Low-fat or low-carb? Are low-fat diets or low-carb diets...

Low-fat or low-carb? Are low-fat diets or low-carb diets more effective for weight loss? A sample of 77 subjects went on a low-carbohydrate diet for six months. At the end of that time, the sample mean weight loss was 4.7 kilograms with a sample standard deviation of 7.2 kilograms. A second sample of 79 subjects went on a low-fat diet. Their sample mean weight loss was 2.6 kilograms with a standard deviation of 5.9 kilograms. a. Construct a 98% confidence interval for the difference in mean weight loss between the low-fat and low-carb diets. b. A dietitian claims that both diets are equally effective. Does the confidence interval contradict this claim?

Transcript

00:01 Hello students, as per the given question we need to compare the mean weight loss between the low fat and the low carb diet and determine if there are equally efficient or not.

00:11 So, the given data is for low carb diet it is given that sample size n1 is equals to 77, sample a mean weight loss which is x1 bar is equals to 4 .7 kilograms and sample standard deviation s1 is equals to 7 .2 kilograms and for low fat diet it is n2 is equals to 79 and sample mean weight loss is x2 bar is equals to 2 .6 kgs and s2 is equals to 5 .9 kgs.

01:02 So, let us construct for the first bit a where we need to construct 98 percent confidence interval for difference in mean weight loss.

01:14 So, in order to calculate that first we need to calculate the standard error where standard error is given by under root of s1 square by n1 plus s2 square by n2.

01:39 So, this is the formula for standard error where we need to substitute the values which gives standard error is equals to under root of 1 .5092 after substituting the values and simplifying that and the standard error is equals to 1 .2279 when rounded to 4 decimals.

02:02 So, in the next step let us find the degrees of freedom and calculate the critical value.

02:12 So, the degrees of freedom is minimum of n1 minus 1 comma n2 minus 1.

02:22 So, the minimum value of 77 minus 1 which is 76 comma 79 minus 1 which is 78 which is 76.

02:37 So, the critical degrees of freedom is equals to 76 degrees of freedom and the critical value is for alpha by 2 which is 0 .01 as it is two tailed test for degrees of freedom 76 it is approximately equals to 2 .621.

03:02 This is extracted from the t table values and for mean error is given by t star into standard error which is 2 .621 into 1 .2279 which is approximately equals to 3 .2244 when rounded to 4 decimal values...

Question

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