Question

For each of the four problems below use the online applications in the notes to calculate the hypothesis test. In your response provide all of the steps listed above. For the test statistic and p-value you do not need to show your work you only need to give the values provided by the applications. 1. A study was conducted to see if there was a difference in how men and women feel about animal testing. A random sample of 55 women was selected and asked to rank how they felt about animal testing on a 10 point scale where a score of 1 represented strong opposition and a score of 10 represented strong support. The mean score of the sample was 4.5 with a standard deviation of 0.7. A random sample of 75 men was selected and asked to rank how they felt about animal testing on a 10 point scale where a score of 1 represented strong opposition and a score of 10 represented strong support. The mean score of the sample was 5.2 with a standard deviation of 0.9. Test the claim that there is a difference in how men and women feel about animal testing. Use a-0.01. 2. EKU would like to know if fewer seniors pan on staying in Kentucky after graduation compared to freshman. A random sample of 100 seniors are surveys and 71 responded that they plan on staying in Kentucky after graduation. A random sample of 100 freshman are surveys and 82 responded that they plan on staying in Kentucky after graduation. Use q=0.05.

Answer 1

• Here we are to test
  
  Ho : Amen = Mwomen v/s H1 : Amen + Mwomen
  
  We use the test statistic:
  
  Степ - I women TE men2 + comen nmen nomen
  
  
  Under Null,
  $\large T \sim t_{n_{men}+n_{women}-2}$
  
  Given Information
  
  Imen = 5.2
  
  $\large s_{men}=0.9$
  $\large n_{men}=75$
  
  $\large \overline{x_{women}}=4.5$
  $\large s_{women}=0.7$
  $\large n_{women}=55$
  
  
  $\large \therefore T=\frac{5.2-4.5}{\sqrt{\frac{0.9^2}{75}+\frac{0.7^2}{55}}}=4.9861$
  
  p-value
  $\large P(t_{128}<|T|)$
  $\large =P(t_{128}<-4.9861)+P(t_{128}>4.9861)$
  $\large =0.00000098 +0.00000098$
  
  = 0.00000196 <0.01
  
  
  Since the p-value is less than 0.01, we reject H0 at 1% level of significance, and conclude that there is a difference between a man and a woman's outlook towards the scenario.
  
  
  
  
• Here we are to test
  $\small H_0:p_{senior}=p_{freshma} \text{ v/s }H_1:p_{senior} < p_{freshman}$
  We use the test statistic:
  $\large T=\frac{\widehat{p_{senior}}-\widehat{p_{freshman}}}{\sqrt{\frac{\widehat{p_{senior}}(1-\widehat{p_{senior}})}{n_{senior}}+\frac{\widehat{p_{freshman}}(1-\widehat{p_{freshman}})}{n_{freshman}}}}$
  
  Under Null,
  $\large T \sim t_{n_{senior}+n_{freshman}-2}$
  
  GIven Information
  $\large \widehat{p_{senior}}=\frac{71}{100}=0.71$
  $\large \widehat{p_{freshman}}=\frac{82}{100}=0.82$
  $\large n_{freshman}=100$
  $\large n_{senior}=100$
  
  $\large \therefore T=\frac{0.71-0.82}{\sqrt{\frac{0.71(1-0.71)}{100}+\frac{0.82(1-0.82)}{100}}}=-1.850$
  
  p-value
  $\large P(T<-t_{198})$
  $\large =P(-T>t_{198})$
  $\large =P(t_{198}<-T)$
  $\large =P(t_{198}<-1.85)$
  $\large =0.03289461<0.05$
  
  Hence, we reject Null Hypothesis at 5% level of significance, and conclude that, there is significant difference between the seniors and freshmen.
  
  
  

I hope this clarifies your doubt. If you're satisfied with the solution, hit the Like button. For further clarification, comment below. Thank You. :)