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Hypothesis Testing Using a Single Sample

Part I: Means

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A P-value is the probability of observing a sample statistic as extreme or more extreme than the one observed under the assumption that the null hypothesis is true.

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Step 1: A claim is made regarding the population mean. The claim is used to determine the null and alternative hypotheses. Again, the hypothesis can be structured in one of three ways:

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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

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

Properties of the t-Distribution

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

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

Properties of the t-Distribution

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

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.

Properties of the t-Distribution

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

Properties of the t-Distribution

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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.

Properties of the t-Distribution

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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.

Properties of the t-Distribution

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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).

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

Setup: Clearly indicate/describe the parameter of interest to the study.

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

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Step 3: Compute the test statistic

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

t0 x 0

s

n

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

P-Value Approach

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P-Value Approach

Two-Tailed

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P-Value Approach

Left-Tailed

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P-Value Approach

Right-Tailed

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Step 5: If the P-value < , reject the null hypothesis. If the P-value ≥ α, fail to reject the null hypothesis

P-Value Approach

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Step 6: State the conclusion in the context of the problem.

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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?

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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.

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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

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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.

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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.

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Parallel Example 2: 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?

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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.

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Assumption of normality appears reasonable.

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No outliers.

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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

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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.

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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.