Practice 2 Solutions

1. Calculate the z-score. Explain the statistical significance at the 5% level.

$$\begin{gathered} 13.2–12.9=0.3 \\ z=\frac{0.3}{2.73}=0.110 \end{gathered}$$

Because $0.11<1.96$, the software update does not show statistical significance at the 5% level.

2. Explain any concerns you have about the marketing claims. 

Sample: Stating that the software is “revolutionary” ignores the lack of statistical significance. 

3. Software updates are important for security, performance, and system stability.
   Practically speaking, should software be updated even if the battery life is not significantly improved?

Sample: The software should be updated even if the battery life is not significantly improved because the update is important to the overall device function. 

4. Calculate z-scores for the observed difference. Does the new medication have statistical significance at the 5% level?

$$\begin{gathered} 3.8–2.1=1.7 \\ z=\frac{1.7}{0.76}=2.237 \end{gathered}$$

Because $2.237>1.96$, the new medication has statistical significance at the 5% level.

5. If you were looking at the research for the pharmaceutical company, would you recommend releasing the medication to a larger group? Justify your response.

Sample: Releasing the medication to a larger group is not recommended because of the significant increase in pain in the treatment group. The medication did not reduce pain as it was intended. 

6. What assumptions about the experiment and participants should be verified before making any final recommendations?

Sample: Researchers should verify that participants know that a lower pain score means less pain. Researchers should also check if the increased pain score of the treatment group was related to side effects from the medication. 

7. Calculate both z-scores for the observed difference. Explain the statistical significance at the 5% level.

$\begin{array}{lcl}\mathrm{Temperature}& & \mathrm{Drinkable}\\ 156–128=28& & 86–43=43\\ z=\frac{28}{7.8}=3.59& & z=\frac{43}{11.2}=3.84\end{array}$

Because both z-scores (temperature and drinkable) are above 1.96, they are statistically significant at the 5% level. 

8. The environmental science club wants to show that reusable cups are more economically as well as environmentally friendly. How many weeks would it take for the reusable cup to pay for itself?

$$\begin{gathered} \left(\$0.15/\mathrm{day}\right)\left(5 \mathrm{days}/\mathrm{week}\right)=\$0.75/\mathrm{week} \\ \frac{\$18 \mathrm{cup}}{\$0.75/\mathrm{week}}=24 \mathrm{weeks} \end{gathered}$$

It would take 24 weeks for the reusable cup to pay for itself.

9. Club poster: “Purchase your reusable travel cup today for only $18! Your coffee stays warmer longer, tastes better, AND is significantly better for the environment.” Does the experiment support the club poster? Explain.  

Sample: The environmental science club’s experiment shows improvement in coffee temperature and taste, but not the environment. They did not measure the environmental impact of the disposable cups, but it is generally accepted that reusable cups are better for the environment.

10. Why do you think it is useful to use two variables (temperature and drinkability) rather than temperature alone?

Sample: Some people might say that they want their coffee to be cooler when they drink it. Showing that the taste is also better in a reusable cup may make people more willing to switch to a reusable cup to improve the taste of their coffee. 

11. Calculate the z-score for the observed difference. Is the difference in the reaction time statistically significant?

$z=\frac{0.07}{0.053}=1.3207$

Because the z-score is less than 1.96, listening to music for teen drivers was not statistically significant.

12. Practically speaking, why might the reaction time still be important for driving safety, especially for teen drivers?

Sample: Teen drivers do not have the driving experience that adult drivers may have. Any additional reaction time for a new driver is important because they are not used to or may not expect driving conditions.

13. Based on your analysis, should driver’s education instructors recommend teen drivers avoid listening to music?

Sample: Based on driving experience, teen drivers should start by NOT listening to music. The results are not statistically significant, but they are practically significant because every bit of additional reaction time is important to a teen (new) driver.

14. Calculate the z-score for the observed difference. Explain the statistical significance at the 5% level.

$$\begin{gathered} 84.6–77.4=7.2 \\ z=\frac{7.2}{1.92}=3.75 \end{gathered}$$

Because 3.75 is much larger than 1.96, handwriting class notes are statistically significant at the 5% level. 

15. Considering what students wrote about distractions, what might explain writing nearly twice as much but receiving a lower test grade?

Sample: Because the students are not fully focused on taking notes and are doing other things, they might be repeating information without realizing it. Their test grades are likely lower because they were not paying attention during class.

16. Suppose you are a student tutor working with a student who only types their notes and is struggling with their tests. What recommendations would you make to the struggling student?

Sample: If I were tutoring a student, I would tell them to put away their computer and start handwriting their notes because it has been shown to greatly improve test scores.