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 13 days following no advertisements, the mean was 23.9 purchasing customers with a standard deviation of 1.9 customers. On 6 days following advertising, the mean was 24.7 purchasing customers with a standard deviation of 1.6 customers. Test the claim, at the 0.01 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 3 of 3: Draw a conclusion and interpret the decision.

One year Terry had the lowest ERA (earned-run average, mean number of runs yielded per nine innings pitched) of any male pitcher at his school, with an ERA of 2.98. Also, Karla had the lowest ERA of any female pitcher at the school with an ERA of 3.37. For the males, the mean ERA was 4.456 and the standard deviation was 0.557. For the females, the mean ERA was 4.166 and the standard deviation was 0.635. Find their respective z-scores. Which player had the better year relative to their peers, Terry or Karla? (Note: In general, the lower the ERA, the better the pitcher.) Terry had an ERA with a z-score of Karla had an ERA with a z-score of (Round to two decimal places as needed.) Which player had a better year in comparison with their peers? OA. Karla had a better year because of a lower z-score. B. Terry had a better year because of a lower z-score. OC. Terry had a better year because of a higher z-score. OD. Karla had a better year because of a higher z-score.

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 13 days following no advertisements, the mean was 23.9 purchasing customers with a standard deviation of 1.9 customers. On 6 days following advertising, the mean was 24.7 purchasing customers with a standard deviation of 1.6 customers. Test the claim, at the 0.01 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 1 of 3: State the…

It seems these days that college graduates who are employed full-time work more than 40-hour weeks. Data are available that can help us decide if this is true. A survey was recently sent to a group of adults selected at random. There were 15 respondents who were college graduates employed full-time. The mean number of hours worked per week by these 15 respondents was 43 hours, with a standard deviation of 8 hours. Assume that the population of hours worked per week by college graduates employed full-time is normally distributed with mean μ. Can we conclude that μ is greater than 40 hours? Use the 0.05 level of significance. State the null hypothesis  H0  and the alternative hypothesis  H1 Find the t-test df & Solve the t-test value  Find the critical value Can we conclude, at the 0.05 level of significance, that the mean number of hours worked per week by college graduates is greater than 40 hours? Yes or No?