2:30 PM webassign.net \"I Sprint ve increase our food intake, we generally gain

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2:30 PM webassign.net "I Sprint ve increase our food intake, we generally gain weight. Nutrition scientists calculate the amount of weight gain that would be associated with a en increase in calories. In one study, 16 nonobese adults, aged 25 to years, were fed 1000 calories per day in excess of the calories needed to intain a stable body weight. The subjects maintained this diet for veeks, so they consumed a total of 56,000 extra calories. According to ory, 3500 extra calories will translate into a weight gain of 1 pound. erefore, we expect each of these subjects to gain ,000/3500 = 16 pounds (Ib). Here are the weights before and after the veek period, expressed in kilograms (kg). Weight before 73.3 634 681 737 917 559 617 578 (a) For each subject, subtract the weight before from the weight after to determine the weight change in kg Subjects 48 Subject Subject 123a select S 1.5 kg (b) Find the mean and the standard deviation for the weight change. (Round your answers to four decimal places.) -4725k -1.7823 (c) Calculate the standard error se and the margin of error me for 95% confidence. (Round your answers to four decimal places.) Report the 95% confidence interval for weight change in a sentence that explains the meaning of the 95%. (Round your answers to four decimal places.) Based on a method that gives correct results 95% of the time, the mean weight change is kgtox kg

Explanation / Answer

Solution:

Part c

We are given

Sample mean = Xbar = 4.725

Sample Standard deviation = S = 1.7823

Sample size = n = 16

Confidence level = 95%

Degrees of freedom = df = n – 1 = 16 – 1 = 15

Critical t-value = 2.1314

(By using t-table or excel)

Standard error = se = S/sqrt(n) = 1.7823/sqrt(16) = 1.7823/4 = 0.445575

Se = 0.4456

Margin of error = me = t*se = 2.1314*0.445575 = 0.949699

Me = 0.9497

Confidence interval = Xbar -/+ Me

Lower limit = Xbar – Me = 4.725 - 0.9497 = 3.7753

Upper limit = Xbar + Me = 4.725 + 0.9497 = 5.6747

Based on a method that gives correct results 95% of the time, the mean weight change is 3.7753 kg to 5.6747 kg.

Part d

Sample mean = Xbar = 4.725*2.2 = 10.3950 pounds

Sample Standard deviation = S = 1.7823*2.2 = 3.9211 pounds

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