Transcript Class Notes

Section 7.2
Tables 7.2-1 and 7.2-2 summarize hypothesis tests about one mean.
(These tests assume that the sample is taken from a normal distribution,
but the tests are robust when the sample is taken from a non-normal
distribution with finite mean and variance.)
The relationship between hypothesis testing and confidence intervals
can be described by saying that
rejecting the hypothesized value of a parameter with significance
level  is expected to correspond to that hypothesized value not
being contained in the 100(1 – )% confidence interval,
not rejecting the hypothesized value of a parameter with significance
level  is expected to correspond to that hypothesized value being
contained in the 100(1 – )% confidence interval.
Confidence intervals have the following properties:
(1) For fixed sample size n, the length of the confidence interval tends
larger
to be __________
as the confidence level 100(1 – )% is increased,
smaller as
and the length of the confidence interval tends to be __________
the confidence level 100(1 – )% is decreased.
(2) For fixed confidence level 100(1 – )%, the length of the
confidence interval tends to be __________
smaller as the sample size n is
increased, and the length of the confidence interval tends to be
larger
__________
as the sample size n is decreased.
Hypothesis tests have the following properties:
(1) For fixed sample size n, as the significance level  is decreased, a
type II error probability  is __________,
increased and as the significance
level  is increased, a type II error probability  is __________.
decreased
(2) For fixed significance level , as the sample size n is increased, a
type II error probability  is __________,
decreased and as the sample size n
is decreased, a type II error probability  is __________.
increased
1. It is assumed that the cereal weight in a “10-ounce box” is N(, 2).
To test H0:  = 10.1 versus H1:  > 10.1 with a 0.05 significance level, a
random sample of size n = 16 boxes is selected where it is observed that
x = 10.4 and s = 0.4.
(a) Calculate the appropriate test statistic, describe the appropriate
critical region, and write a summary of the results.
x – 10.1
10.4 – 10.1
The test statistic is t = ——— = ————— = 3.000
s /n
0.4 /16
The one-sided critical region with  = 0.05 is t  1.753 .
Since t = 3.000 > t0.05(15) = 1.753, we reject H0. We conclude
that the mean weight of cereal per box is greater than 10.1
ounces.
(b) Find the approximate p-value of this hypothesis test.
The p-value of the test is
X – 10.1
P ———  3.000 ;  = 10.1
S /n
slightly less than 0.005.
a t(15) random variable
= P(T  3.000) is
2.
Use the Analyze > Compare Means > One-Sample T Test options
in SPSS to do Text Exercise 7.2-4.
x – 7.5
x – 7.5
(a) The test statistic is t = ——— = ———
s /n
s /10
The two-sided critical region with  = 0.05 is |t|  2.262 .
– 2.262
2.262 =
t0.025(9)
(b)
One-Sample Statistics
N
thick
10
Mean
7.5500
Std. Deviation
.10274
Std. Error
Mean
.03249
One-Sample Test
Test Value = 7.5
thick
t
1.539
df
9
Sig . (2-tailed)
.158
Mean
Difference
.05000
95% Confidence
Interval of the
Difference
Lower
Upper
-.0235
.1235
From the SPSS output, we find t = 1.539 .
Since t = 1.539 < t0.025(9) = 2.262, we fail to reject H0.
We conclude that the mean thickness for vending
machine spearmint gum is not different from 7.5
hundredths of an inch.
(c) From the SPSS output, we find that the limits of the
95% confidence interval for  are 7.4765 and 7.6235.
The hypothesized mean  = 7.5 is contained in these
limits, just as we would expect.
Note that to calculate the p-value of this test, we need to find
P
X – 7.5
———  1.539 ;  = 7.5
S /n
3.
Use the Analyze > Compare Means > One-Sample T Test options
in SPSS to do Text Exercise 7.2-6.
(a) H0:  = 3.4
(b) H1:  > 3.4
x – 3.4
x – 3.4
(c) The test statistic is t = ——— = ———
s /n
s /9
(d) The one-sided critical region with  = 0.05 is t  1.860 .
1.860 =
t0.05(8)
(e)
One-Sample Statistics
N
liters
Mean
3.5556
9
Std. Deviation
.16667
Std. Error
Mean
.05556
One-Sample Test
Test Value = 3.4
liters
t
2.800
df
8
Sig . (2-tailed)
.023
Mean
Difference
.15556
95% Confidence
Interval of the
Difference
Lower
Upper
.0274
.2837
From the SPSS output, we find t = 2.800 .
(f) Since t = 2.800 > t0.05(8) = 1.860, we reject H0.
We conclude that the mean FVC for the
volleyball players is greater than 3.4 liters.
3.-continued
(g) The p-value of the test is
X – 3.4
P ———  2.800 ;  = 3.4
S /n
a t(8) random variable
= P(T  2.800) is
between 0.01 and 0.025, from Table VI in Appendix B.
From the SPSS output, we find the p-value of the test to be
0.023 / 2 = 0.0115.
4.
Use the Analyze > Descriptive Statistics > Descriptives options in
SPSS to do Text Exercise 7.2-8. We shall perform this hypothesis
test about the mean under the assumption that the variance in weight
of home-born babies is 5252.
x – 3315
x – 3315
(a) The test statistic is z = ———— = ————
525 /n
525 /11
The one-sided critical region with  = 0.01 is z  2.326 .
(b)
Descriptive Statistics
N
grams
Valid N (listwise)
11
11
Minimum
2657.00
Maximum
3856.00
Mean
3385.9091
Std. Deviation
336.31606
From the SPSS output, we find x = 3385.91 .
x – 3315
z = ———— = 0.448
525 /11
Since z = 0.448 < z0.01 = 2.326, we fail to reject H0.
We conclude that the mean weight of home-born
babies is not greater than 3315 grams.
(c) The p-value of the test is
X – 3315
P ————  0.448 ;  = 3315 = P(Z  0.448) =
525 /n
1 – (0.448) = 1 – 0.6736 = 0.3264 .
Variance
113108.5
5.
Note that a hypothesis test concerning the mean difference between
two random variables measured on the same units is essentially a
one-sample test about a mean when the data consist of differences
between paired observations. Use the Analyze > Descriptive
Statistics > Paired-Samples T Test options in SPSS to do Text
Exercise 7.2-14.
Paired Samples Statistics
Pair
1
dist_a
dist_b
Mean
256.1765
251.4118
N
17
17
Std. Deviation
18.70239
14.93343
Std. Error
Mean
4.53600
3.62189
Paired Samples Test
Paired Differences
Pair 1
dist_a - dist_b
Mean
4.76471
Std. Deviation
9.08659
Std. Error
Mean
2.20382
95% Confidence
Interval of the
Difference
Lower
Upper
.09281
9.43660
d–0
d–0
The test statistic is t = —— = ———
sd /n
sd /17
t
2.162
df
16
Sig . (2-tailed)
.046
The one-sided critical region with  = 0.05 is t  1.746 .
From the SPSS output, we find t = 2.162 .
Since t = 2.162 > t0.05(16) = 1.746, we reject H0.
We conclude that the mean distance is greater
with brand A golf balls than with brand B.
The p-value of the test is
D–0
P ———  2.162 ; D = 0
SD /n
a t(16) random variable
= P(T  2.162) is
between 0.01 and 0.025, from Table VI in Appendix B.
From the SPSS output, we find the p-value of the test to be
0.046 / 2 = 0.023.