Download 8.5 Testing a Claim about a Mean.Standard Deviation Not Known

8.5-Testing a Claim about a Mean: Standard
Deviation Not Known:
Objectives:
1. Find P-values with a student t distribution.
2. Test a claim about a mean when the population standard deviation is NOT known.
Overview:
This section presents methods for testing a claim about a population mean when we do not know the
value of σ. The methods of this section use the Student t distribution introduced earlier.
Basic Methods of Testing Claims about a Population Mean:
The objective is to test a claim about a population mean using a formal method of hypothesis testing.
Notation:
n = sample size or number of trials
x = sample mean
µ x = population mean of all sample means from samples of size n
s = known value of the sample standard deviation
Requirements:
1) The sample is a simple random sample.
2) The value of the population standard deviation σ is NOT known.
3) Either or both of these conditions is satisfied: The population is normally distributed or n > 30.
Test Statistic for Testing a Claim about a Mean:
t=
x − µx
s
n
P-values and Critical Values:
Found in Table A-3
Degrees of freedom (df) = n – 1
Important Properties of the Student t Distribution:
1. The Student t distribution is different for different sample sizes (see the following graph,
for the cases n = 3 and n = 12).
2. The Student t distribution has the same general symmetric bell shape as the standard normal
distribution but it reflects the greater variability (with wider distributions) that is expected with
small samples.
3. The Student t distribution has a mean of t = 0 (just as the standard normal distribution has a mean
of z = 0).
4. The standard deviation of the Student t distribution varies with the sample size and is greater
than 1 (unlike the standard normal distribution, which has a s = 1).
5. As the sample size n gets larger, the Student t distribution gets closer to the normal distribution.
Conclusions:
Example: People have died in boat accidents because an obsolete estimate of the mean weight of men
was used. Using the weights of the simple random sample of men we obtain these sample statistics:
n = 40 and x = 172.55 lb, and s = 26.33 lb. Using
Us
a 0.05 significance level, test the claim that men have
a mean weight greater than 166.3 lb.
lb Use the Traditional Method.
Solution: Requirements are satisfied: simple random sample σ is not known and sample size
n = 40 (n > 30)
Step 1: Original claim is that men have a mean weight greater than 166.3 lb.
lb µ > 166 .3 .
Step 2: Opposite
pposite of original claim is µ ≤ 166 .3
Step 3: Because µ > 166 .3 does not contain equality it is H1. Therefore, H0: µ ≤ 166 .3 is the null
hypothesis and original claim and H1: µ > 166 .3 is the alternative hypothesis.
hypothesis
Step 4: Significance level is α = 0.05
0.
Step 5: The claim is about the population mean, so the relevant statistic is the sample mean x = 172.55
with s = 26.33 lbs and sample size n = 40.
Step 6: Calculate z:
t=
x − µ x 172.55 − 166.3
=
= 1.501
s
26.33
n
40
df = n – 1 = 39, area of 0.05, one--tail yields t = 1.685.
Step 7: Because the value of the test statistic t = 1.501 does not fall in the critical region bounded by
t = 1.685, we fail to reject the null hypothesis.
Step 8: Here is the correct conclusion: Because we fail to reject the null hypothesis, we conclude that
there is not sufficient evidence to support a conclusion that the population mean is greater than 166.3 lb
lb.
Example: Assume that a simple random sample has been selected from a normally distributed
population. Test the claim that for the population of female college students, the mean weight is given
by µ = 132 lbs. Sample data are summarized as n = 20, x = 137 lbs and s = 14.2 lbs. Use a significance
level of α = 0.1 .Use the Traditional Method.
Solution: Requirements are satisfied: simple random sample σ is not known and sample size
n = 20 is from a normally distributed population.
Step 1: Original claim is that the mean weight of a population of female college students is 132 lbs.
µ = 132 .
Step 2: Opposite of original claim is µ ≠ 132
Step 3: Because µ ≠ 132 does not contain equality it is H1. Therefore, H0: µ = 132 is the null
hypothesis and original claim and H1: µ ≠ 132 is the alternative hypothesis.
Step 4: Significance level is α = 0.1
Step 5: The claim is about the population mean, so the relevant statistic is the sample mean x = 137
with s = 14.2 lbs and sample size n = 20.
Step 6: Calculate z:
t=
x − µ x 137 − 132
=
= 1.575
s
14.2
n
20
df = n – 1 = 19, area of 0.1, two-tail yields t = 1.328.
Step 7: Because the value of the test statistic t = 1.575 does fall in the critical region bounded by
t = 1.328, we reject the null hypothesis.
Step 8: Here is the correct conclusion: Because we reject the null hypothesis, we conclude that there is
sufficient evidence to warrant rejection of the claim that the mean weight of a population of female
college students is 132 lbs.
Example: Assume that a simple random sample has been selected from a normally distributed
population. Test the claim that the mean age of the prison population in one city is less than 26 years.
Sample data are summarized as n = 25, x = 24.4 and s = 9.2 years. Use a significance level of α = 0.05 .
Use the Traditional Method.
Solution: Requirements are satisfied: simple random sample σ is not known and sample size
n = 25 is from a normally distributed population.
Step 1: Original claim is that the mean age of the prison population in one city is less than 26 years.
µ < 26 .
Step 2: Opposite of original claim is µ ≥ 26
Step 3: Because µ < 26 does not contain equality it is H1. Therefore, H0: µ ≥ 26 is the null
hypothesis and original claim and H1: µ < 26 is the alternative hypothesis.
Step 4: Significance level is α = 0.05
Step 5: The claim is about the population mean, so the relevant statistic is the sample mean x = 24.4
with s = 9.2 and sample size n = 25.
Step 6: Calculate z:
t=
x − µ x 24.4 − 26
=
= −0.870
s
9.2
n
25
df = n – 1 = 24, area of 0.05, one-tail yields t = 1.711 (or t = -1.711).
Step 7: Because the value of the test statistic t = -0.870 does not fall in the critical region bounded by
t = -1.685, we fail to reject the null hypothesis.
Step 8: Here is the correct conclusion: Because we fail to reject the null hypothesis, we conclude that
there is not sufficient evidence to support a conclusion that the prison population in one city is less than
26 years.