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 3 decimals.
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 5a
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 5b
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 5c
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 6a
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 6b
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 6c
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 6d
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 7a
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 7b
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 7c
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%