Question

The data in NutritionStudy include information on nutrition and health habits of a sample of 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 given that identify themselves as smokers.
Required:
Clearly state the null and alternative hypotheses. Your answer should be an expression.

Answer 1

Answer:

H₀: p = 0.20.

Hₐ: p ≠ 0.20.

Step-by-step explanation:

The question is:

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%.