Answer :
Answer:
We conclude that the mean amount of time college students spend on homework per week is less than the mean amount of time last year.
Step-by-step explanation:
We are given that according to a research study, college students spent 19.7 hours doing homework per week last year, on average. A random sample of 21 college students was surveyed and the mean amount of time per week each college student spent on homework was 19.1. This data has a sample standard deviation of 1.9.
We have to test if the mean amount of time college students spend on homework per week is less than the mean amount of time last year or not.
Let, NULL HYPOTHESIS, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 19.7 {means that the mean amount of time college students spend on homework per week is higher than or equal to the mean amount of time last year}
ALTERNATE HYPOTHESIS, [tex]H_1[/tex] : [tex]\mu[/tex] < 19.7 {means that the mean amount of time college students spend on homework per week is less than the mean amount of time last year}
The test statistics that will be used here is One-sample t-test;
T.S. = [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean amount of time per week = 19.1
s = sample standard deviation = 1.9
n = sample of students = 21
So, test statistics = [tex]\frac{19.1 - 19.7}{\frac{1.9}{\sqrt{21} } }[/tex] ~ [tex]t_2_0[/tex]
= -1.447
Now, at 10% significance level t table gives critical value of -1.325. Since our test statistics is less than the critical value of t so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.
Therefore, we conclude that the mean amount of time college students spend on homework per week is less than the mean amount of time last year.