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A study was conducted to determine if there was a difference in the driving ability of students from West University and East University by sending a survey to a sample of 100 students at both universities. Of the 100 sampled from West University, 15 reported they were involved in a car accident within the past year. Of the 100 randomly sampled students from East University, 12 students reported they were involved in a car accident within the past year.

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Answer:

There is no significant evidence which shows that there is a difference in the driving ability of students from West University and East University, assuming a significance level 0.1

Step-by-step explanation:

Let p1 be the proportion of West University students who involved in a car accident within the past year

Let p2 be the proportion of East University students who involved in a car accident within the past year

Then

[tex]H_{0}: [/tex]p1=p2

[tex]H_{a}: [/tex]p1≠p2

The formula for the test statistic is given as:

z=[tex]\frac{p1-p2}{\sqrt{\frac{p*(1-p)*(n1+n2)}{n1*n2} } }[/tex]  where

  • p1 is the sample proportion of West University students who involved in a car accident within the past year (0.15)
  • p2 is the sample proportion of East University students who involved in a car accident within the past year (0.12)
  • p is the pool proportion of p1 and p2 ([tex]\frac{15+12}{100+100}=0.135[/tex])
  • n1 is the sample size of the students from West University (100)
  • n2 is the sample size ofthe students from East University (100)

Then we have z=[tex]\frac{0.03}{\sqrt{\frac{0.135*0.865*(100+100)}{100*100} } }[/tex] ≈ 0.6208

Since this is a two tailed test, corresponding p-value for the test statistic is ≈ 0.5347.

Assuming significance level 0.1, The result is not significant since 0.5347>0.1. Therefore we fail to reject the null hypothesis at 0.1 significance

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