SOLUTION: 7. A researcher wants to test the null hypothesis Ho p = 0.79 against the alternative H: p > 0.79. A sample of 1200 observations is inputted into a computer software and returns th

Question 1186606: 7. A researcher wants to test the null hypothesis Ho p = 0.79 against the alternative H: p > 0.79. A sample of 1200 observations is inputted into a computer software and returns the result: z = 3 and the one - tailed p - value = 0.0013. What is your interpretation of this p-value, using alpha 0.05?
answer the problem using the steps in hypothesis testing.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to interpret the p-value in this hypothesis test:
**1. State the Hypotheses:**
* **Null Hypothesis (H₀):** p = 0.79
* **Alternative Hypothesis (H₁):** p > 0.79 (This is a right-tailed test.)
**2. Level of Significance (alpha):**
* α = 0.05
**3. Test Statistic and P-value:**
* The test statistic is given as z = 3.
* The one-tailed p-value is given as 0.0013.
**4. Decision:**
* **Compare the p-value to alpha:** The p-value (0.0013) is less than the level of significance (0.05).
* **Decision:** Since the p-value is less than alpha, we reject the null hypothesis.
**5. Interpretation of the p-value:**
The p-value (0.0013) represents the probability of observing a sample proportion as extreme as, or more extreme than, the one obtained (resulting in a z-score of 3 or greater), *assuming that the true population proportion is actually 0.79* (i.e., assuming the null hypothesis is true). In other words, if the true proportion voting for the candidate was 79%, there is only a 0.13% chance of observing a sample where 3 or more standard deviations above the mean.
**6. Conclusion:**
Because the p-value is so small, it provides strong evidence against the null hypothesis. We reject the null hypothesis. There is sufficient evidence at the 0.05 level of significance to conclude that the true population proportion is greater than 0.79. The sample data suggests that the proportion of individuals supporting the alternative hypothesis is significantly higher than 79%.

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