Question

According to a Stanford Business article, Americans may eat fewer calories at restaurants if the calories of the food items are labeled on the menu. To investigate this, researchers compared Starbucks receipts from locations where the menus were labeled to receipts from stores where the menus were not labeled. A randomsample of 30 receipts from stores with the menus labeled had an average number of calories of 225 calories with a standard deviation of 100 calories. A random sample of 40 receipts liom stores withous menus labeled showed an average of 265 calories per receipt with a standard deviation of 75 calories. Does this provide convincing evidehee that the average calories per receipt at Starbucks with a labeled menu is less than at a Starbucks without labeled menus?

Answer 1

To determine if there is convincing evidence that the average calories per receipt at Starbucks with a labeled menu is less than at a Starbucks without labeled menus, we can conduct a two-sample t-test.

Step-by-step explanation:

To determine whether there is convincing evidence that the average calories per receipt at Starbucks with a labeled menu is less than at a Starbucks without labeled menus, we can conduct a hypothesis test. We will compare the two sample means using a two-sample t-test.

We will set up the null and alternative hypotheses as follows:

• Null hypothesis (H0): The average calories per receipt at Starbucks with a labeled menu is greater than or equal to the average calories per receipt at a Starbucks without labeled menus.
• Alternative hypothesis (Ha): The average calories per receipt at Starbucks with a labeled menu is less than the average calories per receipt at a Starbucks without labeled menus.

Next, we calculate the test statistic and p-value. Using the provided information, the test statistic can be calculated as:

t = ((225 - 265) - 0) / sqrt((100^2/30) + (75^2/40)) = -2.761

We can then find the p-value associated with this test statistic using a t-table or a t-distribution calculator. Based on the p-value, if it is less than the significance level (commonly 0.05), we reject the null hypothesis in favor of the alternative hypothesis. This would provide convincing evidence that the average calories per receipt at Starbucks with a labeled menu is less than at a Starbucks without labeled menus.