In a program designed to help patients stop smoking, 191 patients were given sustained care, and $82.7 \%$ of them were no longer smoking after one month. Use a 0.10 significance level to test the claim that $80 \%$ 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. \[ \begin{array}{l} H_{0}: p=0.80 \\ H_{1}: p \neq 0.80 \end{array} \] (Type integers or decimals. Do not round.) Identify the test statistic. \[ \mathrm{z}= \] (Round to two decimal places as needed.)

Step 1 ：Let $p$ denote the population proportion of patients who would no longer be smoking after one month when given sustained care. The null and alternative hypotheses are identified as follows: \[H_{0}: p=0.80\] \[H_{1}: p \neq 0.80\]

Step 2 ：We need to use the formula for the test statistic in a hypothesis test for a proportion. The formula is: \[z = \frac{\hat{p} - p_{0}}{\sqrt{\frac{p_{0} * (1 - p_{0})}{n}}}\] where \(\hat{p}\) is the sample proportion, \(p_{0}\) is the hypothesized population proportion, and \(n\) is the sample size.

Step 3 ：In this case, \(\hat{p} = 0.827\), \(p_{0} = 0.80\), and \(n = 191\).

Step 4 ：We can calculate the test statistic using this formula: \[z = \frac{0.827 - 0.80}{\sqrt{\frac{0.80 * (1 - 0.80)}{191}}}\]

Step 5 ：The test statistic is calculated to be approximately 0.93.

Step 6 ：Final Answer: The test statistic is \(\boxed{0.93}\).