Answer :
To determine whether the sociologist can conclude that religious freshmen who had been discouraged from reading fiction remain faithful at lower rates, we can perform a hypothesis test for the difference in proportions.
Let's define the following:
\( p_1 \) = Proportion of students discouraged from reading fiction who maintained at least a moderate level of faith in their initial religion by senior year.
\( p_2 \) = Proportion of students not discouraged from reading fiction who maintained at least a moderate level of faith in their initial religion by senior year.
We want to test the null hypothesis (\( H_0 \)) that there is no difference in the proportions of students maintaining faith regardless of whether they were discouraged from reading fiction. The alternative hypothesis (\( H_1 \)) is that students discouraged from reading fiction have a lower proportion of maintaining faith compared to those not discouraged.
\[ H_0: p_1 = p_2 \]
\[ H_1: p_1 < p_2 \]
We'll conduct a one-tailed test because we want to determine if the proportion of students discouraged from reading fiction who maintained faith is significantly lower than those not discouraged.
We can use the Z-test for proportions to test this hypothesis. The test statistic is given by:
\[ Z = \frac{(p_1 - p_2) - 0}{\sqrt{p(1-p)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \]
Where:
- \( p = \frac{n_1p_1 + n_2p_2}{n_1 + n_2} \) is the pooled sample proportion.
- \( n_1 \) and \( n_2 \) are the sample sizes.
- \( p_1 \) and \( p_2 \) are the sample proportions.
Let's calculate the test statistic and determine if it falls in the critical region for a one-tailed test at 1% significance level. We'll reject the null hypothesis if the test statistic falls in the critical region. If it doesn't, we fail to reject the null hypothesis.
First, let's calculate the sample proportions \( p_1 \) and \( p_2 \):
\[ p_1 = \frac{188}{612} \approx 0.3072 \]
\[ p_2 = \frac{272}{688} \approx 0.3953 \]
Next, let's calculate the pooled sample proportion \( p \):
\[ p = \frac{612 \times 0.3072 + 688 \times 0.3953}{1300} \approx 0.3520 \]
Now, let's calculate the standard error:
\[ \text{Standard Error} = \sqrt{p(1-p)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \]
\[ \text{Standard Error} = \sqrt{0.3520(1-0.3520)\left(\frac{1}{612} + \frac{1}{688}\right)} \]
\[ \text{Standard Error} \approx \sqrt{0.3520(0.6480)\left(0.0016 + 0.0015\right)} \]
\[ \text{Standard Error} \approx \sqrt{0.3520(0.6480) \times 0.0031} \]
\[ \text{Standard Error} \approx \sqrt{0.3520(0.0020)} \]
\[ \text{Standard Error} \approx \sqrt{0.000704} \]
\[ \text{Standard Error} \approx 0.0265 \]
Now, let's calculate the test statistic \( Z \):
\[ Z = \frac{(0.3072 - 0.3953) - 0}{0.0265} \]
\[ Z \approx \frac{-0.0881}{0.0265} \]
\[ Z \approx -3.32 \]
Now, we need to find the critical value for a one-tailed test at 1% significance level. Since this is a lower-tailed test, we're interested in finding the critical value that corresponds to 1% in the lower tail of the standard normal distribution.
Using a standard normal distribution table or a calculator, we find that the critical value for a one-tailed test at 1% significance level is approximately -2.33.
Since our calculated test statistic \( Z = -3.32 \) is less than the critical value -2.33, it falls in the critical region. Therefore, we reject the null hypothesis \( H_0 \) in favor of the alternative hypothesis \( H_1 \).
So, at 1% significance level, the sociologist can conclude that religious freshmen who had been discouraged from reading fiction remain faithful at lower rates.
