The data in Nutrition Study include information on nutrition and health habits of a sample of 315 people. One of the variables is Smoke, indicating whether a person smokes or not (yes or no). Use technology to test whether the data provide evidence that the proportion of smokers is different from 20% given that 43 identify themselves as smokers. Clearly state the null and alternative hypotheses

In this case we need to test whether the proportion of smokers is different from 20%.

A one-proportion z-test can be used to determine the conclusion for this test.

The hypothesis defined as:

H₀: The proportion of smokers is 20%, i.e. p = 0.20.

Hₐ: The proportion of smokers is different from 20%, i.e. p ≠ 0.20.

The information provided is:

n = 315

X = number of people who identified themselves as smokers = 43

Compute the sample proportion of smokers as follows:

$\hat p=\frac{X}{n}=\frac{43}{315}=0.137$

Compute the test statistic as follows:

$z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.137-0.20}{\sqrt{\frac{0.20(1-0.20)}{315}}}=-2.80$

The test statistic is -2.80.

Compute the p-value as follows:

$p-value=2\times P(Z$

*Use a z-table.

The p-value is 0.00512.

The p-value is quite small. So, the null hypothesis will be rejected at any significance level.

Thus, it can be concluded that the proportion of smokers is different from 20%.

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