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8.32.
a. The hypothesis of interest is two-tailed:
H0 : µ 1 - µ 2 = 0 Ha : µ 1 - µ 2 0
The observed value of the test statistic is
( x − x 2 ) − ( µ1 − µ 2 ) 78,100 − 82,700
z= 1 = = −3.807
s12 s 2 2
6300 2 7100 2
+ +
n1 n2 57 66
The p-value is 2 P(Z > 3.807 ) ≈ 2(0) = 0
Since the p-value is smaller than the significance level 0.1, we can reject H0.
There is sufficient evidence to indicate that the means differ from April to May.
b. In this exercise, H0 would be rejected in either case.
8.33.
The hypothesis of interest is one-tailed:
H0 : µ 1 - µ 2 = 0 Ha : µ 1 - µ 2 < 0
since increased scores imply µ 2 > µ 1.
9(.95) 2 + 9(.56) 2
The observed value of the test statistic with s = 2
= .60805 , is
18
( x − x2 ) − 0 6.82 − 8.17
t= 1 = = −3.871
2 1 1 1 1
s ( + ) .60805( +
n1 n2 10 10
The p-value is P(t > 3.871) < 0.005 for 9 degrees of freedom.
Since the p-value is less than the significance level of 0.05, we can reject H0.
The training course was effective in increasing customer service scores.
8.44.
The hypothesis of interest is
H0 : µ A - µ B = 0 or H0 : µ d = 0
Ha : µ A - µ B > 0 or Ha : µ d > 0
where µ1 is the mean for the current year and µ 2 is the mean for last year. The differences, along
with necessary calculations, are given below:
di .54 -.94 -.23 .30 .28 .76
di .71
d = = = .118333
n 6
( di )2 (.71) 2
di −
2
1.9741 −
sd =
2 n = 6 = .3780167
n −1 5
the test statistic is
d − µd .118333 − 0
t= = = .471
sd .3780167
n 6
The p-value is P (t > .471) > 0.1 for 5 degrees of freedom.
Since the p-value is greater than the significance level of 0.05, we cannot reject Ho.
That is, the data does not indicate that the mean has increased.
8.81.
The parameter of interest is µ , the average daily wage of workers in a given industry. A sample of
n = 40 workers has been drawn from a particular company within this industry and x , the sample
average, has been calculated. The objective is to determine whether this company pays inferior
wages in comparison to the total industry. That is, assume that this sample of forty workers has
been drawn from a hypothetical population of workers. Does this population have as an average
wage µ =23.20, or is µ less than 23.20? Thus, the hypothesis to be tested is
H0 : µ = 23.20 Ha : µ < 23.20
The test statistic is
x − µ 21.20 − 23.20
z≈ = = −2.811
s 4 .5
n 40
and the p-value is
p-value = P[ z < -2.81 ] =.5 - .4975 = .0025
Since = .01 is larger than the p-value, .0025, H0 can be rejected.
We conclude that the company is paying inferior wages.
8.86.
The hypothesis to be tested is
H0 : µ = .05 Ha : µ > .05
and the test statistic is
x − µ .058 − .05
t= = = 2.108
s .012
n 10
The p-value is P(t > 2.108) < 0.05 for 9 degrees of freedom.
Since the p-value is smaller than the significance level of 0.05, we can reject Ho.
We conclude that µ is greater than .05.
The observed level of significance is precisely the p-value = P[ t > 2.108 ], which is between .025
and .05 since t = 2.108 falls between t.025 and t.05.
8.93.
a. The hypothesis of interest is two-tailed:
H0 : µ 1 - µ 2 = 0 Ha : µ 1 - µ 2 0
and the pooled estimate of σ is calculated as
2
(n − 1) s12 − (n2 − 1) s 2 9(16.36) + 9(18.92)
2
s2 = 1 = = 17.64
n1 + n2 − 2 18
The test statistic is then
( x − x2 ) − 0 22.2 − 28.5
t= 1 = = −3.354
2 1 1 1 1
s ( + ) 17.64( + )
n1 n2 10 10
The p-value is: 2 P(t > 3.354) < 2(0.005) = 0.01 for 18 degrees of freedom.
Since the p-value is smaller than the significance level of 0.1, we can reject Ho.
We conclude that there is a difference in the mean servicing times between the two employees.
b. The observed level of significance or the p-value is
p-value = 2P[ t > 3.354 ]
for a two-tailed test with 18 degree of freedom. Since t = 3.354 exceeds the largest tabulated value,
t.005 = 2.878, we have p-value < 2(.005) = .01. (as above)
8.94
The 95% confidence interval is
1 1
( x1 − x 2 ) ± t .025 s 2 ( + )
n1 n2
1 1
(22.2 − 28.5) ± 2.101 17.64( + )
10 10
-6.3 ± 3.946 or -10.246 < ( µ1 − µ 2 ) < -2.354
The interval clearly does not contain 0, which confirms our conclusion above that the mean service
times are not equal.
8.98.
The samples are not independent because the paired reactions for a single person will not be
independent (we are performing the measurements on the same people). However, if the difference
between processing times for the two processes is calculated, an independent random sample of
differences is generated and a paired-difference analysis can be conducted. The hypothesis if
interest is
H0 : µ 2 - µ 1 = 0 or H0 : µ d = 0
Ha : µ 2 - µ 1 0 or Ha : µ d 0
The table of differences, x 2 − x1 , along with the calculation of d and s d , is presented below.
2
di 1 1 2 -1 1 1 0 2
di 7
d = == .875
n 8
( di )2 ( 7) 2
di −
2
13 −
sd =
2 n = 8 = .9821
n −1 7
the test statistic is
d − µd .875 − 0
t= = = 2.497
sd .9821
n 8
The p-value is: 2 P(t > 2.497) < 2(0.025) = 0.05
Since the p-value is smaller than the significance level of 0.05, we can reject Ho.
We can conclude that there is a difference in means between the two processes.
The observed value of t, t = 2.497, falls between t.025 = 2.365 and t.01 = 2.998, so that
2(.01) < p-value < 2(.025) .02 < p-value < .05
8.100.
The hypothesis to be tested is
H0 : µ = 48,000 Ha : µ > 48,000
and the test statistic is
x − µ 51,102 − 48,000
t= = = 2.096
s 5127
n 12
The p-value is P(t > 2.096) < 0.05 for 11 degrees of freedom.
Since the p-value is smaller than the significance level of 0.05, we can reject Ho.
There is evidence in the sample that water consumption has increased.
8.101.
A paired-difference analysis is conducted. The hypothesis of interest is
H0 : µ A - µ B = 0 or H0 : µ d = 0
Ha : µ A - µ B 0 or Ha : µ d 0
The table of differences, x A − x B , along with the calculation of d and s d , is presented below.
2
di 1 2 1 1 0 1 2 2 1 1
di 12
d = = = 1 .2
n 10
( di )2 (12) 2
di −
2
18 −
sd =
2 n = 10 = .4
n −1 9
the test statistic is
d − µ d 1.2 − 0
t= = = 6.00
s .4
n 10
The p-value is 2 P(t > 6) ≈ 2(0) = 0
Since the p-value is smaller than the significance level of 0.05, we can reject Ho.
We can conclude that there is a difference in mean scores for the two methods.
b. A 98% confidence interval for µ A − µ B = µ d is
sd .4
d ± t .01 1.2 ± 2.821 1.2 ± .564
n 10
or .636 < ( µ A − µ B ) <1.764.
Again, the conclusion above is confirmed by the fact that 0 does not belong to this interval.
8.126.
a. The hypothesis to be tested is
H0 : µ 1 - µ 2 = 0 Ha : µ 1 - µ 2 0
and the pooled estimate of σ is calculated as
2
(n1 − 1) s12 − (n2 − 1) s 2 19(4.28) + 19(3.89)
2
s =
2
= = 4.085
n1 + n2 − 2 38
the test statistics is then
( x − x2 ) − 0 43.1 − 44.6
t= 1 = = −2.347
2 1 1 2
s ( + ) 4.085( )
n1 n2 20
The p-value is 2 P(t > 2.347) < 2(0.01) = 0.02 for n1 + n2 − 2 = 38 degrees of freedom.
Since the p-value is smaller than the significance level of 0.05, we can reject Ho.
We conclude that there is a difference in mean productivity for the two work schedules.
b. The observed level of significance is the p-value: p-value = 2P[ t > 2.347 ] for 38 degree of
freedom. Since t = 2.347 lies between t.005 = 2.576, and t.01 = 2.326, we have
2(.005) < p-value < 2(.01) or .01 < p-value < .02, as above.
c. The 95% confidence interval is
1 1
( x1 − x 2 ) ± t .025 s 2 ( + )
n1 n2
1 1
(43.1 − 44.6) ± 1.96 4.085( + )
20 20
-1.5 ± 1.253 or -2.753 < ( ( µ1 − µ 2 ) < -.247
Same conclusion as above.
