocolate Chip Cookies

Solution to the problem

The degrees of freedom for this case are:

$df = n_1 +n_2 -2 = 86+75-2=159$

Since the Confidence is 0.80 or 80%, the value of $\alpha=0.2$ and $\alpha/2 =0.1$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.1,159)".And we see that $t_{\alpha/2}=1.287$

The confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{2}}$

And replacing we got:

$(132-124) -1.287 \sqrt{\frac{27^2}{86} +\frac{33^2}{75}}= 1.828$

$(132-124) +1.287 \sqrt{\frac{27^2}{86} +\frac{33^2}{75}}= 14.172$