A marketing firm asked a random set of married women and married men how much they were willing to spend for jewellery as a present for their spouse. Can the firm conclude, at , that
Each of the two groups has a different willingness to spend? (Use the unequal variances formula.)
$$\begin{array} { | l | c | c | }
\hline & \text { Women } & \text { Men } \
\hline \text { Sample size } & 8 & 11 \
\hline \text { Mean spending amount } & 80 & 55 \
\hline \text { Sample variance } & 50 & 800 \
\hline
\end{array}$$
A) No, because the test value 0.32 is inside the interval (-2.31, 2.31)
B) Yes, because the test value 2.81 is outside the interval (-2.36, 2.36)
C) Yes, because the test value 16.00 is inside the interval (-2.31, 2.31)
D) No, because the test value 16.00 is outside the interval (-2.36, 2.36)

Question

A marketing firm asked a random set of married women and married men how much they were willing to spend for jewellery as a present for their spouse. Can the firm conclude, at  , that
Each of the two groups has a different willingness to spend? (Use the unequal variances formula.)
$$\begin{array} { | l | c | c | } 
\hline & \text { Women } & \text { Men } \
\hline \text { Sample size } & 8 & 11 \
\hline \text { Mean spending amount } & 80 & 55 \
\hline \text { Sample variance } & 50 & 800 \
\hline
\end{array}$$
A) No, because the test value 0.32 is inside the interval (-2.31, 2.31) 
B) Yes, because the test value 2.81 is outside the interval (-2.36, 2.36) 
C) Yes, because the test value 16.00 is inside the interval (-2.31, 2.31) 
D) No, because the test value 16.00 is outside the interval (-2.36, 2.36)

Quizplus · Accepted Answer

We can use the two-sample t-test with unequal variances formula to determine if the means of the two groups are significantly different.

The test statistic is calculated as:

$t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$

Plugging in the values from the table, we get:

$t = \frac{80 - 55}{\sqrt{\frac{50}{8} + \frac{800}{11}}} \approx 2.81$

To determine if this test statistic is significant, we need to compare it to the critical values of the t-distribution with degrees of freedom given by:

$df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{s_1^4}{n_1^2(n_1-1)} + \frac{s_2^4}{n_2^2(n_2-1)}}$

Plugging in the values from the table, we get:

$df \approx 12.96$

Using a two-tailed test with significance level $\alpha = 0.05$, the critical values are:

$t_{0.025, 12.96} \approx \pm 2.36$

Since our test statistic of 2.81 is outside the interval (-2.36, 2.36), we reject the null hypothesis and conclude that there is a significant difference in the mean spending amount between the two groups. Therefore, the marketing firm can conclude that each of the two groups has a different willingness to spend. Hence, the answer is B.

A Marketing Firm Asked a Random Set of Married Women