A nutritionist believes that obesity is more prevelant among American adults tha

Question

A nutritionist believes that obesity is more prevelant among American adults than it was in the past. He discovers that in a study conducted in the year 1994, 380 of 1630 randomly chosen adults were classified as obese. However, in a more recent study, he finds 726 out of 2350 randomly chosen adults were classified as obese. At alpha = 0.05, do these studies provide evidence to support the nutrionist's clam that the proportion of obese adults has significantly increaed since 1994?

A) State the hypothesis and the claim

B) Give the forumula and find the standard test statistic

C) Find and state the rejection region

D) Make a decision

E) Interpret the results

Explanation / Answer

A) Here the hypothesis can be stated as below:

Null Hypothesis = H0 = There is no change in the proportion of obese adults since 1994 i.e. P = 380/1630 = 0.2331288

Alternative Hypothesis = HA = Proportion of obese adults has increased since 1994 i.e. P > 0.2331288

B) For this analysis, the significance level is 0.05. The test method will be one sample z-test.

We need to calculate the standard deviation () and compute the z-score test statistic (z) as follows

= sqrt[ P * ( 1 - P ) / n ] where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

= sqrt[ 0.2331288 * (1 - 0.2331288) / (2350) = 0.008722184

z = (p - P) / = [ (726/2350) - 0.2331288 ] / 0.008722184 = 8.691329

C) At alpha = 0.05, the critical value can be seen from the z-table which is 1.645

So if our test statisitc is greater than 1.645, we can reject the null hypothesis.

D) As our test statistic is 8.69 which is greater than critical value of 1.645, we can reject the null hypothesis.

E) This means that the proportion of obese adults has significantly increased since 1994.

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