Answered

You wish to test the following claim $\left(H_a\right)$ at a significance level of $\alpha=0.01$.

$\begin{array}{l}

H_o: p=0.81 \\

H_a: p<0.81

\end{array}$



You obtain a sample of size $n=393$ in which there are 300 successful observations. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution.



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 =

Verified Answer :

  • Calculate the sample proportion: \(\hat{p} = \frac{300}{393} \approx 0.7634\).
  • Calculate the test statistic: \(z = \frac{0.7634 - 0.81}{0.0198} \approx \boxed{-2.357}\).
  • Determine the p-value for the left-tailed test: \(P(Z < -2.357) \approx \boxed{0.0092}\).
  • Compare the p-value to \(\alpha = 0.01\): Since \(0.0092 < 0.01\), reject the null hypothesis.

Explanation

  1. Identify the data and hypothesis test parameters

To test the hypothesis \(H_0: p = 0.81\) versus \(H_a: p < 0.81\), we first identify the parameters: the null hypothesis proportion is \(p_0 = 0.81\), the sample size is \(n = 393\), and the number of successes is \(x = 300\). The sample proportion is calculated as \(\hat{p} = \frac{300}{393} \approx 0.76336\). We will use the normal approximation to the binomial distribution to determine the test statistic and p-value.

  1. Calculate the test statistic (z-score)

The standard error for the sampling distribution of the proportion under the null hypothesis is given by the formula \(SE = \sqrt{\frac{p_0(1 - p_0)}{n}}\). Substituting the values:
\[SE = \sqrt{\frac{0.81 \times (1 - 0.81)}{393}} = \sqrt{\frac{0.81 \times 0.19}{393}} = \sqrt{\frac{0.1539}{393}} \approx 0.01979\]
The test statistic \(z\) is calculated using \(z = \frac{\hat{p} - p_0}{SE}\):
\[z = \frac{0.7633587786 - 0.81}{0.019789} \approx -2.3569\]
Rounding to three decimal places, the test statistic is \(-2.357\).

  1. Determine the p-value

The p-value is the probability of obtaining a test statistic as extreme as, or more extreme than, the observed value under the null hypothesis. Since this is a left-tailed test (\(H_a: p < 0.81\)), the p-value is \(P(Z < -2.357)\). Using the standard normal distribution table or calculation, we find:
\[p\text{-value} \approx 0.0092\]
Rounding to four decimal places, the p-value is \(0.0092\).

  1. Compare the p-value to the significance level and conclude

We compare the calculated p-value of \(0.0092\) to the significance level \(\alpha = 0.01\). Because \(0.0092 < 0.01\), the p-value is less than the significance level. Therefore, we reject the null hypothesis \(H_0\). There is sufficient evidence at the \(\alpha = 0.01\) level to support the claim that the population proportion is less than \(0.81\).

Examples

Imagine you are a quality control analyst testing whether a manufacturing process meets a specific standard. By calculating the test statistic and p-value from a random sample, you can make an informed decision on whether to reject the claim that the process is performing at the expected level. For instance, if you test a batch of 393 components and find only 300 are functional, comparing this result against a target proportion of 0.81 helps ensure production quality standards are maintained. If the p-value is below your chosen threshold (e.g., 0.01), you know it is statistically significant and requires attention.

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