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
[tex]\overline x_1[/tex] = 1,880 cal
s₁ = 148 cal
n₁ = 10
Bakery B
[tex]\overline x_2[/tex] = 1,711 cal
s₂ = 192 cal
n₂ = 10
[tex]\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}}}[/tex]
df = n₁ + n₂ - 2
∴ df = 10 + 18 - 2 = 26
From the t-table, we have, for two tails, [tex]t_c[/tex] = 1.706
[tex]\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}}[/tex]
[tex]\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469[/tex]
[tex]\hat \sigma[/tex] ≈ 178
Therefore, we get;
[tex]\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}}[/tex]
Which gives;
[tex]169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}[/tex]
Therefore, by rounding to the nearest integer, we have;
The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289
