Answer :
Answer:
(1) Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] <= 35 minutes
(2) Calculated value of the test statistic is 1.6
(3) We conclude that homework was actually more time consuming
Step-by-step explanation:
We are given that the time spent by DCSI3710 students on trying to certify any HLS homework is known to have a mean of 35 minutes.
(1) Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] <= 35 minutes {means that the homework takes less than or equal to 35 minutes}
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 35 minutes {means that the homework was actually more time consuming as it takes more than 35 minutes}
(2) We are provided that the sample of 16 students had a mean completion time of 39 minutes and the sample standard deviation was 10 minutes.
So, the test statistics used here will be;
T.S. = [tex]\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, X bar = sample mean = 39 minutes
s = sample standard deviation = 10 minutes
n = sample of students = 16
So, test statistics = [tex]\frac{39-35}{\frac{10}{\sqrt{16} } }[/tex] ~ [tex]t_1_5[/tex]
= 1.6
So, calculated value of the test statistic in this case is 1.6 .
(3) We are given that p-value associated with the test statistic is 0.08 or 8% . And the significance level is given as 0.1 or 10%.
It is stated that If p-value is less than significance level ⇒ Reject [tex]H_0[/tex]
If p-value is more than significance level ⇒ Do not Reject [tex]H_0[/tex]
Since here, p-value is less than the significance level as 8% < 10% so we conclude that null hypothesis must be rejected.
Therefore, we conclude that homework was actually more time consuming.