STAT217 Midterm practice solutions
Question 1
a) Ho: m £ 1000
Ha: m > 1000
Reject Ho if Z > 1.645
Z = (1120 – 1000)/(250/sqrt(14)) =
1.8
Reject Ho. Conclude the modifications significantly increased the distance traveled.
b) P-value = P(Z > 1.8) = 0.5 – 0.4641 = 0.0359 = 3.59%. Since we rejected Ho, we would not reject Ho if the level of significance is no greater than the p-value. Thus, we would not reject Ho for levels of significance between 1% and 3.59%.
c) Reject Ho if (xbar – 1000)/(250/sqrt(14)) > 1.645
Reject Ho if xbar > 1000 + 1.645(250)/sqrt(14)
Reject Ho if xbar > 1,109.9112
Power = P(xbar > 1,109.9112 | m = 1120)
Power = P(Z > (1,109.9112 – 1120)/(250/sqrt(14))
Power = P(Z > -0.15) = 0.5 + 0.0596 = 0.5596 = 55.96%
Question 2
a) Ho: p ³ 2%
Ha: p < 2%
Reject Ho if Z < -2.326
Z = (0.012 – 0.02)/sqrt(0.02(0.98)/500) = -1.2778
Do not reject Ho. Conclude the percentage of companies exceeding water pollution limits is not significantly less than 2%.
b) Reject Ho if phat < 0.02 - 1.645sqrt(0.02(0.98)/500) or phat < 0.0097
Power = P(phat < 0.0097 | p =
0.012)
= P(Z < (0.0097-0.012)/sqrt(0.012(0.988)/500))
= P(Z < -0.47) = 0.5 – 0.1808 = 0.3192 = 31.92%
Question 3
a) Choose downtown as group 1 since it has the larger mean.
Ho: m1 = m2
Ha: m1 ¹ m2
df = (153.98082/12 + 29.83392/12)2/((153.98082/12)/11 + (29.83392/12)/11) = 11.8 which we round to 11. Reject Ho if t < -2.201 or > 2.201
t = (503.0833 – 476.3333)/sqrt(153.98082/12 + 29.83392/12) = 0.5908
Do not reject Ho. Conclude there is no significant difference in average daily rates between downtown and suburban hotels.
b) Under the general rule of thumb, since the p-value is greater than 10%, we do not reject Ho.
c) Lower limit = (503.0833 – 476.3333) – 2.201sqrt(153.98082/12 + 29.83392/12)
= 26.75 – 99.66 = -72.91
Upper limit = 26.75 + 99.66 = 126.41
-72.91 < m1 - m2 < 126.41
Since the hypothesis difference of zero falls inside the 95% confidence interval, we do not reject Ho at a 5% level of significance.
Question 4
We choose system B as group 1 in order to set up a right tail test.
Ho: d £ 0
Ha: d > 0
First we compute the differences:
|
Person |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
System A |
5 |
7 |
7 |
4 |
6 |
7 |
7 |
7 |
7 |
6 |
|
System B |
8 |
7 |
6 |
7 |
6 |
8 |
9 |
8 |
8 |
8 |
|
d = B-A |
3 |
0 |
-1 |
3 |
0 |
1 |
2 |
1 |
1 |
2 |
Since there are two differences of zero, n = 8. Reject Ho if T- £ 6
We rank the absolute values of the differences from lowest to highest:
|
Observation |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
|
|d| |
1 |
1 |
1 |
1 |
2 |
2 |
3 |
3 |
|
Sign |
- |
+ |
+ |
+ |
+ |
+ |
+ |
+ |
|
Rank |
2.5 |
2.5 |
2.5 |
2.5 |
5.5 |
5.5 |
7.5 |
7.5 |
T- = 2.5. Reject Ho. Conclude that system B is rated significantly better than system A.
Question 5
a) Ho: m ³ 45
Ha: m < 45
Reject Ho if t < -1.895
t = (40.1343 – 45)/(5.8271/sqrt(7)) = -2.2093
Reject Ho. Conclude that the average weekly income of a panhandler is significantly less than $45.
b) Since we do not reject Ho if the p-value is at least as large as the level of significance, we would not reject Ho for levels of significance between 1% and 3.46%.
Question 6
a) Choose region 1 as group 1.
Ho: m1 £ m2
Ha: m1 > m2
Z = (67.5 – 64.4)/sqrt(7.82/100 + 8.22/100) = 2.74
P-value = P(Z > 2.74) = 0.5 – 0.4969 = 0.0031 = 0.31%
Since the p-value < 1%, reject Ho. Conclude that the average score for region 1 is significantly higher than that of region 2.
b) Lower limit = (67.5 – 64.4) – 1.96 sqrt(7.82/100 + 8.22/100) = 3.1 – 2.2 = 0.9
Upper limit = 3.1 + 2.2 = 5.3
0.9 < m1 - m2 < 5.3
Question 7
a) Have d = operator 1 – operator 2. Then the differences are:
|
Day |
Monday |
Tuesday |
Wednesday |
Thursday |
Friday |
Saturday |
|
Operator 1 |
108 |
115 |
118 |
116 |
110 |
115 |
|
Operator 2 |
106 |
118 |
116 |
115 |
114 |
112 |
|
difference |
2 |
-3 |
2 |
1 |
-4 |
3 |
Ho: md = 0
Ha: md ¹ 0
Reject Ho if t < -2.571 or > 2.571
t = (0.1667 – 0)/(2.9269/sqrt(6)) = 0.1395
Do not reject Ho. Conclude there is no significant difference in average production between the two operators.
b) Lower limit = 0.1667 – 2.571(2.9269)/sqrt(6) = 0.1667 – 3.0721 = -2.9
Upper limit = 0.1667 + 3.0721 = 3.2
-2.9 < md < 3.2. Since md = 0 falls inside the 95% confidence interval, we do not reject Ho at a 5% level of significance.
Question 8
a) Choose the 25 and under group as group 1.
Ho: p1 – p2 £ 10%
Ha: p1 – p2 > 10%
Reject Ho if Z > 1.645.
Z = [(0.75 – 0.6) – 0.1]/sqrt(0.75(0.25)/1200 + 0.6(0.4)/1500) = 2.81
Reject Ho. Conclude that the difference between the two age groups in the percentage that use MP3s on a regular basis is significantly more than 10%.
b) P-value = P(Z > 2.81) = 0.5 – 0.4975 = 0.0025 = 0.25%. Since the p-value is less than 1%, we reject Ho.
Question 9
a)
Ho: s ³ 1300
Ha: s <
1300
Reject Ho if C2
< 3.0535
C2
= 11(1161.894)2/13002 = 8.787
Do not reject
Ho. Conclude there is not significantly less variability in daily sales.
b) Lower
limit for s2
= 11(1161.894)2/26.7568 = 554,998.144
Upper limit for s2
= 11(1161.894)2/2.6032 = 5,704,507.66
Lower limit for s =
sqrt(554,998.144) = 744.98
Upper limit for s =
sqrt(5,704,507.66) = 2,388.41
With 99% confidence, the standard deviation for daily sales ranges between $744.98 and $2,388.41.
Question 10
a) We choose magazine A as group 1.
Ho: m1 £ m2
Ha: m1 > m2
d.f. = 7 + 5 – 2 = 10. Reject Ho if t > 1.812
sp2 = [6(3.98812) + 4(6.54222)]/10 = 26.6629
t = (19.2857 – 12.6)/sqrt(22.6629/7 + 22.6629/5) = 2.2113
Reject Ho. Conclude magazine A has significantly more ads on average than magazine B.
b) In order to reject Ho, the p-value must be less than the level of significance. Thus, we should choose the level of significance between 2.57% and 10%.
c) Lower limit = (19.2857 – 12.6) – 2.228 sqrt(22.6629/7 + 22.6629/5) = 6.6857 – 6.7364 = -0.1
Upper limit = 6.6857 + 6.7364 = 13.4
-0.1 < m1 - m2 < 13.4
Question 11
We choose the richer neighbourhood as group 1 since both sample sizes are equal and we want to see if the richer neighbourhood is greater than the poorer neighbourhood.
Ho: median 1 £ median 2
Ha: median 1 > median 2
Reject Ho if T1 ³ 50
Here is the ranking of the data:
|
Observation |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
|
Value |
4 |
5 |
5 |
5 |
8 |
9 |
9 |
17 |
18 |
18 |
18 |
19 |
|
Group |
P |
R |
P |
P |
P |
P |
P |
R |
R |
R |
R |
R |
|
Rank |
1 |
3 |
3 |
3 |
5 |
6.5 |
6.5 |
8 |
10 |
10 |
10 |
12 |
T1 = 3 + 8 + 10 + 10 + 10 + 12 = 63
Reject Ho. Conclude that people in the richer neighbourhood eat out more often on average than those in the poorer neighbourhood.