Suppose we test H0 : p = .3 versus Ha : p ≠ .3 and that a random sample of n = 100 gives a sample proportion p = .20.
a: Test H0 versus Ha at the .01 level of significance by using critical values. What do you conclude?
b: Find the p-value for this test.
c: Use the p-value to test H0 versus Ha by setting α equal to .10, .05, .01, and .001. What do you conclude at each value of a?
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To solve this problem, we need to follow the steps for hypothesis testing and calculate critical values, p-values, and compare them to the significance level (α).
a: Test H0 versus Ha at the .01 level of significance by using critical values. What do you conclude?
To conduct the hypothesis test using critical values, we need to determine the critical region based on the significance level (α) and compare it with the test statistic.
Here, H0: p = 0.3 (null hypothesis) versus Ha: p ≠ 0.3 (alternative hypothesis), where p is the population proportion.
The test statistic for this hypothesis test is the z-test, which is calculated as (p̂ - p) / sqrt(p(1-p)/n), where p̂ is the sample proportion, p is the hypothesized population proportion under the null hypothesis, and n is the sample size.
Given p̂ = 0.20, p = 0.3, and n = 100, we can calculate the test statistic as follows:
z = (0.20 - 0.30) / sqrt(0.3 * (1 - 0.3) / 100)
= (-0.10) / sqrt(0.21 / 100)
= (-0.10) / sqrt(0.0021)
≈ -4.714
Next, we need to find the critical values associated with a two-tailed test at the 0.01 level of significance. We can find these values using a standard normal distribution table or a calculator. For a two-tailed test, we divide the significance level by 2, making it 0.01 / 2 = 0.005 for each tail.
Checking the critical values at the 0.005 level using a standard normal distribution table, we find that the critical value at each tail is approximately ±2.576.
Since the test statistic (-4.714) is outside the critical region (-2.576 to 2.576), we reject the null hypothesis (H0) in favor of the alternative hypothesis (Ha) at the 0.01 level of significance. The evidence suggests that the population proportion is not equal to 0.3.
b: Find the p-value for this test.
The p-value is the probability of observing a test statistic as extreme as the one calculated (or even more extreme) given the null hypothesis is true. To find the p-value, we compare the test statistic to the standard normal distribution.
For a two-tailed test, we need to find the probability of observing a test statistic as extreme as -4.714 (in the left tail) and as extreme as +4.714 (in the right tail) under the standard normal distribution.
Using a standard normal distribution table or a calculator, we find that the area to the left of -4.714 is very close to 0 and the area to the right of +4.714 is also very close to 0. Therefore, the p-value is approximately 0. This means that the data provide strong evidence against the null hypothesis.
c: Use the p-value to test H0 versus Ha by setting α equal to .10, .05, .01, and .001. What do you conclude at each value of α?
To use the p-value to test H0 versus Ha at different significance levels (α), we compare the p-value with the chosen α level. If the p-value is less than or equal to the α level, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
For α = 0.10: Since the p-value (approximately 0) is less than or equal to α (0.10), we reject the null hypothesis. The data provide strong evidence against the null hypothesis.
For α = 0.05: Since the p-value (approximately 0) is less than α (0.05), we reject the null hypothesis. The data provide strong evidence against the null hypothesis.
For α = 0.01: Since the p-value (approximately 0) is less than α (0.01), we reject the null hypothesis. The data provide strong evidence against the null hypothesis.
For α = 0.001: Since the p-value (approximately 0) is less than α (0.001), we reject the null hypothesis. The data provide strong evidence against the null hypothesis.
In summary, at every chosen level of significance, the p-value (approximately 0) is less than the α level, indicating that we reject the null hypothesis. The data provide strong evidence to suggest that the population proportion is not equal to 0.3.