MGMT2263 Homework sheet 1 solutions

MGMT2263

Worksheet #1 Solutions

Question 1a
Ho: m > 300
Ha: m < 300
Reject the null hypothesis if the test statistic is less than -1.645
Z = (283 - 300)/58/sqrt(60) = -2.27
Reject the null hypothesis.
Conclude the site receives fewer than 300 hits per day on average.

Question 1b
p-value = P(Z < -2.27) = 0.5 - 0.4884 = 0.0116. If the level of significance were 1%, the p-value would be greater than that and we would not reject Ho.

Question 1c

Reject Ho if Z < -1.645

Reject Ho if (xbar – 300)/(58/sqrt(60)) < -1.645

Reject Ho if xbar < 300 – 1.645(58)/sqrt(60)

Reject Ho if xbar < 287.6826

Power = P(reject Ho | Ha true)

Power = P(xbar < 287.6826 | m = 283)

Power = P[Z < (287.6826 – 283)/(58/sqrt(60))]

Power = P(Z < 0.63) = 0.5 + 0.2357 = 0.7357

Question 1d

We know we reject Ho if xbar < 287.6826. In order for the power to be 99%, the Z value must be 2.326 since P(Z < 2.326) = 0.99. Then 2.326 = (287.6826 - m)/58/sqrt(60)) from which we get

m = 287.6826 – 2.326(58)/sqrt(60) = 270.3 after rounding to 1 decimal.

Question 2a
Lower limit = 625.4 - 1.645(20)/sqrt(40) = 625.4 - 5.2 = 620.2
Upper limit = 625.4 + 5.2 = 630.6
With 90% confidence, the average chips life is between 620.2 and 630.6 hours.

Question 2b
The null and alternative hypotheses are:
Ho: m = 633
Ha: m ¹ 633
Since the hypothesized mean of 633 does not fall inside the 90% confidence interval, we would reject Ho at a 10% level of significance.

Question 2c
Z = (625.4 - 633)/20/sqrt(40) = -2.40
p-value = 2P(Z > 2.4) = 2(0.0082) = 0.0164. So, 1.64% is the highest level of significance we could have chosen in which we would not reject the null hypothesis.

Question 2d

Accept Ho if –1.645 < Z < 1.645

Accept Ho if –1.645 < (xbar – 633)/(20/sqrt(40)) < 1.645

Accept Ho if 633 – 1.645(20)/sqrt(40) < xbar < 633 + 1.645(20)/sqrt(40)

Accept Ho if 627.7981 < xbar < 638.2019

b = P(accept Ho | Ha true)

b = P(627.7981 < xbar < 638.2019 | m = 625)

b = P[(627.7981 – 625)/(20/sqrt(40)) < Z < (638.2019 – 625)/(20/sqrt(40))]

b = P(0.88 < Z < 4.17) = 0.5 – 0.3106 = 0.1894

Power = 1 – 0.1894 = 0.8106 = 81.06%

Question 3a
Ho: m < 175
Ha: m > 175
Reject the null hypothesis if the test statistic is greater than 1.88
Z = (182 - 175)/28/sqrt(70) = 2.09
Reject the null hypothesis.
Conclude the average student weighs significantly more than 175 pounds.

Question 3b
Lower limit = 182 - 1.96(28)/sqrt(70) = 182 - 6.56 = 175.44
Upper limit = 182 + 6.56 = 188.56
With 95% confidence, the average student weighs between 175.44 and 188.56 pounds.

Question 3c

Reject Ho if Z > 1.88

Reject Ho if (xbar – 175)/(28/sqrt(70)) > 1.88

Reject Ho if xbar > 175 + 1.88(28)/sqrt(70)

Reject Ho if xbar > 181.2917

Power = P(reject Ho | Ha true)

Power = P(xbar > 181.2917 | m = 182)

Power = P[Z > (181.2917 – 182)/(28/sqrt(70))]

Power = P(Z > -0.21) = 0.5 + 0.0832 = 0.5832 = 58.32%

Question 3d

Reject Ho if Z > 1.282

Reject Ho if (xbar – 175)/(28/sqrt(70)) > 1.282

Reject Ho if xbar > 175 + 1.282(28)/sqrt(70)

Reject Ho if xbar > 179.2904

Power = P(reject Ho | Ha true)

Power = P(xbar > 179.2904 | m = 182)

Power = P[Z > (179.2904 – 182)/(28/sqrt(70))]

Power = P(Z > -0.81) = 0.5 + 0.2910 = 0.7910 = 79.1%

Question 4a
Ho: m = 16
Ha: m ¹ 16
We need t with 9 degrees of freedom. So the critical value is 2.262
t = (15.89 - 16)/0.3071/sqrt(10) = -1.1326
Do not reject the null hypothesis.
Conclude the machine does not need adjusting.

Question 4b
If the level of significance were 10%, the critical value would be 1.833. However, the test statistic would still be inside the acceptance region, even at a 10% level of significance. That means the p-value is greater than 10%. So, there is no level of significance between 1% and 10% that could have been chosen in which the opposite conclusion would be reached.

Question 5

Ho: median £ 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- £ 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 6a
Ho: s = 13
Ha: s ¹ 13
The degrees of freedom is 24. Reject Ho if the test statistic is either less than 12.4012 or greater than 39.3641.
test stat = 24(9.8)²/13² = 13.6388
Do not reject the null hypothesis.
Conclude the standard deviation has not significantly changed from 13%.

Question 6b
Lower limit = sqrt[24(9.8)²/39.3641] = 7.6521
Upper limit = sqrt[24(9.8)²/12.4012] = 13.6333
Since the hypothesized standard deviation of 13 falls inside the 95% confidence interval, we would not reject the null hypothesis at a 5% level of significance.

Question 6c
If we were to test at a 10% level of significance, the left-hand critical value changes to 13.8484. In this case, the test statistic of 13.6388 falls inside the rejection region.

Question 7a
Ho: p < 0.5
Ha: p > 0.5
Reject the null hypothesis if the test statistic is greater than 1.645.
Z = (40/75 - 0.5)/sqrt(0.5(0.5)/75) = 0.58
Do not reject the null hypothesis.
Conclude that not more than half of claims are due to speeding.

Question 7b
p-value = P(Z > 0.58) = 0.281. Since this is greater than 10%, we would not reject the null hypothesis under the general rule of thumb.

Question 7c
Lower limit = 40/75 - 1.96sqrt(40/75)(35/75)/75) = 0.5333 - 0.1129 = 0.4204
Upper limit = 0.5333 + 0.1129 = 0.6462
With 95% confidence, the average percentage of claims associated with speeding ranges from 42.04% to 64.62%

Question 7d

Accept Ho if Z < 1.645

Accept Ho if (phat – 0.5)/sqrt(0.5(0.5)/75) < 1.645

Accept Ho if phat < 0.5 + 1.645sqrt(0.5(0.5)/75)

Accept Ho if phat < 0.595

b = P(accept Ho | Ha true)

b = P(phat < 0.595 | p = 0.53)

b = P[Z < (0.595 – 0.53)/sqrt(0.53(0.47)/75)]

b = P(Z < 1.13) = 0.5 + 0.3708 = 0.8708 = 87.08%

Question 8a

Ho: p = 0.25

Ha: p ¹ 0.25

Reject Ho if Z < -2.055 or Z > 2.055

Z = (0.295 – 0.25)/sqrt(0.25(0.75)/800)) = 2.94

Reject Ho

Conclude the percentage of executives using this PDA has significantly changed from 25%.

Question 8b

P-value = 2P(Z > 2.94)

P(Z > 2.94) = 0.5 – 0.4984 = 0.0016

P-value = 2(0.0016) = 0.0032 = 0.32%

Since the p-value is less than 1%, we reject Ho under the general rule of thumb.

Question 8c

Accept Ho if –2.055 < Z < 2.055

Accept Ho if –2.055 < (phat – 0.25)/sqrt(0.25(0.75)/800) < 2.055

Accept Ho if 0.25 – 2.055 sqrt(0.25(0.75)/800) < phat < 0.25 + 2.055 sqrt(0.25(0.75)/800)

Accept Ho if 0.2185 < phat < 0.2815

b = P(accept Ho | Ha true)

b = P(0.2185 < phat < 0.2815 | p = 0.3)

b = P[(0.2185 – 0.3)/sqrt(0.3(0.7)/800) < Z < (0.2815 – 0.3)/sqrt(0.3(0.7)/800)]

b = P(-5.03 < Z < -1.14) = 0.5 – 0.3729 = 0.1271

Power = 1 – 0.1271 = 0.8729 = 87.29%

From the textbook

8.32 page 252

Ho: m > 30
Ha: m < 30

Z = (29.17 – 30)/(4/sqrt(100)) = -2.08

P-value = P(Z < -2.08) = 0.5 – 0.4812 = 0.0188 = 1.88%. The p-value of 1.88% is less than α = 5%. Reject Ho and conclude the average service time is less than 30 seconds.

8.47 page 257

Part a

Ho: m = 4
Ha: m ¹ 4

Z = (3.3 – 4)/(0.7/sqrt(800)) = -27.8859

P-value = 2P(Z > 27.8859) = 0. Under the general rule of thumb, the p-value is less than 1%. Reject Ho and conclude the average number of visits is significantly different than 4. Since the sample mean is less than 4 and the sample mean is an unbiased estimator of the population mean, we estimate m < 4.

Part b

Ho: m = 4
Ha: m ¹ 4

Z = (4.3 – 4)/(0.66/sqrt(500)) = 10.1639

P-value = 2P(Z > 10.1639) = 0. Under the general rule of thumb, the p-value is less than 1%. Reject Ho and conclude the average number of visits is significantly different than 4. Since the sample mean is greater than 4 and the sample mean is an unbiased estimator of the population mean, we estimate m > 4.

8.52 page 260

Part a

Ho: m < 3.5
Ha: m > 3.5

Part b

t = (6 – 3.5)/(1.8257/sqrt(7)) = 3.6229

The degrees of freedom are 6. The 1% critical value is 3.143 and the 0.5% critical value is 3.707. Therefore, 0.5% < p-value < 1%. For levels of significance of 10%, 5% and 1%, reject Ho. Conclude the mean bad debt ratio is significantly higher for Ontario. Only for the 0.1% level of significance would we not reject Ho. There is strong evidence to support the banking official’s claim.

8.73 page 266

Part a

Ho: p > 95%
Ha: p < 95%

Part b

Z = (0.79 – 0.95)/sqrt(0.95(0.05)/400) = -14.68

P-value = P(Z < -14.68) = 0. Regardless of the level of significance reject Ho. There is very strong evidence that the manufacturer’s claim is false.

Part c

Suppose that the company sells 1 million sets per year and the average repair cost under warranty is $250.

5% repair rate -> 50,000 sets x $250 = $12.5 million

21% repair rate -> 210,000 sets x $250 = $52.5 million

This is a difference of $40 million. I would suspect most people would consider this a matter of practical importance.