Question

A nutrition expert claims that the average American is overweight. To test his claim, a random sample of 26 Americans was selected, and the difference between each person's actual weight and idea weight was calculated. For this data, we have x¯=16.1x¯=16.1 and s=30s=30. Is there sufficient evidence to conclude that the expert's claim is true? Carry out a hypothesis test at a 6% significance level.

A. The value of the standardized test statistic:

Note: For the next part, your answer should use interval notation. An answer of the form (−∞,a)(−∞,a) is expressed (-infty, a), an answer of the form (b,∞)(b,∞) is expressed (b, infty), and an answer of the form (−∞,a)∪(b,∞)(−∞,a)∪(b,∞) is expressed (-infty, a)U(b, infty).

B. The rejection region for the standardized test statistic:

C. The p-value is

D. Your decision for the hypothesis test:

A. Do Not Reject H0H0.
B. Reject H0H0.
C. Do Not Reject H1H1.
D. Reject H1H1.

Also, can you add in how I would solve this on a TI-83 calculator? Thanks!

Answer 1

Solution:

Given:

Sample Size = n = 26

the difference between each person's actual weight and idea weight was calculated.

Sample mean of differences =

Sample standard deviation = s = 30

Claim: the average American is overweight.

That is we have to test the hypothesis:

Vs

significance level = 6% = 0.06

We use following steps in TI 83 / TI 84 ( steps are same on both calculators)

Press STAT and select TESTS

Select T Test

Select Stats

Under Stats enter numbers:

Click on Calculate and pres Enter

Part A. The value of the standardized test statistic:

t = 2.736

B. The rejection region for the standardized test statistic:

Since this is right tailed test, rejection region would be:
(b,∞)

So we need to find value b = t critical value.

df = n - 1 = 26 - 1 = 25

Significance level is 0.06

Thus use following steps in TI 84:

Press 2ND and VARS

select invT(

Since this is right tailed test enter Area = 1 - 0.06

Thus

Click on Paste and press Enter two times.

t critical value = 1.6097895

t critical value = 1.610

Thus rejection region for the standardized test statistic is: ( 1.610 , ∞)

If TI 83 does not have invT( , use Excel command:

=T.INV( 1-0.06,25)

Part C. The p-value is:

From TI 83/Ti 84 calculator:

we have: P-value = 0.0056

Part D. Your decision for the hypothesis test:

Since P-value = 0.0056 < 0.06 significance level , we reject H0.

Thus correct answer is:

B. Reject H0