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(2 points) Independent random samples, each containing 70 observations, were selected from two populations. The samples from populations 1 and 2 produced 53 and 43 successes, respectively.
Test [tex]H_0: (p_1 - p_2) = 0[/tex] against [tex]H_a: (p_1 - p_2) \ \textgreater \ 0[/tex]. Use [tex]\alpha = 0.02[/tex].
(a) The test statistic is [tex]\square[/tex].
(b) The [tex]P[/tex]-value is [tex]\square[/tex].
(c) The final conclusion is: A. We can reject the null hypothesis that [tex](p_1 - p_2) = 0[/tex] and accept that [tex](p_1 - p_2) \ \textgreater \ 0[/tex]. B. There is not sufficient evidence to reject the null hypothesis that [tex](p_1 - p_2) = 0[/tex].
To solve this hypothesis testing problem, we need to follow these steps:
1. Calculate the sample proportions [tex]\( p_1 \)[/tex] and [tex]\( p_2 \)[/tex] - [tex]\( p_1 \)[/tex] is the proportion of successes in sample 1: [tex]\( \frac{53}{70} = 0.757 \)[/tex] - [tex]\( p_2 \)[/tex] is the proportion of successes in sample 2: [tex]\( \frac{43}{70} = 0.614 \)[/tex]
2. Calculate the pooled sample proportion [tex]\( p_{\text{pool}} \)[/tex] - The pooled proportion combines successes and sample sizes from both populations: [tex]\( \frac{53 + 43}{70 + 70} = 0.686 \)[/tex]
3. Calculate the standard error of the difference in sample proportions - The standard error (SE) is given by the formula: [tex]\[
SE = \sqrt{p_{\text{pool}} \cdot (1 - p_{\text{pool}}) \cdot \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}
\][/tex] Substituting the values: [tex]\[
SE = \sqrt{0.686 \cdot (1 - 0.686) \cdot \left( \frac{1}{70} + \frac{1}{70} \right)} = 0.078
\][/tex]
4. Calculate the test statistic (z-score) - The test statistic [tex]\( z \)[/tex] is calculated as: [tex]\[
z = \frac{p_1 - p_2}{SE} = \frac{0.757 - 0.614}{0.078} = 1.821
\][/tex]
5. Determine the P-value for the test statistic - The P-value is the probability that under the null hypothesis, the test statistic is at least as extreme as the observed z-value in the direction of the alternative hypothesis: [tex]\[
P = 1 - \Phi(z) = 1 - \Phi(1.821) = 0.0343
\][/tex]
6. Compare the P-value to the significance level [tex]\( \alpha = 0.02 \)[/tex] - Since the P-value (0.0343) is greater than [tex]\( \alpha \)[/tex] (0.02), we do not reject the null hypothesis.
Based on these steps:
(a) The test statistic is [tex]\( 1.821 \)[/tex].
(b) The P-value is [tex]\( 0.0343 \)[/tex].
(c) The final conclusion is:
B. There is not sufficient evidence to reject the null hypothesis that [tex]\( (p_1 - p_2) = 0 \)[/tex].
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