Question

In 1965, about 44% of the U.S. adult population had never smoked cigarettes. A national health survey of 1360 U.S. adults (selected randomly) during 2020 revealed that 626 had never smoked cigarettes. Using α = 0.05, test whether there has been a change since 1965 in the proportion of U.S. adults that have never smoked cigarettes. State the hypotheses, list and check the conditions, calculate the test statistic, find the p-value, and make a conclusion in a complete sentence related to the scenario.

Answer 1

Using the z-distribution, it is found that since the p-value is greater than 0.05, there is not enough evidence to conclude that there has been a change since 1965 in the proportion of U.S. adults that have never smoked cigarettes.

What are the hypothesis tested?

At the null hypothesis, it is tested if the proportion is still of 44%, that is:

$H_0: p = 0.44$

At the alternative hypothesis, it is tested if the proportion is now different of 44%, that is:

$H_1: p \neq 0.44$

What is the test statistic?

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

• $\overline{p}$ is the sample proportion.

• p is the proportion tested at the null hypothesis.

• n is the sample size.

For this problem, the parameters are:

$p = 0.44, n = 1360, \overline{p} = \frac{626}{1360} = 0.4603$

Hence the test statistic is:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.4603 - 0.44}{\sqrt{\frac{0.44(0.56)}{1360}}}$

z = 1.51

What is the p-value and the conclusion?

Using a z-distribution calculator, for a two-tailed test, as we are testing if the proportion is different of a value, with z = 1.51, the p-value is of 0.1310.

Since the p-value is greater than 0.05, there is not enough evidence to conclude that there has been a change since 1965 in the proportion of U.S. adults that have never smoked cigarettes.

More can be learned about the z-distribution at https://brainly.com/question/16313918

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