STAT217
Tutorial sheet #2 Solutions
Question 1
Ho: m £ 10
Ha: m > 10
Reject Ho
if Z > 2.3263
Z = (11.25
– 10)/(3/sqrt(40)) = 2.6352
Reject Ho.
Conclude
the average setup time is significantly greater than 10 minutes. They should
not buy the equipment.
Question 2
P-value = P(Z > 2.6352) = 1 – P(Z < 2.6352) = 1 – 0.9958 =
0.0042. Since the p-value is less than 1%, this provides strong support for Ha
and we reject Ho.
Question 3
Reject Ho
if Z > 2.3263
Reject Ho
if (xbar – 10)/(3/sqrt(40)) > 2.3263
Reject Ho
if xbar > 10 + 2.3263(3)/sqrt(40)
Reject Ho
if xbar > 11.1035
Power = P(reject Ho | Ha true)
Power = P(xbar > 11.1035 | m = 12)
Power = P(Z > (11.1035 – 12)/(3/sqrt(40)))
Power = P(Z > -1.89) = 1 – P(Z < -1.89) = 1 – 0.0294 = 0.9706.
Based on
the power, it appears management made the correct decision to not buy the
equipment.
Question 4
Ho: m ³ 22
Ha: m < 22
Reject Ho
if t < -1.7613
t = (20.75
– 22)/(3.4/sqrt(15)) =
-1.4239
Do not
reject Ho.
Conclude the
average battery life is not significantly less than 22 hours.
Question 5
P-value = P(t < -1.4239) = 0.0882 based on 14 degrees of freedom.
Since we did not reject Ho in question 4, we would reject Ho for levels of
significance between 8.82% and 10%.
Question 6
Ho: p £ 35%
Ha: p > 35%
Reject Ho if Z > 1.6449
p-hat = 84/200 = 0.42
Z = (0.42 – 0.35)/sqrt(0.35*0.65/200) = 2.0755
Reject Ho
Conclude that the percentage of Ipod users in the 18-34 age range has significantly increased.
Question 7
P-value = P(Z > 2.0755) = 1 – P(Z < 2.0755) = 1 – 0.981 = 0.019.
Since this p-value is greater than 1%, we would have not have rejected Ho.
Question 8
Reject Ho
if Z > 1.6449
Reject Ho
if (phat – 0.35)/sqrt(0.35(0.65)/200) >
1.6449
Reject Ho
if phat > 0.35 + 1.6449 sqrt(0.35(0.65)/200)
Reject Ho
if phat > 0.4055
Power = P(reject Ho | Ha true)
Power = P(phat > 0.4055 | p = 0.42)
Power = P(Z > (0.4055 – 0.42)/sqrt(0.42(0.58)/200))
= P(Z > -0.4155) = 1 – P(Z < -0.4155) = 1 – 0.3389 = 0.6611
Question 9
Ho: p = 0.5
Ha: p ¹ 0.5
Reject Ho
if Z < -1.96 or > 1.96
Z = (0.55 –
0.5)/sqrt(0.5(0.5)/400)
= 2
Reject Ho.
Conclude
the percentage of men watching HNIC is significantly different than 50%. Thus,
the report is not true.
Question 10
P-value = 2P(Z > 2) = 2[1 – P(Z < 2)] = 2[1 – 0.9773] = 0.0454. Since we rejected Ho, we would not
reject Ho for levels of significance between 1% and 4.54%.
Question 11
Since this
is a 2-tail test, we will compute Beta and then the power.
Accept Ho
if –1.96 < Z < 1.96
Accept Ho
if –1.96 < (phat – 0.5)/sqrt(0.5(0.5)/400) < 1.96
0.5 – 1.96 sqrt(0.5(0.5)/400)
< phat < 0.5 + 1.96 sqrt(0.5(0.5)/400)
0.451 < phat < 0.549
B = P(accept Ho | Ha true)
B = P(0.451 < phat < 0.549 | p =
0.6)
B = P((0.451 – 0.6)/sqrt(0.6(0.4)/400)
< Z < (0.549 – 0.6)/sqrt(0.6(0.4)/400))
B = P(-6.0829 < Z < -2.0821) = P(Z < -2.0821) = 0.0187
Power = 1 –
0.0187 = 0.9813
Question 12
Ho: m ³ 20
Ha: m < 20
Since the sample size is 8, the degrees of freedom is 7. Reject Ho if t < -1.8946.
t = (18.5 –
20)/6.3246/sqrt(8) = -0.6708
Do not
reject Ho.
Conclude
that people do not watch significantly less than 20 hours of TV a week.
Question 13
P-value = P(t < -0.6708) = 0.2619 based on 7 degrees of freedom.
Since the p-value is greater than 10%, we do not reject Ho under the general
rule of thumb.
Question 14
Ho: m = 0
Ha: m ¹ 0
Reject Ho if t < -2.5706 or t > 2.5706 based on 5 degrees of freedom.
t = (0.005317 – 0)/(0.022951/sqrt(6)) = 0.5674.
Do not
reject Ho. Conclude this clock system is not significantly different than GMT.
Question 15
Lower limit = 0.005317 – 2.5706(0.02295)/sqrt(6) = 0.005317 – 0.02409 = -0.02
Upper limit = 0.005317 + 0.02409 = 0.03
With 95% confidence, the average difference between the clock and GMT is between
–0.02 and +
0.03 seconds. We would reach the same conclusion at a 5% level of significance
since the hypothesis mean of zero falls inside the 95% confidence interval.
Question 16
P-value = 2P(t > 0.5674)
P(t >
0.5674) = 1 – P(t < 0.5674) = 1 – 0.7025 = 0.2975. P-value = 2(0.2975) =
0.595. Since the p-value is greater than 10%, we do not reject Ho under the
general rule of thumb.
Question 17
Ho: s £ 0.75
Ha: s > 0.75
Reject Ho if c2 > 36.415
c2 = 24(0.9)2/(0.75)2 = 34.56
Do not reject Ho
Conclude
that there is not significantly less consistency for this study.
Question 18
P-value = P(c2 > 34.56) = 1 - P(c2 < 34.56) = 0.0752. Since we did not reject
Ho, we would reject Ho for levels of significance between 7.52% and 10%.
Question 19
Ho: s = 3.50
Ha: s ¹ 3.50
Reject Ho
if c2 < 2.7004 or c2 > 19.0228
c2 = 9(7.1081)2/3.52 =
37.1207
Reject Ho
Conclude
there is a significant change in the variability of oil prices.
Question 20
Lower limit for s2 = 9(7.1081)2/19.0228 = 23.9044
Upper limit for s2 = 9(7.1081)2/2.7004 = 168.3929
Lower limit for s = sqrt(23.9044) = 4.8892
Upper limit for s = sqrt(168.3929) = 12.9766
With 95%
confidence, the standard deviation of oil prices is between $4.89 and $12.98.
We would
reject Ho at a 5% level of significance since the hypothesis standard deviation
of 3.50 falls outside the 95% confidence interval.