Respuesta :
Answer:
[tex] (240-210) -1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=27.953[/tex]
[tex] (240-210) +1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=32.047[/tex]
And the confidence interval for the difference is between:
[tex] 27.953 \leq \mu_1 -\mu_2 \leq 32.047[/tex]
Step-by-step explanation:
We have the following info given:
[tex] \bar X_1 = 240[/tex] sample mean for medium Pizzas from Prim's
[tex] \bar X_2 = 210[/tex] sample mean for medium Pizzas from Pizza Place
[tex] s_1 =8.6[/tex] sample deviation for Prim's
[tex] s_2 =5.7[/tex] sample deviation for Pizza Palca
[tex] n_1 =n_2 = 100[/tex] sample size selected for each case
The confidence interval for the difference of means is given by:
[tex] (\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}[/tex]
And for the 95% confidence we need a significance level of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], the degrees of freedom are given by:
[tex] df= n_1 +n_2 -2= 100+100-2 =98[/tex]
And the critical value would be
[tex] t_{\alpha/2}= 1.984[/tex]
And replacing we got:
[tex] (240-210) -1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=27.953[/tex]
[tex] (240-210) +1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=32.047[/tex]
And the confidence interval for the difference is between:
[tex] 27.953 \leq \mu_1 -\mu_2 \leq 32.047[/tex]
Answer:
E
Step-by-step explanation:
30 plus or minus 1.98 sqrt 8.6^2/100 + 5.7^2/100
