Question

Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A survey found that \(83 \%\) of the men questioned preferred computer-assisted instruction to lecture and \(75 \%\) of the women preferred computer-assisted instruction to lecture. There were 100 randomly selected individuals in each sample. At \(\alpha=0.05\), test the claim that there is no difference in the proportion of men and the proportion of women who favor computer-assisted instruction over lecture. Find the \(95 \%\) confidence interval for the difference of the two proportions.

Short answer

The claim that there is no difference in proportions cannot be rejected. The 95% confidence interval is \((-0.0295, 0.1895)\).

Step-by-step solution

State the Hypotheses

For testing the claim of no difference, set up the null hypothesis as \( H_0: p_1 = p_2 \), where \( p_1 \) is the proportion of men who prefer computer-assisted instruction, and \( p_2 \) is the proportion of women who prefer the same. The alternative hypothesis, \( H_a \), is \( H_a: p_1 eq p_2 \). The claim is that there is no difference, which corresponds to the null hypothesis.

Find the Critical Value(s)

For a two-tailed test with \( \, \alpha = 0.05 \), divide the significance level by two to find \( 0.025 \) per tail. Using the standard normal distribution (z-distribution), the critical values are \( Z_{0.025} \approx \pm 1.96 \).

Compute the Test Value

Calculate the test statistic using the formula for the difference of two proportions: \[ z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{\widehat{p}(1-\widehat{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \] where \( \hat{p}_1 = 0.83 \), \( \hat{p}_2 = 0.75 \), \( \widehat{p} = \frac{x_1 + x_2}{n_1 + n_2} \), \( n_1 = n_2 = 100 \), \( x_1 = 83 \), and \( x_2 = 75 \). Calculate \( \widehat{p} = \frac{83 + 75}{200} = 0.79 \). Substitute into the formula to find \( z = \frac{0.83 - 0.75}{\sqrt{0.79 \times 0.21 \times (\frac{1}{100} + \frac{1}{100})}} \approx 1.488 \).

Make the Decision

Compare the test statistic \( z \approx 1.488 \) with the critical values \( \pm 1.96 \). Since \( 1.488 \) is within the range of \(-1.96\) to \(1.96\), do not reject the null hypothesis.

Summarize the Results

There is insufficient evidence to reject the null hypothesis at the \( \alpha = 0.05 \) significance level. Thus, we do not have enough evidence to claim that there is a difference in the proportions of men and women who prefer computer-assisted instruction.

Calculate the 95% Confidence Interval

The formula for the confidence interval for the difference between two proportions is: \[ (\hat{p}_1 - \hat{p}_2) \pm Z_{\alpha/2} \times \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \] Using \( \hat{p}_1 = 0.83 \), \( \hat{p}_2 = 0.75 \), and \( Z_{0.025} = 1.96 \), calculate the margin of error: \[ ME = 1.96 \times \sqrt{\frac{0.83(0.17)}{100} + \frac{0.75(0.25)}{100}} \approx 0.1095 \] Therefore, the 95% confidence interval is \[ (0.83 - 0.75) \pm 0.1095 = 0.08 \pm 0.1095 = (-0.0295, 0.1895) \].

Key concepts

Understanding the Z-Distribution

In statistics, the z-distribution is a crucial concept, especially when performing hypothesis testing. It's essentially the standard normal distribution and is used to determine how data points compare to a standard mean. In our specific problem, the z-distribution helps find the critical values for a hypothesis test.

Here's why it's important:

• The z-distribution is perfectly symmetrical, centering around zero. This property is vital because it ensures that the test can equally consider variations from the mean, whether they're positive or negative.


• It's used to determine probabilities and critical values when the sample size is sufficiently large. Typically, anything beyond 30 observations is considered adequate to apply the z-distribution safely.


• The critical values you'll find from a z-distribution table or tool, like the 1.96 for a 95% confidence interval in our problem, are essential in deciding whether to accept or reject hypotheses.

Using a z-distribution helps ensure accurate inference about the population from the sample data and is fundamental for effective decision-making in hypothesis tests.

Exploring the Confidence Interval

A confidence interval (CI) gives a range within which we believe the true difference between two population parameters lies. Our exercise specifically explores the 95% confidence interval for the difference in proportions between men and women who prefer computer-assisted instruction.

Understanding CIs involves the following key points:

• The confidence level, in this case, 95%, signifies that if we were to draw numerous samples and compute a CI for each one, we'd expect 95% of those intervals to contain the true difference.


• We calculate it using the observed sample statistics, and it reflects the precision of our estimate. For our problem, the margin of error was calculated as approximately 0.1095, adding and subtracting from the observed difference (0.08), giving the interval (-0.0295, 0.1895).


• The range tells us that the true difference could be slightly negative or positive, providing a sense of uncertainty, but also showing that zero could be a plausible value for the true differencesuggesting no difference between men's and women's preferences.

Confidence intervals give more information than a simple yes/no hypothesis test, offering insights into the possible range of the true population parameter.

Understanding the Two-Tailed Test

In hypothesis testing, a two-tailed test is deployed when we want to check for an effect that could occur in either direction. For instance, with our hypothesis test about preferences between male and female participants for computer-assisted instruction, we were open to finding out if one proportion might be higher or lower than the otherhence the two tails of the test.

Key components of a two-tailed test include:

• Two-tailed tests are appropriate when the null hypothesis states equality (e.g., 0: there is no difference between the groups). We are interested in detecting any deviation, whether positive or negative.


• In a two-tailed test, the alpha level (our significance level) is split between the two tails of the distribution. For a 0.05 significance level, each tail would contain 0.025, which translates to critical z-values of approximately 1.96 and -1.96.


• A two-tailed test provides a balanced approach, which means it rigorously examines the possibility of differences in both directions. This ensures comprehensive testing, but requires that calculated test statistics stay between both critical values in order not to reject the null hypothesis.

Two-tailed tests are incredibly useful for balanced hypothesis testing when the direction of an effect is uncertain or when outcomes in either direction are consequential.