You are conducting a study to see if the accuracy rate for fingerprint identification is significantly different from 0.42. You use a significance level of $\alpha=0.05$.

$\begin{array}{l}

H_0: p=0.42 \\

H_1: p
eq 0.42

\end{array}$

You obtain a sample of size $n=533$ in which there are 202 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.)

$p$-value =

We are conducting a hypothesis test for a population proportion to determine if the accuracy rate of fingerprint identification differs significantly from a hypothesized value of $p_0 = 0.42$.
Given data:

Null Hypothesis: $H_0: p = 0.42$
Alternative Hypothesis: $H_1: p eq 0.42$
Significance level: $\alpha = 0.05$
Sample size: $n = 533$
Observed successes: $x = 202$

First, we calculate the sample proportion $\hat{p}$:
\[ \hat{p} = \frac{x}{n} = \frac{202}{533} \approx 0.378987 \]
Next, we determine the standard error (SE) under the null hypothesis:
\[ SE = \sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.42(1-0.42)}{533}} = \sqrt{\frac{0.42 \cdot 0.58}{533}} \approx 0.021378 \]
The test statistic $z$ is calculated as:
\[ z = \frac{\hat{p} - p_0}{SE} = \frac{0.378987 - 0.42}{0.021378} \approx -1.918439 \]
Rounding to three decimal places, the test statistic is $z = -1.918$.

Calculate the p-value

Now, we calculate the p-value for the two-tailed test. The p-value is the probability of obtaining a sample result as extreme as or more extreme than the observed result, given that the null hypothesis is true.
For a two-tailed test, the p-value is given by:
\[ p\text{-value} = 2 \cdot P(Z > |z|) = 2 \cdot P(Z > |-1.918439|) \approx 2 \cdot P(Z > 1.918439) \]
Using the standard normal distribution, the probability $P(Z > 1.918439) \approx 0.027528$.
Thus, the p-value is:
\[ p\text{-value} = 2 \cdot 0.027528 = 0.055056 \]
Rounding to four decimal places, the p-value is $0.0551$.

Conclusion based on the significance level

Finally, we compare the p-value to the significance level $\alpha = 0.05$ to make a decision about the null hypothesis.
Since the p-value ($0.0551$) is greater than the significance level $\alpha = 0.05$, we fail to reject the null hypothesis $H_0$. There is not sufficient evidence to conclude that the accuracy rate is significantly different from 0.42.

Examples

Hypothesis testing is widely used in quality control and scientific research to make data-driven decisions. For example, if a machine is set to produce items with a 95% defect-free rate, the manager can test a sample to ensure the actual performance hasn't significantly drifted from that target, ensuring quality standards are maintained. This specific process of calculating a test statistic and p-value allows businesses and researchers to objectively determine if observed deviations in their data are due to random chance or represent a real change.