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 < -2.327
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%
Beta = 100% - 31.92% = 68.08%
Question 3
a)
Ho:
m ³ 45
Ha: m < 45
Reject Ho if t < -1.943
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)
P-value
= P(t < -2.2093) based on 6 df.
The 2.5% critical value is –2.447 and the 5% critical value is –1.943. Therefore,
2.5% < p-value < 5%. Since the p-value is between 1% and 10%, the results
are inconclusive under the general rule of thumb.
Question 4
a) Ho: s ³ 1300
Ha: s < 1300
Reject Ho if C2 < 3.053
C2 = 11(1161.894)2/13002 = 8.787
Do not reject Ho. Conclude there is
not significantly less variability in daily sales.
b)
P-value
= P(C2 < 8.787) based on 11 d.f.
The 90% critical value is 5.578. Since P(C2 > 5.578) = 90%, this means P(C2 < 5.578) = 10%. Therefore, the p-value is
greater than 10% and under the general rule of thumb, we would not reject Ho.
c) Lower limit for s2 = 11(1161.894)2/26.757 = 554,993.9956
Upper limit for s2 = 11(1161.894)2/2.603 = 5,704,945.962
554,993.9956 < s2 < 5,704,945.962
745 < s < 2,389
With 99% confidence, the standard deviation for
daily sales ranges between $745 and $2,389.
Question 5
Ho: median £ 90
Ha: median > 90
Reject Ho if T- £ 2
|
Observation |
1 |
2 |
3 |
4 |
5 |
6 |
|
| d | |
45 |
60 |
60 |
60 |
90 |
630 |
|
Sign |
- |
- |
+ |
+ |
+ |
+ |
|
Rank |
1 |
3 |
3 |
3 |
5 |
6 |
T- = 1 + 3 = 4
Do not reject Ho.
Conclude the average amount of time people twitter is not significantly more than 90 minutes per week.
Question 6
a)
We
first need to determine the equality of variances. Since magazine B has the
larger standard deviation, we choose this as group 1.
Ho: s1 = s2
Ha: s1 ¹ s2
Numerator df
= 5 – 1 = 4
Denominator df = 7 – 1 = 6
Reject Ho if F > 6.2272. Note that the lower
critical value is not needed.
F = 6.54222/3.98812 =
2.6910
Do not reject Ho.
Conclude no significant difference in the
variances. Assume that s1 = s2 and proceed with the t test
assuming equal variances.
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(26.6629/7
+ 26.6629/5) = 2.2113
Reject Ho. Conclude magazine A has
significantly more ads on average than magazine B.
a)
P-value
= P(t > 2.2113) based on 10 df.
The 5% critical value is 1.812 and the 2.5% critical value is 2.228. Therefore,
2.5% < p-value < 5%. If the level of significance were 1%, since the
p-value > 2.5%, by inference, p-value > alpha and we would not reject Ho.
b)
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 7
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
With 95% confidence, the average score for
region 1 is 0.9 to 5.3 marks higher than the average score for region 2.
Question 8
a)
We
first need to determine the equality of variances. Since downtown has the
larger standard deviation, we choose this as group 1.
Ho: s1 = s2
Ha: s1 ¹ s2
Numerator df
= denominator df = 11. You will notice that there is
no 11 df in the numerator of
the F table. Therefore, we take the average of the 10 df and 12 df with 11 df in the denominator. The critical value is (3.5257 +
3.4296)/2 = 3.47765
Reject Ho if F > 3.47765. Note that the
lower critical value is not needed.
F = 153.98082/29.83392 =
26.6387
Reject Ho.
Conclude a significant difference in the
variances. Assume that s1 ¹ s2 and proceed with the t test
assuming unequal variances.
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)
P-value
= 2P(t > 0.5908) based on 11 df.
The 10% critical value is 1.363. Therefore, P(t >
0.5908) > 10% and the p-value > 20%. 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 9
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.
Question 10
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 11
a)
Choose
the women as group 1.
Ho: p1 £ p2
Ha: p1 > p2
Reject Ho if Z > 1.645.
Phat = (1242 + 1066)/(1350
+ 1350) = 2308/2700 = 0.8548
Z = (0.92 – 0.7896)/sqrt(0.8548(0.1452)/1350 +
0.8548(0.1452)/1350) = 9.6164
Reject Ho. Conclude that the percentage of
women who make impulse purchases is significantly higher than that of men.
b)
P-value
= P(Z > 9.6164) = 0 for all intents and purposes.
Since the p-value is less than 1%, we reject Ho.
c)
Lower
limit = (0.92 – 0.7896) – 1.96sqrt(0.92(0.08)/1350 + (0.7896)(0.2104)/1350) =
0.1304 – 0.0261 = 0.1043
Upper limit = 0.1304 + 0.0261 = 0.1565
0.1043 < p1 – p2 < 0.1565
With 95% confidence, the percentage of women who make impulse purchases
is 10.43% to 15.65% higher than the corresponding percentage of men.
Question 12
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.