You wish to test the following claim (H_a) at a significance level of alpha= 0.10.

H_o : p = 0.59
H_a : p > 0.59

You obtain a sample of size n=217 in which there are 135 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 = ___

The p-value is...
A. less than (or equal to) alpha
B. greater than alpha

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

As such, the final conclusion is that...
A. There is sufficient evidence to warrant rejection of the claim that the population proportion is greater than 0.59.
B. There is not sufficient evidence to warrant rejection of the claim that the population proportion is greater than 0.59.
C. The sample data support the claim that the population proportion is greater than 0.59.
D. There is not sufficient sample evidence to support the claim that the population proportion is greater than 0.59.

1 answer

To conduct the hypothesis test and calculate the test statistic and p-value, we proceed as follows:

  1. Identify the sample proportion (p̂): \[ \hat{p} = \frac{x}{n} = \frac{135}{217} \approx 0.6212 \]

  2. Identify the null hypothesis proportion (p_0): \[ p_0 = 0.59 \]

  3. Calculate the standard deviation of the sampling distribution under the null hypothesis: \[ \sigma_{\hat{p}} = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.59(1 - 0.59)}{217}} = \sqrt{\frac{0.59 \times 0.41}{217}} \approx \sqrt{\frac{0.2419}{217}} \approx \sqrt{0.001113} \approx 0.0333 \]

  4. Calculate the test statistic (z): \[ z = \frac{\hat{p} - p_0}{\sigma_{\hat{p}}} = \frac{0.6212 - 0.59}{0.0333} \approx \frac{0.0312}{0.0333} \approx 0.9364 \] (rounded to three decimal places: \( z \approx 0.936 \))

  5. Find the p-value: The p-value is calculated using the standard normal distribution. Since this is a one-tailed test, we find \( P(Z > z) \): \[ p\text{-value} = P(Z > 0.936) \approx 1 - P(Z \leq 0.936) \approx 1 - 0.8258 \approx 0.1742 \] (rounded to four decimal places: \( p\text{-value} \approx 0.1742 \))

  6. Compare the p-value with alpha:

    • \( \alpha = 0.10 \)
    • Since \( 0.1742 > 0.10 \), we conclude that the p-value is greater than alpha.
  7. Decision based on p-value: Since the p-value is greater than alpha, we fail to reject the null hypothesis.

  8. Final conclusion: The conclusion is that: There is not sufficient evidence to warrant rejection of the claim that the population proportion is greater than 0.59.

Final Answers:

  • Test statistic = 0.936
  • p-value = 0.1742
  • The p-value is: B. greater than alpha
  • This test statistic leads to a decision to: C. fail to reject the null
  • As such, the final conclusion is that: D. There is not sufficient sample evidence to support the claim that the population proportion is greater than 0.59.