MGMT2263 Tutorial Sheet 1 Solutions
1) Ho: m ³ 125
Ha: m < 125
Reject Ho if p-value < 1%. Do not reject Ho if p-value > 10%.
Z = (110.03 – 125)/20.5/sqrt(12) = -2.53
p-value = P(Z < -2.53) = 0.5 – 0.4943 = 0.0057
Since the p-value < 1%, reject Ho
Conclude that customers spend less than $125 on average.
2) Reject Ho if Z < -1.645
Reject Ho if (xbar – 125)/(20.5/sqrt(12)) < -1.645
Reject Ho if xbar < 125 – 1.645(20.5)/sqrt(12)
Reject Ho if xbar < 115.2652
Power = P(reject Ho | Ha true)
Power = P(xbar < 115.2652 | m = 110)
Power = P[Z < (115.2652 – 110)/(20.5/sqrt(12))]
Power = P(Z < 0.89) = 0.5 + 0.3133 = 0.8133 = 81.33%
3) We know that we reject Ho if xbar < 115.2652.
In order for the power to be 99%, the Z value must be 2.326 since P(Z < 2.326) = 0.99.
Then, (115.2652 - m)/(20.5/sqrt(12)) = 2.326
m = 115.2652 – 2.326(20.5)/sqrt(12) = 101.5 after rounding. The alternative mean would need to be $101.50
4) Ho: m = 0
Ha: m ¹ 0
Reject Ho if Z < -1.96 or Z > 1.96
Z = (0.02 – 0)/0.0832/sqrt(40) = 1.52
Do not reject Ho
Conclude there is no significant difference between this clock system and GMT.
5) Lower limit = 0.02 – 1.96(0.0832)/sqrt(40) = 0.02 – 0.0258 = -0.0058
Upper limit = 0.02 + 0.0258 = 0.0458
Since the hypothesized mean of 0 falls inside the 95% confidence interval, we do not reject Ho at a 5% level of significance.
6) p-value = 2P(Z > 1.52) = 2(0.0643) = 0.1286. Since the p-value is greater than 10%, there is no level of significance between 1% and 10% that could have been chosen in which the opposite conclusion would have been reached.
7) Accept Ho if –1.96 < Z < 1.96
Accept Ho if –1.96 < (xbar – 0)/(0.0832/sqrt(40)) < 1.96
Accept Ho if –1.96(0.0832)/sqrt(40) < xbar < 1.96(0.0832)/sqrt(40)
Accept Ho if –0.0258 < xbar < 0.0258
b = P(Accept Ho | Ha true)
b = P(–0.0258 < xbar < 0.0258 | m = 0.02)
b = P[(-0.0258 – 0.02)/(0.0832/sqrt(40)) < Z < (0.0258 – 0.02)/(0.0832/sqrt(40))]
b = P(-3.48 < Z < 0.44) = 0.4998 + 0.17 = 0.6698 = 66.98%
8) Ho: m ³ 20
Ha: m < 20
Since the sample size is 8, the degrees of freedom is 7. Reject Ho if p-value < 1%. Do not reject Ho if p-value > 10%.
t = (18.5 – 20)/6.3246/sqrt(8) = -0.6708
p-value = P(t < -0.6708). From the t table, we see that the closest critical value for 7 degrees of freedom is the 10% critical value of 1.415. This means P(t < -1.415) = 10%. Since our test statistic is to the right of –1.415, this means P(t < -0.6708) > 10%. Therefore we do not reject Ho.
We conclude people do not watch less than 20 hours of TV a week on average.
If we use KPK for the analysis, here is the output:
|
t Test for Population
Mean |
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Number of Observations |
8 |
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Sample Standard Deviation |
6.324555 |
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Sample Mean |
18.500000 |
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Ho:m ³ 20 |
Ha:m < 20 |
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T* |
-0.670820 |
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P[T £
T*] |
0.261922 |
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T Critical, a =
0.05 |
-1.894578 |
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95% CI for Pop. Mean |
13.212543 |
to |
23.787457 |
From the output, we see the exact p-value of 0.261922, confirming that the p-value is greater than 10%.
9) Ho: s = 10
Ha: s ¹ 10
As with the t test in question 5, the degrees of freedom is 7. Reject Ho if c2 <1.6899 or >16.0128.
c2 = 7(6.3246)2/(10)2 = 2.8
Do not reject Ho
Conclude there is no significant difference in the standard deviations of the group and the general population.
10) Lower limit = sqrt[7(6.3246)2/16.0128] = 4.1816
Upper limit = sqrt[7(6.3246)2/1.6899] = 12.8722
Since the hypothesized standard deviation of 10 falls inside the 95% confidence interval, we do not reject Ho at a 5% level of significance.
11) Since 6.3246 < 10, p-value = 2P(c2 < 2.8). From the chi-square table with 7 degrees of freedom, we see that 2.8 is between the 95% critical value of 2.1683 and the 90% critical value of 2.8331. This means P(c2 > 2.1683) = 0.95 and P(c2 > 2.8331) = 0.9. Conversely, this means P(c2 < 2.1683) = 0.05 and P(c2 < 2.8331) = 0.1. So, P(c2 < 2.8) is between 5% and 10% (although closer to 10% than 5%). By extension, the p-value = 2P(c2 < 2.8) is between 10% and 20%. Since the p-value is greater than 10%, we would not reject Ho under the general rule of thumb.
If we use KPK for the analysis, here is the output:
|
Chi-Square Test for Population
Standard Deviation |
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Number of Observations |
8 |
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Sample Standard Deviation |
6.324555 |
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Ho:s = 10 |
Ha:s ¹
10 |
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c2* |
2.8 |
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2 * P[c2 ³ ½c2*½] two tail |
0.194266 |
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Left tail c2 Critical, a =
0.05 |
1.689864 |
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Right tail c2 Critical, a = 0.05 |
16.012774 |
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95% CI for Pop. Standard Deviation |
4.181631 |
to |
12.872211 |
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95% CI for Pop. Variance |
17.486040 |
to |
165.693804 |
From the output, we see the exact p-value of 0.194266, confirming that the p-value is greater than 10%.
12) Ho: p £ 50%
Ha: p > 50%
Reject Ho if Z > 1.645
p-hat = 220/400 = 0.55
Z = (0.55 – 0.5)/sqrt(0.5*0.5/400) = 2
Reject Ho
Conclude the percentage of household with an income above $50,000 exceeds 50%.
13) Reject Ho if Z > 1.645
Reject Ho if (phat – 0.5)/sqrt(0.5(0.5)/400) > 1.645
Reject Ho if phat > 0.5 + 1.645sqrt(0.5(0.5)/400)
Reject Ho if phat > 0.5411
Power = P(reject Ho | Ha true)
Power = P(phat > 0.5411 | p = 0.55)
Power = P(Z > (0.5411 – 0.55)/sqrt(0.55(0.45)/400))
Power = P(Z > -0.36) = 0.5 + 0.1406 = 0.6406 = 64.06%
14) Lower limit = 0.55 – 1.96sqrt(0.55(0.45)/400) = 0.55 – 0.049 = 50.1%
Upper limit = 0.55 + 0.049 = 59.9%
If we use KPK for the analysis, here is the output:
|
Z Test for One
Proportion |
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Sample Proportion |
0.550000 |
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Number of Observations |
400 |
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Ho:p £ 0.5 |
Ha:p > 0.5 |
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Z* |
2.000000 |
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P[Z ³
Z*] |
0.022750 |
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Z Critical, a =
0.05 |
1.644853 |
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95% CI for Pop. Proportion |
0.501186 |
to |
0.598814 |