Factors that inhibit learning in marketing. What factors inhibit the learning process in the classroom? To answer this question, researchers at Murray State University surveyed 40 students from a senior-level marketing class (Marketing Education Review). Each student was given a list of factors and asked to rate the extent to which each factor inhibited the learning process in courses offered in their department. A 7-point rating scale was used, where 1 = “not at all” and 7 = “to a great extent.” The factor with the highest rating was instructor related: “Professors who place too much emphasis on a single right answer rather than overall thinking and creative ideas.” Summary statistics for the student ratings of this factor are\(\overline x = 4.70\),\(s = 1.62\)

a. Conduct a test to determine if the true mean rating for this instructor-related factor exceeds 4. Use\(\alpha = 0.05\).Interpret the test results.

b. Examine the results of the study from a practical view, and then discuss why “statistically significant” does not always imply “practically significant.”

c. Because the variable of interest, rating, is measured on a 7-point scale, it is unlikely that the population of ratings will be normally distributed. Consequently, some analysts may perceive the test, part a, to be invalid and search for alternative methods of analysis. Defend or refute this argument

Step by step solution

01

Given information

The sample mean is 4.70, and the sample standard deviation is 1.62.

Also, it is given that the survey was taken from 40 students.

02

Concept

When the sample size exceeds 30, the z-test statistic is used to test the hypothesis.

The test statistic for the z- test is as follows:

\(z = \frac{{\overline x - \mu }}{{\frac{\sigma }{{\sqrt n }}}}\)

03

Testing the hypothesis

a.

Null hypothesis:

\({H_0}:\mu = 4\)

The true mean rating for this instructor-related factor does not exceed 4.

Alternative hypothesis:

\({H_a} :\mu > 4\)

That is, the true mean rating for this instructor-related factor exceeds 4.

Consider,

\(\sigma = 1.62,\overline x = 4.70\,and\,n = 40\)

The test statistic is calculated below:

\(z = \frac{{\overline x - \mu }}{{\frac{\sigma }{{\sqrt n }}}}\)

\(\begin{aligned}z &= \frac{{4.70 - 4}}{{\frac{{1.62}}{{\sqrt {40} }}}}\\ &= \frac{{0.7}}{{0.256}}\\ &= 2.73\end{aligned}\)

The test is a right tail with \(\alpha = 0.05\)

From the standard normal table, the required value of\({z_{0.05}} = 1.645\)

The value of z is 2.73, and the value\({z_{0.05}}\)is 1.645.

Here,\(z\left( {2.73} \right) > {z_{0.05}}\left( {1.645} \right)\)

Therefore, by the condition, if z is more significant than\({z_{0.05}}\), then reject the null hypothesis. Thus, it can be concluded that there is evidence to reject the null hypothesis at\(\alpha = 0.05\)\(\)

Hence, the true mean rating for this instructor-related factor exceeds 4 at\(\alpha = 0.05\).

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