11. ackerman and goldsmith (2011) report that students who study from a screen (smartphone, tablet, or…

11. 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.\na. assuming a two - tailed test, state the null and alternative hypotheses in a sentence that includes the two variables being examined.\nb. 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.
Answer
Explanation:
Step1: State the null and alternative hypotheses
The null hypothesis (H_0): There is no significant difference between the exam scores of students studying from an electronic screen and the general population. In symbols, (H_0:\mu = 77) The alternative hypothesis (H_1): There is a significant difference between the exam scores of students studying from an electronic screen and the general population. In symbols, (H_1:\mu\neq77)
Step2: Compute the z - statistic
The formula for the z - statistic in a one - sample z - test is (z=\frac{M-\mu}{\frac{\sigma}{\sqrt{n}}}) Given (M = 72.5), (\mu=77), (\sigma = 8), and (n = 16) First, calculate (\frac{\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 z - values
For a two - tailed test with (\alpha = 0.05), the critical z - values are (z=\pm1.96)
Step4: Make a decision
Since (|z|=|-2.25| = 2.25>1.96), we reject the null hypothesis.
Answer:
a. (H_0:) The exam scores of students studying from an electronic screen are equal to the scores of the general population ((\mu = 77)). (H_1:) The exam scores of students studying from an electronic screen are not equal to the scores of the general population ((\mu\neq77)) 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 had exam scores that are significantly different from those for the general population.