You wish to test the following claim (
H
a
) at a significance level of
α
=
0.005
.
H
o
:
p
1
=
p
2

H
a
:
p
1
<
p
2
You obtain 64 successes in a sample of size
n
1
=
506
from the first population. You obtain 113 successes in a sample of size
n
2
=
610
from the second population. 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 standardized test statistic for this sample? (Report answer accurate to three decimal places.)

The problem asks to conduct a hypothesis test for comparing two population proportions. Specifically, we want to test the null hypothesis:

\[
H_0: p_1 = p_2
\]

against the alternative hypothesis:

\[
H_a: p_1 < p_2
\]

### Step 1: Calculate the sample proportions
First, calculate the sample proportions for each population.

For the first population:

\[
\hat{p}_1 = \frac{x_1}{n_1} = \frac{64}{506} \approx 0.1265
\]

For the second population:

\[
\hat{p}_2 = \frac{x_2}{n_2} = \frac{113}{610} \approx 0.1852
\]

### Step 2: Pooled proportion
Since the null hypothesis assumes \( p_1 = p_2 \), we calculate the pooled sample proportion \( \hat{p} \), which is the overall proportion of successes from both samples combined:

\[
\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{64 + 113}{506 + 610} = \frac{177}{1116} \approx 0.1587
\]

### Step 3: Standard error
The standard error of the difference in sample proportions is calculated using the pooled proportion:

\[
SE = \sqrt{\hat{p} \cdot (1 - \hat{p}) \cdot \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}
\]

Substitute the values:

\[
SE = \sqrt{0.1587 \cdot (1 - 0.1587) \cdot \left(\frac{1}{506} + \frac{1}{610}\right)}
\]

\[
SE = \sqrt{0.1587 \cdot 0.8413 \cdot \left(0.001976 + 0.001639\right)}
\]

\[
SE = \sqrt{0.1587 \cdot 0.8413 \cdot 0.003615} \approx \sqrt{0.0004818} \approx 0.02195
\]

### Step 4: Test statistic
Now, compute the standardized test statistic \( z \) using the formula:

\[
z = \frac{\hat{p}_1 - \hat{p}_2}{SE}
\]

Substitute the values:

\[
z = \frac{0.1265 - 0.1852}{0.02195} \approx \frac{-0.0587}{0.02195} \approx -2.674
\]

### Conclusion:
The standardized test statistic is approximately:

\[
z \approx -2.674
\]

Would you like more details, or do you have any questions? Here are some follow-up questions to expand on this:

1. How would you calculate the p-value for this test statistic?
2. What is the critical value for a left-tailed test at \( \alpha = 0.005 \)?
3. How does the continuity correction affect the test statistic?
4. How do you interpret the result of this test?
5. What assumptions are necessary for using the normal distribution as an approximation?

**Tip:** Always check the assumptions (sample size and normality) when using the normal approximation for binomial tests.

You wish to test the following claim (Ha) at a significance level of α=0.005. Ho: p1 = p2 Ha: p1 < p2 You obtain 64 successes in a sample of size n1 = 506 from the first population. You obtain 113 successes in a sample of size n2 = 610 from the second population. 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 standardized test statistic for this sample? (Report answer accurate to three decimal places.)

Conduct a hypothesis test comparing two population proportions. Test the null hypothesis Ho: p1 = p2 against Ha: p1 < p2 using a significance level of α = 0.005. Includes step-by-step calculations of sample proportions, pooled proportion, standard error, and the Z-test statistic. Suitable for statistics students at the undergraduate level.

Math Problem Statement

You wish to test the following claim ( H a ) at a significance level of α

0.005 . H o : p 1

H a : p 1 < p 2 You obtain 64 successes in a sample of size n 1

506 from the first population. You obtain 113 successes in a sample of size n 2

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