1. Ages of College Students The dean of students wants to see whether there is a
ID: 3315877 • Letter: 1
Question
1. Ages of College Students The dean of students wants to see whether there is a significant difference in the ages of residential and commuting students. She selects a sample of 50 students from each group. The ages are as follows:
> summary(residents) Min. 1st Qu. Median Mean 3rd Qu. Max. sd
18.000 19.000 21.500 22.120 25.000 32.000 3.685
> summary(commuters) Min. 1st Qu. Median Mean 3rd Qu. Max. sd
18.000 19.000 21.500 22.760 25.000 36.000 4.702
The above values are for the sample of 50 students, not the population.
b. The definition of “kid” is “a human being below the age of 18 years.” We, as researchers working for the dean, would like to find out whether the average residential student is a “kid.”
i. What kind of question are we asking (two-sided/non-directional or one-sided/directional, and if one- sided/directional, state which side or direction)?
ii. State the null and alternative hypothesis.
iii. Write out and calculate the appropriate test statistic.
iv. What is the distribution of the test statistic under the null hypothesis (so-called the “null distribution”)?
v. Obtain the p-value (using R or the table—you can approximate if using the table).
vi. Using a significance level of = 0.05, state your conclusions.
Explanation / Answer
i. What kind of question are we asking (two-sided/non-directional or one-sided/directional, and if one- sided/directional, state which side or direction)?
this is a one directional 1 tail left sided test , as we are interested in knowing if the value is less than 18 years
ii. State the null and alternative hypothesis.
Ho : The average value of the residents is not less than 18 years
H1 : The average value of the residents is less than 18 years
iii. Write out and calculate the appropriate test statistic.
the test stat is
t = (xbar-mu)/(sd/sqrt(n)), n = 50 and xbar = 22.12 from the summary stats of residents
= (22.12-18)/(3.685/sqrt(50)) = 7.90
the df is n-1 = 50-1 = 49
v. Obtain the p-value (using R or the table—you can approximate if using the table).
> t<- 7.90
> n <- 50
> pt(-abs(t),df=n-1)
[1] 1.358743e-10
as the p value is less than 0.05 , hence we reject null hypothesis in favor of alternate hypothesis to conclude that The average value of the residents is less than 18 years
vi. Using a significance level of = 0.05, state your conclusions
as the p value is less than 0.05 , hence we reject null hypothesis in favor of alternate hypothesis to conclude that The average value of the residents is less than 18 years
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