Question

1. ackerman and goldsmith (2011) report that students who study from a screen (smartphone, tablet, or computer) tended to have lower quiz scores than students who studied the same material from printed pages. to test this finding, a professor identifies a sample of ( n = 16 ) students who used the electronic version of the course textbook and determines that this sample had an average score of ( m = 72.5 ) on the final exam. during the previous three years, the final exam scores for the general population of students taking the course averaged ( mu = 77 ) with a standard deviation of ( sigma = 8 ) and formed a roughly normal distribution. the professor would like to use the sample to determine whether students studying from an electronic screen had exam scores that are significantly different from those for the general population.

a. assuming a two - tailed test, state the null and alternative hypotheses in a sentence that includes the two variables being examined.
b. using the standard four - step procedure, conduct a two - tailed hypothesis test with ( alpha =.05 ) to evaluate the effect of studying from an electronic screen.

Explanation:

Step1: State the null and alternative hypotheses

The null hypothesis \(H_0\) is that there is no difference in exam scores between students studying from an electronic screen and the general population. So, \(H_0:\mu = 77\). The alternative hypothesis \(H_1\) is that there is a difference in exam scores, \(H_1:\mu
eq77\)

Step2: Compute the test - statistic

The formula for the \(z\) - test statistic is \(z=\frac{M - \mu}{\sigma/\sqrt{n}}\)
Given \(M = 72.5\), \(\mu=77\), \(\sigma = 8\), \(n = 16\)
First, calculate \(\sigma/\sqrt{n}=\frac{8}{\sqrt{16}}=\frac{8}{4} = 2\)
Then, \(z=\frac{72.5 - 77}{2}=\frac{- 4.5}{2}=-2.25\)

Step3: Determine the critical region

For a two - tailed test with \(\alpha = 0.05\), the critical values are \(z=\pm1.96\)

Step4: Make a decision

Since \(z=-2.25\lt - 1.96\), we reject the null hypothesis

Answer:

a. Null hypothesis: The population mean exam score of students studying from an electronic screen is equal to the population mean exam score of the general population (\(H_0:\mu = 77\)). Alternative hypothesis: The population mean exam score of students studying from an electronic screen is not equal to the population mean exam score of the general population (\(H_1:\mu
eq77\))

b. We reject the null hypothesis. There is sufficient evidence at the \(\alpha = 0.05\) level to conclude that students studying from an electronic screen have exam scores that are significantly different from those of the general population.