Compute by hand and upload your work You are a teacher and want to analyze if your morning class has significantly different scores than your afternoon class. You have 12 students in both of your classes. Compute a t-test for independent samples on these scores from the last quiz you administered and then use the eight steps to test the null hypothesis that there is no difference between the two conditions. Do the two groups differ?Morning class Afternoon class87 8698 7375 7988 5676 7285 7092 8756 5989 6485 7784 7492 72.  How would you interpret:: t(21) = 28.90, p < .05?

A local ice cream shop collects data on the temperature of the day and how many ice cream cones were sold that day.  What type of data was collected?     Bivariate and Quantitative     Bivariate and Categorical     Univariate and Quantitative     Univariate and Categorical

Based from the photo, This is the step1, so step 2 and step 3 and Step 4 must be answered. (Step 2: Test Statistic -Pearson's r correlation coefficient AND Step 3: Critical Values, and Step 4: Solve For the test stat) so based it from the photo. The table shows the number of hours spend by 30 students in studing and their test scores. Solve for the Pearson’s r and interpret the results. Use 5% significance level. Step 1: Null and Alternative Hypothesis Ho: r = 0; There is no significant relationship between the number of hours spent studying and the students test scores. Ha: r ≠ 0; There is a significant relationship between the number of hours spent studying and the students test scores. Step 2: Test Statistic -Pearson's r correlation coefficient Step 3: Critical Values, Step 4: Solve For the test stat PLEASE SHOW ME COMPLETE SOLUTIONS AND CORRECT ANSWERS THANK YOU!! It was a pleasure to me. HIGHLIGHT STEP 4: (step 4 is must)!!

Suppose that 75%75% of college students prefer pepperoni on their pizzas. You took a sample of ?=150n=150 at your college and found that 70%70% said they want pepperoni on their pizzas. You want to know if your college is different from the rest at the 10%10% level. What is the ?P‑value for this test?