You are conducting a study to see if the accuracy rate for fingerprint identification is significantly less than 0.85 . You use a significance level of $\alpha=0.002$.
\[
\begin{array}{l}
H_{0}: p=0.85 \\
H_{1}: p< 0.85
\end{array}
\]

You obtain a sample of size $n=167$ in which there are 136 successes.
What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic $=$

What is the p-value for this sample? (Report answer accurate to four decimal places.) $\mathrm{p}$-value $=$

The $p$-value is...
less than (or equal to) $\alpha$
greater than $\alpha$

This test statistic leads to a decision to...
reject the null
accept the null
fail to reject the null

Step 1 ：Calculate the sample proportion (p̂) which is the number of successes divided by the sample size: \(p̂ = \frac{136}{167} = 0.8144\)

Step 2 ：Substitute the values into the formula for the test statistic: \(Z = \frac{(0.8144 - 0.85)}{\sqrt{(0.85(1 - 0.85)) / 167}}\)

Step 3 ：Simplify the denominator: \(Z = \frac{-0.0356}{\sqrt{0.1275 / 167}}\)

Step 4 ：Further simplify the denominator: \(Z = \frac{-0.0356}{0.0242}\)

Step 5 ：Calculate the test statistic: \(Z = -1.471\)

Step 6 ：Calculate the p-value, which is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. Since this is a one-tailed test, find the area to the left of the test statistic on the standard normal distribution. Using a Z-table or a calculator, the p-value is 0.0708

Step 7 ：Compare the p-value to the significance level (α = 0.002). Since the p-value is greater than α, we fail to reject the null hypothesis

Step 8 ：The final answers are: Test statistic = \(\boxed{-1.471}\), p-value = \(\boxed{0.0708}\), and the decision is to \(\boxed{\text{fail to reject the null hypothesis}}\)