Chapter

© 2010 Pearson Prentice Hall. All rights reserved

Chapter

Hypothesis Tests Regarding a Parameter

10

© 2010 Pearson Prentice Hall. All rights reserved

Section

Hypothesis Tests for a Population Mean- Population Standard Deviation Unknown

10.3

© 2010 Pearson Prentice Hall. All rights reserved 10-3

Objectives

1. Test the hypotheses about a population mean with unknown


To test hypotheses regarding the population mean assuming the population standard deviation is unknown, we use the t-distribution rather than the Z-distribution. When we replace with s,

follows Student’s t-distribution with n-1 degrees of freedom.

x s

n


1. The t-distribution is different for different degrees of freedom.

Properties of the t-Distribution



2. The t-distribution is centered at 0 and is symmetric about 0.




2. The t-distribution is centered at 0 and is symmetric about 0.

3. The area under the curve is 1. Because of the symmetry, the area under the curve to the right of 0 equals the area under the curve to the left of 0 equals 1/2.



4. As t increases (or decreases) without bound, the graph approaches, but never equals, 0.



4. As t increases (or decreases) without bound, the graph approaches, but never equals, 0.

5. The area in the tails of the t-distribution is a little greater than the area in the tails of the standard normal distribution because using s as an estimate of introduces more variability to the t-statistic.



6. As the sample size n increases, the density curve of t gets closer to the standard normal density curve. This result occurs because as the sample size increases, the values of s get closer to the values of by the Law of Large Numbers.



Objective 1

• Test hypotheses about a population mean with unknown


Testing Hypotheses Regarding a Population Mean with σ Unknown

To test hypotheses regarding the population mean with unknown, we use the following steps, provided that:

1. The sample is obtained using simple random sampling.

2. The sample has no outliers, and the population from which the sample is drawn is normally distributed or the sample size is large (n≥30).


Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways:


Step 2: Select a level of significance, , based on the seriousness of making a Type I error.


Step 3: Compute the test statistic

which follows Student’s t-distribution with n-1 degrees of freedom.

t0 x 0

s

n


Step 4: Use Table VI to determine the critical value using n-1 degrees of freedom.

Classical Approach


Classical Approach

Two-Tailed


Classical Approach

Left-Tailed


Classical Approach

Right-Tailed


Step 5: Compare the critical value with the test statistic:

Classical Approach


Step 4: Use Table VI to estimate the P-value using n-1 degrees of freedom.

P-Value Approach


P-Value Approach

Two-Tailed


P-Value Approach

Left-Tailed


P-Value Approach

Right-Tailed


Step 5: If the P-value < , reject the null hypothesis.

P-Value Approach


Step 6: State the conclusion.


Parallel Example 1: Testing a Hypothesis about a Population Mean, Large Sample

Assume the resting metabolic rate (RMR) of healthy males in complete silence is 5710 kJ/day. Researchers measured the RMR of 45 healthy males who were listening to calm classical music and found their mean RMR to be 5708.07 with a standard deviation of 992.05.

At the =0.05 level of significance, is there evidence to conclude that the mean RMR of males listening to calm classical music is different than 5710 kJ/day?


Solution

We assume that the RMR of healthy males is 5710 kJ/day. This is a two-tailed test since we are interested in determining whether the RMR differs from 5710 kJ/day.

Since the sample size is large, we follow the steps for testing hypotheses about a population mean for large samples.


Solution

Step 1: H0: =5710 versus H1: ≠5710

Step 2: The level of significance is =0.05.

Step 3: The sample mean is = 5708.07 and the sample standard deviation is s=992.05. The test statistic is

t0 5708.07 5710

992.05 45 0.013

x


Solution: Classical Approach

Step 4: Since this is a two-tailed test, we determine the critical values at the =0.05 level of significance with n-1=45-1=44 degrees of freedom to be approximately -t0.025= -2.021 and

t0.025=2.021.

Step 5: Since the test statistic, t0=-0.013, is between

the critical values, we fail to reject the null hypothesis.


Solution: P-Value Approach

Step 4: Since this is a two-tailed test, the P-value is the area under the t-distribution with n-1=45-1=44 degrees of freedom to the left of -t0.025= -0.013

and to the right of t0.025=0.013. That is, P-value

= P(t < -0.013) + P(t > 0.013) = 2 P(t > 0.013). 0.50 < P-value.

Step 5: Since the P-value is greater than the level of significance (0.05<0.5), we fail to reject the null hypothesis.


Solution

Step 6: There is insufficient evidence at the =0.05 level of significance to conclude that the mean RMR of males listening to calm classical music differs from 5710 kJ/day.


Parallel Example 3: Testing a Hypothesis about a Population Mean, Small Sample

According to the United States Mint, quarters weigh 5.67 grams. A researcher is interested in determining whether the “state” quarters have a weight that is different from 5.67 grams. He randomly selects 18 “state” quarters, weighs them and obtains the following data.

5.70 5.67 5.73 5.61 5.70 5.67

5.65 5.62 5.73 5.65 5.79 5.73

5.77 5.71 5.70 5.76 5.73 5.72

At the =0.05 level of significance, is there evidence to conclude that state quarters have a weight different than 5.67 grams?


Solution

We assume that the weight of the state quarters is 5.67 grams. This is a two-tailed test since we are interested in determining whether the weight differs from 5.67 grams.

Since the sample size is small, we must verify that the data come from a population that is normally distributed with no outliers before proceeding to Steps 1-6.


Assumption of normality appears reasonable.


No outliers.


Solution

Step 1: H0: =5.67 versus H1: ≠5.67

Step 2: The level of significance is =0.05.

Step 3: From the data, the sample mean is calculated to be 5.7022 and the sample standard deviation is s=0.0497. The test statistic is

t0 5.7022 5.67

.0497 182.75


Solution: Classical Approach

Step 4: Since this is a two-tailed test, we determine the critical values at the =0.05 level of significance with n-1=18-1=17 degrees of freedom to be -t0.025= -2.11 and t0.025=2.11.

Step 5: Since the test statistic, t0=2.75, is greater than

the critical value 2.11, we reject the null hypothesis.


Solution: P-Value Approach

Step 4: Since this is a two-tailed test, the P-value is the area under the t-distribution with n-1=18-1=17 degrees of freedom to the left of -t0.025= -2.75

and to the right of t0.025=2.75. That is, P-value

= P(t < -2.75) + P(t > 2.75) = 2 P(t > 2.75). 0.01 < P-value < 0.02.

Step 5: Since the P-value is less than the level of significance (0.02<0.05), we reject the null hypothesis.


Solution

Step 6: There is sufficient evidence at the =0.05 level of significance to conclude that the mean weight of the state quarters differs from 5.67 grams.


Summary

Which test to use?

Provided that the population from which the sample is drawn is normal or that the sample size is large,

• if is known, use the Z-test procedures from Section 10.2

• if is unknown, use the t-test procedures from this Section.

Documents

Chapter