A diet doctor claims that the average North American is more than 20 pounds overweight. To test his claim, a random sample of 20 North Americans was weighed, and the difference between their actual and ideal weights was calculated. The data are listed here. Do these data allow us to infer at the 5% signif- icance level that the doctor’s claim is true?

Complete question:

A diet doctor claims that the average North American is more than 20 pounds overweight. To test his claim, a random sample of 20 North Americans was weighed, and the difference between their actual and ideal weights was calculated. The data are listed here. Do these data allow us to infer at the 5% signif- icance level that the doctor’s claim is true?

16 23 18 41 22 18 23 19 22 15 18 35 16 15 17 19 23 15 16 26

See explanation below.

Step-by-step explanation:

From the given data, we have:

∑x = 16 + 23 + 18 + 41 + 22 + 18 + 23 + 19 + 22 + 15 + 18 + 35 + 16 + 15 + 17 19 + 23 + 15 + 16 + 26 = 417

The sample mean x' =

(x - x')² = 23.5225, 4.6225, 8.1225, 406.0225, 1.3225, 8.1225, 4.6225, 3.4225, 1.3225, 34.2225, 8.1225, 200.2225, 23.5225, 34.2225, 14.8225, 3.4225, 4.6225, 34.2225, 23.5225, 26.5225

∑(x-x')² = 868.55

Degree of freedom, df = 20 - 1 = 19

For the standard deviation, we have:

Standard deviation = 6.76

We now have the following:

Sample size, n = 20

Mean, u = 20

Sample mean, x' = 20.85

Standard deviation, s.d = 6.76

Significance level = 5% = 0.05

Here, the null and alternative hypotheses are:

H0 : u = 20

H1 : u > 20

This is a right tailed test.

For the test statistic, we have:

T statistics = 0.56

The pvalue for a right tailed test, df=19, t=0.56,

p-value = P(t>0.56) = 1-p(t≤0.056)

= 1 - 0.7090

= 0.291

Since p-value, 0.291 is greater than significance level, 0.05, we fail to reject null hypothesis, H0.

Conclusion:

There is not enough evidence to conclude that the average North American is more than 20 pounds overweight.