Question

The accompanying table lists a random selection of usual travel times to​ school, in​ minutes, for 40 secondary school students in country A. A second selection of usual travel times to​ school, in​ minutes, was randomly selected for 40 students in country B. Complete parts​ a) and​ b).

Sample Country A   Sample Country B
45   29
5   10
4   9
14   30
50   5
21   8
21   7
19   15
21   10
21   36
24   16
36   10
14   24
29   22
19   20
11   26
45   29
10   10
2   25
60   7
24   15
19   19
5   25
16   15
5   10
15   26
18   5
30   2
40   4
20   25
11   20
30   14
10   48
14   21
20   20
10   12
14   19
16   5
10   14
24   11

Calculate the test statistic.

t = _ ?

​(Round to two decimal places as​ needed.)

Calculate the degrees of freedom.

df = _ ?

Determine the​ P-value.

P-value = _ ?

​(Round to four decimal places as​ needed.)

Make a conclusion.

Comparing the​ P-value to the level of​ significance, a = ​0.05, the decision is to (reject/fail to reject) the null hypothesis. There (is/is not) sufficient evidence to conclude that​ students' travel times in each country are (same/different).

Construct a​ 95% confidence interval for​(μ1−μ2​).

( _ , _ )

​b) Are your​ P-value and confidence level in part​ a) trustworthy?

yes or no

Answer 1

X :- Represent sample country A

Y :- Represent sample country B

Mean X̅ = Σ Xi / n
X̅ = 822 / 40 = 20.55
Sample Standard deviation SX = √ ( (Xi - X̅ )2 / n - 1 )
SX = √ ( 6751.9 / 40 -1 ) = 13.1577

Mean Y̅ = ΣYi / n
Y̅ = 678 / 40 = 16.95
Sample Standard deviation SY = √ ( (Yi - Y̅ )2 / n - 1 )
SY = √ ( 3765.9 / 40 -1) = 9.8266

Test Statistic :-
t = (X̅1 - X̅2) / SP √ ( ( 1 / n1) + (1 / n2))



t = ( 20.55 - 16.95) / 11.6122 √ ( ( 1 / 40) + (1 / 40 ))
t = 1.39

Degree of freedom =  n1 + n1 - 2 = 40 + 40 - 2 = 78

P - value = P ( t > 1.3864 ) = 0.1696 ( From t table )

Reject null hypothesis if P value < α = 0.05 level of significance
P - value = 0.1696 > 0.05 ,hence we fail to reject null hypothesis
Conclusion :- Fail to Reject Null Hypothesis

Comparing the​ P-value to the level of​ significance, a = ​0.05, the decision is to (fail to reject) the null hypothesis. There (s not) sufficient evidence to conclude that​ students' travel times in each country are (different).

Confidence interval is :-
( X̅1 - X̅2 ) ± t( α/2 , n1+n2-2) SP √( (1/n1) + (1/n2))
t(α/2, n1 + n1 - 2) = t( 0.05/2, 40 + 40 - 2) = 1.991
( 20.55 - 16.95 ) ± t(0.05/2 , 40 + 40 -2) 11.6122 √ ( (1/40) + (1/40))
Lower Limit = ( 20.55 - 16.95 ) - t(0.05/2 , 40 + 40 -2) 11.6122 √( (1/40) + (1/40))
Lower Limit = -1.5698
Upper Limit = ( 20.55 - 16.95 ) + t(0.05/2 , 40 + 40 -2) 11.6122 √( (1/40) + (1/40))
Upper Limit = 8.7698
95% Confidence Interval is ( -1.5698 , 8.7698 )
Since value 0 lies in the interval   ( -1.5698 , 8.7698 ) , hence we fail to reject null hypothesis.

​b) Are your​ P-value and confidence level in part​ a) trustworthy?

Yes, because both method landed on the same result, ( Fail to reject null ).