A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 10 days following no advertisements, the mean was 23.6 purchasing customers with a standard deviation of 0.9 customers. On 11 days following advertising, the mean was 24.9 purchasing customers with a standard deviation of 1.5 customers. Test the claim, at the 0.10 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2.
Step 2 of 3 : Compute the value of the test statistic. Round your answer to three decimal places.

Step 1 ：We are given that a small business is testing the effectiveness of its radio advertising. They decide to compare the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played versus those days following days when no radio advertisements are played.

Step 2 ：For 10 days following no advertisements, the mean was 23.6 purchasing customers with a standard deviation of 0.9 customers. On 11 days following advertising, the mean was 24.9 purchasing customers with a standard deviation of 1.5 customers.

Step 3 ：We are asked to test the claim that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. This is a two-sample t-test problem.

Step 4 ：The test statistic for a two-sample t-test is given by the formula: $$t=\frac{({\overline{X}}_1-{\overline{X}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$$ where: ${\overline{X}}_1$ and ${\overline{X}}_2$ are the sample means, ${\mu }_1$ and ${\mu }_2$ are the population means, $s_1^2$ and $s_2^2$ are the sample variances, $n_1$ and $n_2$ are the sample sizes.

Step 5 ：In this case, we are testing the null hypothesis that ${\mu }_1={\mu }_2$, so the term $({\mu }_1-{\mu }_2)$ in the numerator is zero. The given data can be plugged into the formula to calculate the test statistic.

Step 6 ：Given: $n_1=10$, ${\overline{X}}_1=23.6$, $s_1=0.9$, $n_2=11$, ${\overline{X}}_2=24.9$, $s_2=1.5$

Step 7 ：Substituting these values into the formula, we get the test statistic $t=-2.4327961883099527$

Step 8 ：Rounding this to three decimal places, we get the final answer: $\boxed{{\displaystyle -2.433}}$