## Related questions with answers

In a new experimental teaching method, when students have questions in class, they send them to the instructor via a laptop computer. From time to time, the instructor pauses to read the questions and to provide the answers, without revealing the identities of the students who are asking the questions. This new method is supposed to eliminate the reluctance of students to ask questions for fear of revealing their lack of understanding in public. In a study of this new method, $67$ male students and $72$ female students participated. Treat these as if they were simple random samples. Of the male students, $38$ said that they liked the new method better than the traditional one in which students raise their hands to ask questions. Of the female students, only $22$ said they liked the new method better.

**a.** Construct a $95\%$ confidence interval for the difference between the proportions of male and female students who like the new method better.

**b.** An educator claims that the proportion of students who like the new method better is the same for males and females. Does the confidence interval contradict this claim?

Solution

Verified## (a.)

To construct a $95\%$ confidence interval for the difference between the proportions of male and female who like the new method better than the traditional one we will determine the value of point estimate and margin of error.

$\text{Point estimate} \pm \text{Margin of error}$

$\^p_1 - \^p_2 \pm z_{\alpha/2}\sqrt{\frac{\^p_1(1-\^p_1)}{n_1}+\frac{\^p_2(1-\^p_2)}{n_2}}$

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