STAT500 HW#9 Solution
1)
n = 378, p0 = 0.108; because n*p0 = 278*(0.108) = 30.024> 5, n*(1 - p0) = 378(1 0.108) = 247.976 > 5, thus the one-proportion z-test can be used.
Ho: π  0.108; Ha: π < 0.108
α = 0.01; one-sided left-tailed test
ˆ = 15/278=0.054
0.054-0.108
Test statistic z =
= -2.901 < - z0.01 = -2.326
(0.108*0.892)/278
Reject Ho (since z < -2.326). There is sufficient evidence to conclude that in 1995,
families whose household had a Bachelor degree or more had a lower percentage
earning incomes below the poverty level than national percentage of 10.8%.
2)
p0 = 0.791; because n*p0 = 200*(0.791) = 158.2 > 5, n*(1 - p0) = 200*(1 - 0.791)
= 41.8 > 5, thus the one-proportion z-test can be used.
Ho: π = 0.791; Ha: π  0.791
α = 0.02; two-sided test
ˆ = 151/200 = 0.76
0.76-0.791
Test statistic z =
= -1.078
(0.791*0.209)/200
P-value= 2*P (z > |-1.078|) = 2*0.1405 = 0.281 >  =0.05
We fail to reject Ha (since p-value >  = 0.05). Hence, there is insufficient
evidence to conclude that this year’s percentage of home buyers purchasing
single-family houses is different from the 1995 figure of 79.1%.
3)
The probability plot suggests that the population may deviate slightly from a Normal
distribution, but not too much. Hence, one can still use a t-distribution since the
1
sample size is 29, not far from 30. Note that we cannot reject the null which is
observed values are normally distributed since p-value (=0.069) is larger than 0.05.
Note that even though the alpha level set for the testing mean problem is at α = 0.01,
the alpha level we need to compare to is the default 0.05. Since here we are checking
whether the sample may come from a normal distribution. The alpha level is not
impacted by the alpha level set for testing mean problem.
Ho: µ ≥ 30 vs. Ha: µ< 30
The Minitab output is summarized in the table below. Since p-value = 0.000 < α =
0.01, the data provides sufficient evidence to reject the null hypothesis. Therefore, we
should conclude that the mean value of the amount of weight lost in a program during
six months should be less than 30 pounds.
One-Sample T: PoundsLost
Test of μ = 30 vs < 30
99% Upper
Variable
N
Mean
PoundsLost 29 19.48
StDev SE Mean
12.83
Bound
2.38
25.36
T
P
-4.41 0.000
4)
(Note: it is also okay to use z table to compute the p-value since n=33)
Ho: µ> 20 vs. Ha: µ<20
y   0 16.2  20
t

 - 2.695
s/ n
8.1 / 33
The rejection region for this test would be a t-value less than t0.05, df=32=-1.69. Since
the calculated t-value is -2.695, we reject the null hypothesis, and thus there is
sufficient data to conclude that the population of US commercial jets has a mean age
less than 20 years.
5)
(Note: it is NOT okay to use z table since n=26)
Ho: µ= 29 vs. Ha: µ≠29; two-tailed test, α=0.02,
t=
y - u0 33.02-29
=
= 2.512
s / n 8.16/ 26
2
Since t0.02/2, df=25 = 2.485. The rejection region for this problem would be |t| >
t0.02/2,df=25 = 2.485. Since the observed t-value is 2.512, we reject Ho. We conclude that
there is sufficient evidence to doubt the supervisor’s claim.
6)
(Note: it is also okay to use z table to compute the p-value since n=30)
With n=30, there is no need to check the probability plot since large sample size
implies that we can use the one sample t-test.
Ho: µ= 675 vs. Ha: µ > 675; two-tailed test
y - m 712.0-675
=
t=
= 2.35 > t0.05,29=1.699 (rejection region)
s / n 86.4 / 30
P-value= P (t > 2.35) =0.0123< α = 0.05.Therefore we can reject Ho and conclude Ha.
Minitab output to verify above by hand calculations:
One-Sample T: Voters
Test of μ = 675 vs > 675
Variable
Voters
N
Mean StDev SE Mean 95% Lower Bound
30 712.0
86.4
15.8
685.2
7)
Ho:  =48
Ha: m > 48
Difference = 1
α = 0.01, s = 1.5, Power = 80%
Power and Sample Size
1-Sample t Test
Testing mean = null (versus > null)
Calculating power for mean = null + difference
α = 0.01 Assumed standard deviation = 1.5
Sample Target
Difference
1
Size
Power Actual Power
26
0.8
0.812333
Thus, 26 bottles of detergent are needed.
3
T
2.35
P
0.013