Question

Construct a 90% confidence interval for μ1-μ2 with the sample statistics for mean calorie content of two​ bakeries' specialty pies and confidence interval construction formula below. Assume the populations are approximately normal with equal variances.

Answer 1

Answer:

The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (49) < μ₁ - μ₂ < (289)

Step-by-step explanation:

The given data are;

Bakery A

$\overline x_1$ = 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

$\overline x_2$ = 1,711 cal

s₂ = 192 cal

n₂ = 10

$\left (\bar{x}_1-\bar{x}_{2} \right ) - t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2} \right ) + t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}$

df = n₁ + n₂ - 2

∴ df = 10 + 18 - 2 = 26

From the t-table, we have, for two tails, $t_c$ = 1.706

$\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}$

$\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469$

$\hat \sigma$ ≈ 178

Therefore, we get;

$\left (1,880-1,711 \right ) - 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}< \mu _{1}-\mu _{2}< \left (1,880-1,711 \right ) + 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}$

Which gives;

$169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}$

Therefore, by rounding to the nearest integer, we have;

The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289