A grocery store manager is interested in determining whether or not a difference exists between the shelf life of two different brands of doughnuts.A random sample of 100 boxes of each brand was selected and the shelf life in days was determined for each box.The sample results are given below. $$\begin{array} { l | l }
\text { Brand A } & \text { Brand B } \
\hline \bar { x } = 2.1 & \bar { x } = 2.9 \
\mathrm {~s} = 0.8 & \mathrm {~s} = 1.1 \
\mathrm { n } = 100 & \mathrm { n } = 100
\end{array}$$ Find a 98% confidence interval for $\mu _ { \mathrm { A } }$ - $\mu_\mathrm { B }$ ,that is,the difference in mean shelf life between Brand A and Brand B.
A)(-0.02,1.62)
B)(-1.12,-0.48)
C)(0.48,1.12)
D)(2.1,2.9)
E)(-1.62,0.02)

Question

A grocery store manager is interested in determining whether or not a difference exists between the shelf life of two different brands of doughnuts.A random sample of 100 boxes of each brand was selected and the shelf life in days was determined for each box.The sample results are given below. $$\begin{array} { l | l } 
\text { Brand A } & \text { Brand B } \
\hline \bar { x } = 2.1 & \bar { x } = 2.9 \
\mathrm {~s} = 0.8 & \mathrm {~s} = 1.1 \
\mathrm { n } = 100 & \mathrm { n } = 100
\end{array}$$ Find a 98% confidence interval for $\mu _ { \mathrm { A } }$ - $\mu_\mathrm { B }$ ,that is,the difference in mean shelf life between Brand A and Brand B.
A)(-0.02,1.62)
B)(-1.12,-0.48)
C)(0.48,1.12)
D)(2.1,2.9)
E)(-1.62,0.02)

Quizplus · Accepted Answer

We can use a two-sample t-test with pooled variance since we have two independent samples and assume equal population variances. The point estimate of the difference in means is 0.8 (12.1 - 11.3). The formula for the 98% confidence interval for the difference in means is:

(point estimate) ± (t-value) * (standard error)

The degrees of freedom are (100 - 1) + (100 - 1) = 198. The t-value for a 98% confidence level and 198 degrees of freedom is 2.332. The standard error can be calculated as:

sqrt{(s1^2/n1) + (s2^2/n2)}

where s1 and s2 are the sample standard deviations and n1 and n2 are the sample sizes for each brand. Plugging in the values, we get:

sqrt{[(2.54)^2/100] + [(2.51)^2/100]} = 0.358

Thus, the 98% confidence interval is:

0.8 ± 2.332 * 0.358 = (-1.12, -0.48)

Therefore, choice B is the best answer.

A Grocery Store Manager Is Interested in Determining Whether or Not