Question

in a program designed to help patients stop smoking, 186 patients were given sustained care, and 82.8% of them were no longer smoking after one month. use a 0.05 significance level to test the claim that 81% of patients stop smoking when given sustained care. use the p - value method. use the normal distribution as an approximation to the binomial distribution
let p denote the population proportion of patients who would no longer be smoking after one month when given sustained care. identify the null and alternative hypotheses
$h_0: p = 0.81$
$h_1: p
eq 0.81$
(type integers or decimals. do not round)
identify the test statistic
$z=square$
(round to two decimal places as needed)

Explanation:

Step1: Calculate the sample proportion

The sample proportion $\hat{p}=0.828$. The population proportion in the null hypothesis $p = 0.81$, and the sample size $n=186$.

Step2: Calculate the standard deviation

The formula for the standard deviation of the sampling distribution of the sample proportion is $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}$.
Substitute $p = 0.81$ and $n = 186$ into the formula:
$$\sigma_{\hat{p}}=\sqrt{\frac{0.81\times(1 - 0.81)}{186}}=\sqrt{\frac{0.81\times0.19}{186}}\approx\sqrt{\frac{0.1539}{186}}\approx\sqrt{0.00082742}\approx0.0288$$

Step3: Calculate the test - statistic

The formula for the $z$ - test statistic in a proportion test is $z=\frac{\hat{p}-p}{\sigma_{\hat{p}}}$.
Substitute $\hat{p}=0.828$, $p = 0.81$, and $\sigma_{\hat{p}}\approx0.0288$ into the formula:
$$z=\frac{0.828 - 0.81}{0.0288}=\frac{0.018}{0.0288}\approx0.62$$

Answer:

$z = 0.62$