STAT217
Worksheet #2
Solutions
Question 1
a) The
first thing we have to do is figure out the rejection region in terms of xbar. We reject the null hypothesis if
Power = P(reject
Ho | Ha true)
Power = P(xbar > 26.028125 | m = 26)
Power = P(Z > (26.028125 - 26)/(5/sqrt(64))) = P(Z > 0.045)
If we use the Z table, we round 0.045 to 0.05.
P(Z > 0.05) = 0.5 0.0199 = 0.4801
b) If the power is 90%, this means the Z value used to calculate the power is 1.282 since P(Z > -1.282) = 0.9. From part a, we reject Ho if xbar > 26.021825.
-1.282 = (26.021825 - m)/(5/sqrt(64))
m = 26.021825 + 1.282(5)/sqrt(64) = 26.823075 which we round to 28.6231.
c) If the power is at least 95%, we need Z < -1.645 since P(Z > -1.645) = 0.95.
That means (xbar
- 26)/(5/sqrt(n)) < -1.645 or xbar
< 26 - 8.225/sqrt(n). Note that we get 8.225 by multiplying 1.645 by 5.
But, we also have to take into consideration that the level of significance is
5%. We reject the null hypothesis if Z > 1.645 or
This means that 25 + 8.225/sqrt(n) < xbar < 26
- 8.225/sqrt(n). Since it is n that we are interested in solving for, we want
to solve 25 + 8.225/sqrt(n) < 26 - 8.225/sqrt(n). Solving
for n, n > 271.
Question 2
a) Ho: m Ł 1200
Ha: m > 1200
Reject Ho if p-value < 1%. Do not reject Ho if p-value > 10%.
Z = (1250 1200)/(135/sqrt(50)) = 2.62
P-value = P(Z > 2.62) = 0.5 0.4956 = 0.0044.
P-value < 1%
Reject Ho
Conclude coffee shops spend significantly more than $1200 per week on average in buying whole beans.
b) Reject Ho if Z > 2.326
Reject Ho if (xbar 1200)/(135/sqrt(50)) > 2.326
Reject Ho if xbar > 1200 + 2.326(135)/sqrt(50)
Reject Ho if xbar > 1244.4077
Power = P(reject Ho | Ha true)
Power = P(xbar > 1244.4077 | m = 1250)
Power = P(Z > (1244.4077 1250)/ (135/sqrt(50))
Power = P(Z > -0.29) = 0.5 + 0.1141 = 0.6141
c) Reject Ho if Z > 1.282
Reject Ho if (xbar 1200)/(135/sqrt(50)) > 1.282
Reject Ho if xbar > 1200 + 1.282(135)/sqrt(50)
Reject Ho if xbar > 1224.4758
Power = P(reject Ho | Ha true)
Power = P(xbar > 1224.4758 | m = 1250)
Power = P(Z > (1224.4758 1250)/ (135/sqrt(50))
Power = P(Z > -1.34) = 0.5 + 0.4099 = 0.9099
The power increases from 61.41% to 90.99% which is a difference of 29.58%.
Question 3
a) Ho: p > 0.02
Ha: p < 0.02
Reject Ho if Z < -1.645
Z = (0.012 - 0.02)/sqrt(0.02(0.98)/500) = -1.28
Do not reject Ho.
Conclude that the percentage of people who watch the show is not significantly below 2%.
b) P-value = P(Z < -1.28) = 0.5 0.3997 = 0.1003. Since the p-value is marginally above 10%, this provides strong support for Ho. Thus, we do not reject Ho.
c) Reject Ho if Z < -1.645
Reject Ho if (phat 0.02)/ sqrt(0.02(0.98)/500) < -1.645
Reject Ho if phat < 0.02 - 1.645sqrt(0.02(0.98)/500)
Reject Ho if 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
d) Reject Ho if Z < -1.282
Reject Ho if (phat 0.02)/ sqrt(0.02(0.98)/500) < -1.282
Reject Ho if phat < 0.02 - 1.282sqrt(0.02(0.98)/500)
Reject Ho if phat
< 0.012
Power = P(phat < 0.012 |
p = 0.012)
= P(Z < (0.012 - 0.012)/sqrt(0.012(0.988)/500))
= P(Z < 0) = 0.5
e) If the power is 90%, we need Z = 1.282 since P(Z < 1.282) = 0.9
We reject Ho if phat < 0.012 from part d.
(0.012 p)/sqrt(p(1-p)/500) = 1.282
0.012 p = 1.282 sqrt(p(1-p)/500)
Squaring both sides:
0.000144 0.024p + p2 = 1.643524p(1-p)/500
Multiplying both sides by 500:
0.072 12p + 500p2 = 1.643524p 1.643524p2
Bringing everything over to the left side:
501.643524p2 13.643524p + 0.072 = 0
Solving for the lower root using the quadratic equation, we find p = 0.007165 = 0.7165%. Note that we cannot use the upper root since the left side of the initial equation would be less than zero.
Question 4
a) Ho: m = 25
Ha: m š 25
Reject Ho if t < -2.365 or t > 2.365 (7 degrees of freedom)
t = (20.25 25)/(4.7132/sqrt(8)) = -2.8505
Reject Ho
Conclude that the average number of hours per week that semi-retired people work is
signficantly different than 25 hours.
b) P-value = 2P(t > 2.8505) based on 7 degrees of freedom. The 2.5% critical value is 2.365 and the 1% critical value is 2.998. Therefore, P(t > 2.8505) is between 1% and 2.5% leading to the p-value being between 2% and 5%. Since the p-value is greater than 2%, it would be greater than a level of significance of 1.5% leading to not rejecting Ho. We would conclude that the average number of hours per week that semi-retired people work is not significantly from 25 hours.
c) Lower limit = 20.25 2.365(4.7132)/sqrt(8) = 20.25 3.94 = 16.31
Upper limit = 20.25 + 3.94 = 24.19.
With 95% confidence, semi-retired people work between 16.3 and 24.2 hours per week. We would reject Ho at a 5% level of significance since the hypothesis mean of 25 hours falls outside the 95% confidence interval.
Question 5
Ho: median Ł 52
Ha: median > 52
There is one value of 52 in the data. We remove the value which decreases the sample size to 6.
Reject Ho if T- Ł 2.
We subtract the median of 52 from the remaining values:
|
Value |
43 |
59 |
68 |
68 |
70 |
40 |
|
Median |
52 |
52 |
52 |
52 |
52 |
52 |
|
d = Value - Median |
-9 |
7 |
16 |
16 |
18 |
-12 |
We then compute the test statistic:
|
Observation |
1 |
2 |
3 |
4 |
5 |
6 |
|
|d| |
7 |
9 |
12 |
16 |
16 |
18 |
|
Sign |
+ |
- |
- |
+ |
+ |
+ |
|
Rank |
1 |
2 |
3 |
4.5 |
4.5 |
6 |
T- = 2 + 3 = 5
Do not reject Ho.
Conclude the median is not significantly greater than 52 and that the new teaching method does not significantly increase the grade on basic arithmetic.
Question 6
a) Ho: s = 1200
Ha: s š 1200
Reject Ho if C2 < 6.262 or C2 > 27.488
C2 = 15(1050)2/12002 = 11.4844
Do not reject Ho.
Conclude the wage spread of small factories is not significantly different than that of large factories.
b) Lower limit of s2 = 15(1050)2/27.4884 = 601,617.4095
Upper limit of s2 = 15(1050)2/6.2621 = 2,640,887.242
We compute the square roots of the two limits to find the confidence interval limits for s.
Lower limit of s = sqrt(601,617.4095) = 775.64
Upper limit of s = sqrt(2,640,887.242) = 1,625.08
If we wanted to interpret this interval, we would say that with 95% confidence, the standard deviation of the monthly wages of small factories ranges from $775.64 to $1,625.08. If we used this confidence interval to test the hypothesis in part a, we would not reject Ho at a 5% level of significance since the hypothesis standard deviation of $1200 falls inside the 95% confidence interval.
c) Since the sample standard deviation is less than the hypothesis standard deviation, the p-value 2P(C2 < 11.4844) based on 15 degrees of freedom. The 90% critical value is 8.547. Since P(X2 > 8.547) = 90%, this means P(X2 < 8.547) = 10%. Thus, P(C2 < 11.4844) is greater than 10% leading to the p-value being greater than 20%. Since the p-value is greater than 10%, under the general rule of thumb, this provides strong support for Ho and we do not reject Ho.
From the textbook
Section 7-3
Question 27 (page 397)
Reject Ho if (xbar 25)/(2.4/sqrt(100)) > 1.645 or if xbar > 25.3948
Power = P(xbar > 25.3948 | ľ = 25.2) = P(Z > (25.3948 25.2)/(2.4/sqrt(100))) = P(Z > 0.81) = 0.5 0.291 = 0.209.
Question 28
If the alternative mean is 25.5, Power = P(Z > (25.3948 25.5)/(2.4/sqrt(100))) = P(Z > -0.44) = 0.5 + 0.17 = 0.67
If we increase the level of significance to 10%, then reject Ho if (xbar 25)/(2.4/sqrt(100)) > 1.282 or if xbar > 25.30768
Then power = P(xbar > 25.30768 | ľ = 25.2) = P(Z > (25.30768 25.2)/(2.4/sqrt(100))) = P(Z > 0.45) = 0.5 0.1736 = 0.3264. The first solution increases the power more.
Question 29
For the power to be 95%, the Z value used in the power calculations is -1.645 = (25.3948 ľA)/(2.4/sqrt(100)) from which we get ľA = 25.3948 + 1.645(2.4)/sqrt(100) = 25.7896.
Section 7-4
Question 22 (page 408)
Ho: m = 3.39
Ha: m š 3.39
Reject Ho if t < -2.132 or t > 2.132
T = (3.675 3.39)/(0.6573/sqrt(16)) = 1.7344
Do not reject Ho and conclude the vitamin supplement does not appear to have an effect on birth weight.
P-value = 2P(t > 1.7344) based on 15 degrees of freedom. The 5% critical value is 1.753 and the 10% critical value is 1.341. Then, P(t > 1.7344) is between 5% and 10% leading to the p-value being between 10% and 20%. Since the p-value is greater than 10%, it is also greater than the 5% level of significance leading to not reject Ho.
Lower limit = 3.675 2.132(0.6573)/sqrt(16) = 3.675 0.3503 = 3.3247
Upper limit = 3.675 + 0.3503 = 4.0253
3.3247 < ľ < 4.0253
Since the hypothesis mean of 3.39 lies inside the 95% confidence interval, we do not reject Ho at a 5% level of significance.
Section 7-5
Question 17 (page 418)
Ho: p ≥ 0.5
Ha: p < 0.5
Reject Ho if Z < -1.282
Phat = 32/71 = 0.4507; Z = (0.4507 0.5)/sqrt(0.5(0.5)/71) = -0.83
Do not reject Ho. The results to not suggest that the nicotine patch therapy is not effective.
P-value = P(Z < -0.83) = 0.5 0.2967 = 0.2033 = 20.33%. Since the p-value is greater than the 10% level of significance, we do not reject Ho.
Section 7-6
Question 3 (page 426)
Ho: s = 43.7
Ha: s≠ 43.7
Reject Ho if c2 < 57.153 or c2 > 106.629
c2 = 80(52.32)/43.72 = 114.5857
Reject Ho and conclude the errors have a standard deviation that is significantly different than 43.7 feet. Since the new production method appears to have more variation than the old method, it appears to be worse.
P-value = 2P(c2 > 114.5857). The 1% critical value is 112.329 and the 0.1% critical value is 116.321. So, P(c2 > 114.5857) is between 0.1% and 1% leading to the p-value being between 0.2% and 2%. Since the p-value is less than 2%, it is less than the 5% level of significance leading to rejecting Ho.
Lower limit for s2 = 80(52.32)/106.629 = 2,052.1922
Upper limit for s2 = 80(52.32)/57.153 = 3,828.7264
Taking the square roots of the limits, we have 45.3 < s < 61.9. Since the hypothesis standard deviation of 43.7 falls outside the 95% confidence interval, we reject Ho at a 5% level of significance.
Section 12-3
Question 9 (page 697)
Ho: median ≤ 2.5
Ha: median > 2.5
Subtracting 1.77 from each value gives differences of -0.03, 0, 0.13, 0.03, 0.11, 0.1, 0.02, 0.04 and 0.12. Subtracting d=0 from the differences gives n=8. Reject Ho if T- ≤ 6.
|
Observation |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
|
|d| |
0.02 |
0.03 |
0.03 |
0.04 |
0.1 |
0.11 |
0.12 |
0.13 |
|
Sign |
+ |
- |
+ |
+ |
+ |
+ |
+ |
+ |
|
Rank |
1 |
2.5 |
2.5 |
4 |
5 |
6 |
7 |
8 |
T- = 2.5. Reject Ho and conclude the median weight is greater than 1.77 kg.