Question

3. Eat This, Not That! Examined nutrition information for one serving of chocolate chip cookies for 18 brands of chocolate chip cookies and found the number of calories per serving. (Accessed from https://www.eatthis.com/best-worst-chocolate-chip-cookies/ on 7/30/20) 160 120 130 160 150 150 150 160 140 140 140 130 160 160 210 140 200 130 (a) Create a 97% confidence interval for the average calorie content of a serving of chocolate chip cookies. (Be sure to show all work as done in class.) [5 pts) (b) A meal plan claims that you will consume 148 calories from a serving of chocolate chip cookies. Test this claim at the 5% significance level. Show all work. i. Define the parameter(s) and state the hypotheses: [3 pts] ii. Find the observed sample statistic (using your calculator) and the test statistic (using a formula). Use proper notation for the sample statistic ( I or ), and clearly identify your test statistic as either a 2-score or a t-score. [4pt]

We do not use critical values at all in our course so please avoid using them when giving me explanations, as they can cause confusion on my end between your explanation and what im learning on my course

here is the link in the paper though it seems the data is already provided so it might not be required, just posting it in case you'd like to click on it https://www.eatthis.com/best-worst-chocolate-chip-cookies/

Answer 1

3. a)

Letī denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, 23.073 D = 151.667, s2 = 532.352941, s = 23.073, n = 18, standard error 5.438305 5.438 V18 The critical value is tcritical t 1-0.97 - t0.015,17 2.368055 2.368 ,17 At 97 percent confidence level, The margin of error is: 2 S 23.073 23.073 ME * t 1-0.97 * +0.015,17 * 2.368 = 12.874608 ~12.875 vn ,17 2 V18 ✓18 a 97 percent confidence interval of population mean u is S * t 1-0.97 = TE ✓n ,17 2 23.073 = 151.667 + *t0.015,17 V18 23.073 = 151.667 + * 2.368 V18 = 151.667 – 12.875 = (138.792058, 164.541275) ~ (138.79, 164.54)

b) parameter of interest : average calories from a serving of chocolate chip cookies = $\mu$

Let u denotes the population mean. Hypotheses : To test null hypothesis H. :p =148 against alternative hypothesis H1 :p1 # 148 Let denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, S 23.07 T = 151.6667, s2 = 532.3529, s = 23.0728, n = 18, standard error of mean 5.4383 in V18 The test statistic can be written as: Vnī – 148) t= S which under H, follows at distribution with n-1 df. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tobs | > to.025,n–1 = 2.109816 0.4 0.3 0.2 0.1 0.0- t Now, Value of the test statistic: The value of the test statistic tobs = 18 (151.6667 – 148) 23.0728 = 0.674230 ~ 0.67 associated degrees of freedom = 17 The critical value = Et 0.05 = £t0.025,17 = +2.109816 +2.1098 2 3n-1 P-value = P(|tn-1| > tobs) = 2 * P(t17 > 0.674230) = 2* (1 – 0.745385) = 0.509231 ~ 0.5092 Conclusion : Since p-value > 0.05 and tobs tcritical = 2.109816, so We fail to reject H, at 0.05 level of significance Hence, we can conclude that the population mean is not significantly different from 148.