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Suppose a 95% confidence interval for the average amount of weight loss on a diet program for males is between 13. 4 and 18. 3 pounds. These results were based on a sample of 42 male participants who were deemed to be overweight at the start of the 4-month study. What is the standard error of the sample mean?.



Answer :

The standard error of the sample mean is 1.21.

Given;

Suppose a diet regimen for men results in an average weight loss of between 13. 4 and 18. 3 pounds, according to a 95% confidence interval. These findings were based on a group of 42 men who were classified as overweight at the beginning of the four-month trial.

A 95% confidence interval for a population mean is (13.4, 18.3)

Upper limit = 18.3

Lower limit = 13.4

Since population SD is unknown, this interval is constructed using the t distribution.

n = 42

c = 0.95

∴ α = 1 - c = 1 - 0.95 = 0.05

α/2 = 0.025

Also, d.f = n - 1 = 42 - 1 = 41

∴ ta/2.d.f =  ta/2.n-1  =  t0.025,41  =  2.02 . . . . use t table

Now,

The margin of error = (Upper limit - Lower limit)/2

                                 = (18.3 - 13.4)/2

                                 = 2.45

But,

Margin of error = ta/2.d.f-  * (s / \sqrt{} n)

Margin of error = ta/2.d.f-  * Standard error

2.45 = 2.02 * Standard error

Standard error = 1.2129

To learn more about Margin of error click here:

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