Answer:
<em>H₀</em>: <em>p</em> = 0.20.
<em>Hₐ</em>: <em>p</em> ≠ 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 <em>z</em>-test can be used to determine the conclusion for this test.
The hypothesis defined as:
<em>H₀</em>: The proportion of smokers is 20%, i.e. <em>p</em> = 0.20.
<em>Hₐ</em>: The proportion of smokers is different from 20%, i.e. <em>p</em> ≠ 0.20.
The information provided is:
<em>n</em> = 315
<em>X</em> = number of people who identified themselves as smokers = 43
Compute the sample proportion of smokers as follows:

Compute the test statistic as follows:

The test statistic is -2.80.
Compute the <em>p</em>-value as follows:

*Use a <em>z</em>-table.
The <em>p</em>-value is 0.00512.
The <em>p</em>-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%.