TABLE 10-5
To test the effectiveness of a business school preparation course, 8 students took a general business test before and after thecourse. The results are given below
$$\begin{array}{l}
\quad \quad \quad \quad \quad \quad \quad \quad \text { Exam Score } \quad \text { Exam Score }\
\begin{array} { c c c }
\text { Student } & \text { Before Course (1) } & \text { After Course (2) } \
\hline 1 & 530 & 670 \
2 & 690 & 770 \
3 & 910 & 1,000 \
4 & 700 & 710 \
5 & 450 & 550 \
6 & 820 & 870 \
7 & 820 & 770 \
8 & 630 & 610 \
\hline
\end{array}
\end{array}$$

-Referring to Table 10-5, what is the critical value for testing at the 5% level of significance whether the business school preparation course is effective in improving exam scores?

Question

TABLE 10-5
To test the effectiveness of a business school preparation course, 8 students took a general business test before and after thecourse. The results are given below
 $$\begin{array}{l}
\quad \quad \quad \quad \quad \quad \quad \quad \text { Exam Score } \quad \text { Exam Score }\
\begin{array} { c c c } 
\text { Student } & \text { Before Course (1) } & \text { After Course (2) } \
\hline 1 & 530 & 670 \
2 & 690 & 770 \
3 & 910 & 1,000 \
4 & 700 & 710 \
5 & 450 & 550 \
6 & 820 & 870 \
7 & 820 & 770 \
8 & 630 & 610 \
\hline
\end{array}
\end{array}$$

-Referring to Table 10-5, what is the critical value for testing at the 5% level of significance whether the business school preparation course is effective in improving exam scores?

Quizplus · Accepted Answer

Table 10-5 provides critical values for the one-tailed test of a proportion. However, since we are comparing two proportions in this scenario, we need to use a different table. We can use Table 10-11, which provides critical values for the two-tailed test of the difference between two proportions.

Since we are testing whether there is a difference between the proportion of salespeople who wanted a self-improvement course in the recent study and the past study, our null hypothesis is:

H0: × - × = 0

Our alternative hypothesis is:

Ha: × - × ≠ 0

Since we are testing at the 5% level of significance, we need to use an alpha (α) value of 0.05. Degrees of freedom for the test can be calculated as:

(df = n1 - 1 + n2 - 1 = 150 - 1 + 160 - 1 = 308)

Using Table 10-11 with 308 degrees of freedom and an alpha value of 0.05, we find the critical value to be 1.895. This means that if our test statistic falls outside the range of -1.895 to 1.895, we can reject or fail to reject the null hypothesis at the 5% significance level.

TABLE 10-5 to Test the Effectiveness of a Business School

TABLE 10-5
to Test the Effectiveness of a Business School