Transcript Section 10.4: Hypothesis Tests for a Population Mean

Section 10.4: Hypothesis
Tests for a Population Mean
Finding P-values for a t test
1. Upper-tailed Test:
1. Ha: μ > hypothesized value
2. P-value computed: area in the upper tail
2. Lower-tailed Test:
1. Ha: μ < hypothesized value
2. P-value computed: area in the lower tail
3. Two-tailed Test:
1. Ha: μ ≠ hypothesized value
2. P-value computed: sum of area in two tails
One-Sample t Test for a Population
Mean
Null Hypothesis: H0: μ = hypothesized value.
Test Statistic:
x  hypothesiz ed value
t
s
n
One-Sample t Test for a Population
Mean
Alternative Hypothesis:
P-value:
Ha: μ > hypothesized value
Area to right of calculated t
under t curve with df=n-1
Area to left of calculated t
under t curve with df=n-1
Ha: μ < hypothesized value
Ha: μ ≠ hypothesized value
(1) 2(area to right of t) if t is
positive, or
(2) 2(area to left of t) if t is
negative
•
Assumptions:
1. x bar and s are the sample mean and
sample standard deviation, respectively,
from a random sample
2. The sample size is large (generally greater
than 30) or the population distribution is at
least approximately normal.
Example
• A study conducted by researchers investigated
whether time perception, a simple indication of a
person’s ability to concentrate, is impaired
during nicotine withdrawal. After a 24-hr smoking
abstinence, 20 smokers were asked to estimate
how much time had passed during a 45-sec
period. Suppose the resulting data on perceived
elapsed time (in seconds) were as shown (these
data are artificial but are consistent with
summary quantities given in the paper:
69 65 72 73 59 55 39 52 67 57
56 50 70 47 56 45 70 64 67 53
From these data, we obtain
n = 20
x bar = 59.30 s = 9.84
• The researchers wanted to determine
whether smoking abstinence had a
negative impact on time perception,
causing elapsed time to be overestimated.
We can answer this question by testing
H0: μ = 45 (no consistent tendency to
overestimate the time elapsed)
Versus
Ha: μ > 45 (tendency for elapsed time to be
overestimated)
• The null hypothesis is rejected only if there
is convincing evidence that μ > 45
• To answer this question, we carry out a
hypothesis test with a significance level of
.05 using the step-by-step procedure
described in 10.3
1. Population Characteristic of Interest:
μ = mean perceived elapsed time for smokers
who have abstained from smoking for 24 hrs
2. Null Hypothesis: H0: μ = 45
3. Alternative Hypothesis: Ha: μ > 45
4. Significance Level: α = .05
x  hypothesiz ed value x  45
5. Test Statistic : t 

s
s
n
n
6. Assumptions: This test requires a random sample and either a large
sample size or a normal population distribution. The authors of the
paper believed that it was reasonable to consider this sample as
representative of smokers in general, so we regard it as if it were a
random sample. Because the sample size is only 20, for the t test to
be appropriate, we must be able to assume that the population
distribution of perceived elapsed times is at least approximately
normal. Is this reasonable? A boxplot shows that there are no
outliers so a t test is reasonable.
7. Computatio ns : n  20, x  59.30, and s  9.84. Therefore,
59.30  45 14.30
t

 6.50
9.84
2.20
20
8. P-value: This is an upper-tailed test (greater than) so the
P-value is the area to the right of the computed t value.
Because df = 20 – 1 = 19, we can use the df = 19
column of the table to find the p-value. Since it is so
high, we find the p-value is ≈ 0.
9. Conclusion: Because p-value ≤ α, we
reject H0 at the .05 level of significance.
There is virtually no chance of seeing a
sample mean (and hence a t value) this
extreme as a result of just chance
variation when H0 is true. There is
convincing evidence that the mean
perceived time elapsed is greater than the
actual time elapsed of 45 sec.