A research article describes a study that investigated the relationship between depression and chocolate consumption. Participants in the study were 931 adults who were not currently taking medication for depression. These participants were screened for depression using a widely used screening test. The participants were then divided into two samples based on the score on the screening test. One sample consisted of people who screened positive for depression, and the other sample consisted of people who did not screen positive for depression. Each of the study participants also completed a food frequency survey. The researchers believed that the two samples were representative of the two populations of interest-adults who would screen positive for depression and adults who would not screen positive. The paper reported that the mean number of servings of chocolate for the sample of people that screened positive for depression was 8.44 servings per month, and the sample standard deviation was 14.82. For the sample of people who did not screen positive for depression, the mean number of servings per month was 5.29, and the standard deviation was 8.78. The paper did not say how many individuals were in each sample, but for the purposes of this exercise, you can assume that the 931 study participants were divided into 311 who screened positive for depression and 620 who did not screen positive. Carry out a hypothesis test to confirm the researchers' conclusion that the mean number of servings of chocolate per month for people who would screen positive for depression is greater than the mean number of chocolate servings per month for people who would not screen positive for depression. (Use α = 0.05. Use μ₁ for people who would screen positive for depression and μ₂ for people who would not screen positive.) State the appropriate null and alternative hypotheses. HH-0 Mai My-M₂-0 Ho: H1-20 Ha H₁₂ = 0 Find the test statistic and P-value. (Use technology to calculate the P-value. Round your test statistic to one decimal place and your P-value to three decimal places.) t = P-value=