A researcher conducted a study with 42 coronary artery bypass surgeries and found the sample complete the surgery to be 2 hours. Suppose the accepted population standard deviation is 1.3 less than the regional average of 2 hours. At a significance level of a=0.05, what is the p-value for this test? Round to 4 decimal places. Provide your answer below:
p-value =
The correct answer and explanation is :
The p-value for this test is approximately 0.0000.
Explanation:
To determine the p-value for this test, we can perform a one-sample z-test. The steps are as follows:
- Formulate the Hypotheses:
- Null Hypothesis (H₀): The mean surgery time is equal to the regional average (μ = 2 hours).
- Alternative Hypothesis (H₁): The mean surgery time is different from the regional average (μ ≠ 2 hours).
- Calculate the Test Statistic: The z-test statistic is calculated using the formula: [ z = \frac{\bar{x} – \mu}{\frac{\sigma}{\sqrt{n}}} ] Where:
- (\bar{x}) is the sample mean
- (\mu) is the population mean
- (\sigma) is the population standard deviation
- (n) is the sample size Given:
- Sample mean ((\bar{x})) = 2 hours
- Population mean ((\mu)) = 2 hours
- Population standard deviation ((\sigma)) = 1.3 hours
- Sample size ((n)) = 42 Plugging in these values: [ z = \frac{2 – 2}{\frac{1.3}{\sqrt{42}}} = 0 ]
- Determine the P-Value: A z-score of 0 corresponds to a p-value of 1. However, since this is a two-tailed test, we need to consider the areas in both tails of the standard normal distribution. The p-value is calculated as: [ \text{p-value} = 2 \times (1 – \Phi(|z|)) ] Where (\Phi) is the cumulative distribution function of the standard normal distribution. Since |z| = 0, (\Phi(0) = 0.5), thus: [ \text{p-value} = 2 \times (1 – 0.5) = 2 \times 0.5 = 1 ]
- Compare with Significance Level: The significance level ((\alpha)) is 0.05. Since the p-value (1) is greater than (\alpha), we fail to reject the null hypothesis.
Conclusion:
Based on the sample data, there is no statistically significant difference between the mean surgery time and the regional average of 2 hours at the 0.05 significance level.