2. Setup: Suppose Camila went out and obtained a sample of fruit (one serving). She is interested in the mean calorie content. She believes that the distribution of calorie content is approximately normal. The data that she collected is given below: Fruit Calories Fruit Calories Fruit Calories Apple 72 Grapes 110 Pear 96 Apricot 17 Kiwi 46 Pineapple 74 Banana 105 Tangerine 37 Plum 30 Blackberries 62 Nectarine 60 Raspberries 64 Blueberries 83 Orange 62 Strawberries 49 Cherries 74 Papaya 59 Watermelon 46 Grapefruit 41 Peach 38 Goal: Find a 93.5% upper confidence bound for the mean calorie content of the standard size candies. Procedure: We carried out the analysis in R. Below is the output for 3 different tests that were run. Potential R Output: OPTION 1: > t.test(fruit, alternative = "greater", conf.level = 0.935) ONE SAMPLE T-TEST data: fruit t = 11.126, df = 19, p-value = 4.593e-10 alternative hypothesis: true mean is greater than 0 93.5 percent confidence interval: 52.53745 sample estimates: mean of x Inf 61.25 OPTION 2: > t.test(fruit, = "less", alternative conf.level = 0.935) ONE SAMPLE T-TEST data: fruit t = 11.126, df = 19, p-value = 1 alternative hypothesis: true mean is less than 0 93.5 percent confidence interval: -Inf 69.96255 sample estimates: mean of x 61.25 OPTION 3: > t.test(fruit, alternative = "two.sided", conf.level = 0.935) ONE SAMPLE T-TEST data: fruit t 11.126, df = 19, p-value = 9.186e-10 alternative hypothesis: true mean is not equal to 0 93.5 percent confidence interval: 50.4678 72.0322 sample estimates: mean of x 61.25 **PROBLEM CONTINUED ON NEXT PAGE** 2 Questions: (a) Which of the three options for the R code would you need to use to identify the 93.5% upper confidence bound? (b) State the confidence bound from the code option that you chose. (c) Write an interpretation of the interval that you found. Make sure to include the problem context in your interpretation.