Question

The owners of Prim's Pizza are concerned that many of their customers are starting to purchase pizza from The Pizza Palace because of its new pizza, which has fewer calories than Prim's. A random sample of 100 medium pizzas from Prim's found a mean of 240 calories and a standard deviation of 8.6 calories. A random sample of 100 medium pizzas from The Pizza Palace found a mean of 210 calories and a standard deviation of 5.7 calories. Which of the following formulas gives a 95% confidence interval for the difference in mean calories between a Prim's Pizza medium pizza and a medium pizza from The Pizza Palace?

Answer 1

Answer:

$(240-210) -1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=27.953$

$(240-210) +1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=32.047$

And the confidence interval for the difference is between:

$27.953 \leq \mu_1 -\mu_2 \leq 32.047$

Step-by-step explanation:

We have the following info given:

$\bar X_1 = 240$ sample mean for medium Pizzas from Prim's

$\bar X_2 = 210$ sample mean for medium Pizzas from Pizza Place

$s_1 =8.6$ sample deviation for Prim's

$s_2 =5.7$ sample deviation for Pizza Palca

$n_1 =n_2 = 100$  sample size selected for each case

The confidence interval for the difference of means is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

And for the 95% confidence we need a significance level of $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$, the degrees of freedom are given by:

$df= n_1 +n_2 -2= 100+100-2 =98$

And the critical value would be

$t_{\alpha/2}= 1.984$

And replacing we got:

$(240-210) -1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=27.953$

$(240-210) +1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=32.047$

And the confidence interval for the difference is between:

$27.953 \leq \mu_1 -\mu_2 \leq 32.047$