You are conducting a study to see if the accuracy rate for fingerprint identification is significantly less than 0.87. You use a significance level of alpha= 0.001

Question

H_0 : p = 0.87

You obtain a sample of size n=501 in which there are 421 successes.

H_1 : p > 0.87

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.)
p-value = __________________

GPT-4o mini · Answer

To calculate the test statistic and the p-value for the given hypothesis test, we follow these steps:

Define the null and alternative hypotheses:
- Null hypothesis ( $H_{0}$ ): $p = 0.87$
- Alternative hypothesis ( $H_{1}$ ): $p < 0.87$ (Note: Your provided alternative hypothesis seems to state $p > 0.87$ , but your question indicates that you're checking if the accuracy rate is less than 0.87, which should be $H_{1} : p < 0.87$ . Please verify if you intended to perform a one-tailed test as described or if it should be two-tailed.)
Sample proportion: Given:
- Sample size ( $n$ ) = 501
- Number of successes ( $x$ ) = 421
Calculate the sample proportion ( $\hat{p}$ ): $\hat{p} = \frac{x}{n} = \frac{421}{501} \approx 0.8403$
Calculate the test statistic: The formula for the test statistic $z$ for a proportion is: $z = \frac{\hat{p} - p_{0}}{\sqrt{\frac{p_{0} (1 - p_{0})}{n}}}$ where $p_{0}$ is the hypothesized population proportion (0.87).

Substituting the values: $z = \frac{0.8403 - 0.87}{\sqrt{\frac{0.87 \times (1 - 0.87)}{501}}}$

First, calculate the standard error: $S E = \sqrt{\frac{0.87 \times 0.13}{501}} \approx \sqrt{\frac{0.1131}{501}} \approx \sqrt{0.000225192} \approx 0.015$

Now compute $z$ : $z = \frac{0.8403 - 0.87}{0.015} \approx \frac{- 0.0297}{0.015} \approx - 1.98$

So, the test statistic is: $z \approx - 1.980$
Calculate the p-value: To find the p-value for a left-tailed test with the calculated $z$ -score, we can use standard normal tables or a calculator to find the area to the left of $z = - 1.980$ : $p-value = P (Z < - 1.980)$

From standard normal distribution tables or software: $p-value \approx 0.0236$

Combining the results, we find:

Test statistic: $z \approx - 1.980$
p-value: $\approx 0.0236$

Final answers:

Test statistic = -1.980

p-value = 0.0236