11.6 Test of a single variance  (Page 39/22)

A test of a single variance assumes that the underlying distribution is normal . The null and alternative hypotheses are stated in terms of the population variance (or population standard deviation). The test statistic is:

where:

• n = the total number of data
• s ² = sample variance
• σ ² = population variance

You may think of s as the random variable in this test. The number of degrees of freedom is df = n - 1. A test of a single variance may be right-tailed, left-tailed, or two-tailed. [link] will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance.

With individual lines at its various windows, a post office finds that the standard deviation for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes.

With a significance level of 5%, test the claim that a single line causes lower variation among waiting times (shorter waiting times) for customers .

Since the claim is that a single line causes less variation, this is a test of a single variance. The parameter is the population variance, σ ² , or the population standard deviation, σ .

Random Variable: The sample standard deviation, s , is the random variable. Let s = standard deviation for the waiting times.

• H ₀ : σ ² = 7.2 ²
• H _a : σ ² <7.2 ²

The word "less" tells you this is a left-tailed test.

Distribution for the test: ${\chi }_{24}^2$ , where:

• n = the number of customers sampled
• df = n – 1 = 25 – 1 = 24

Calculate the test statistic:

${\chi }^2=\frac{(n-1)s^2}{{\sigma }^2}=\frac{(25-1){(3.5)}^2}{{7.2}^2}=5.67$

where n = 25, s = 3.5, and σ = 7.2.

Graph:

Probability statement: p -value = P ( χ ² <5.67) = 0.000042

Compare α and the p -value:

• α = 0.05
• p -value = 0.000042
• α > p -value

Make a decision: Since α > p -value, reject H ₀ . This means that you reject σ ² = 7.2 ² . In other words, you do not think the variation in waiting times is 7.2 minutes; you think the variation in waiting times is less.

Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that a single line causes a lower variation among the waiting times or with a single line, the customer waiting times vary less than 7.2 minutes.

In 2nd DISTR , use 7:χ2cdf . The syntax is (lower, upper, df) for the parameter list. For [link] , χ2cdf(-1E99,5.67,24) . The p -value = 0.000042.

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The FCC conducts broadband speed tests to measure how much data per second passes between a consumer’s computer and the internet. As of August of 2012, the standard deviation of Internet speeds across Internet Service Providers (ISPs) was 12.2 percent. Suppose a sample of 15 ISPs is taken, and the standard deviation is 13.2. An analyst claims that the standard deviation of speeds is more than what was reported. State the null and alternative hypotheses, compute the degrees of freedom, the test statistic, sketch the graph of the p -value, and draw a conclusion. Test at the 1% significance level.

H ₀ : σ ² = 12.2 ²

H _a : σ ² >12.2 ²
df = 14
chi ² test statistic = 16.39

The p -value is 0.2902, so we decline to reject the null hypothesis. There is not enough evidence to suggest that the variance is greater than 12.2 ² .

In 2nd DISTR , use7: χ2cdf . The syntax is (lower, upper, df) for the parameter list. χ2cdf(16.39,10^99,14) . The p -value = 0.2902.

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References

“AppleInsider Price Guides.” Apple Insider, 2013. Available online at http://appleinsider.com/mac_price_guide (accessed May 14, 2013).

Data from the World Bank, June 5, 2012.

Chapter review

To test variability, use the chi-square test of a single variance. The test may be left-, right-, or two-tailed, and its hypotheses are always expressed in terms of the variance (or standard deviation).

Formula review

${\chi }^2=$ $\frac{(n-1)\cdot s^2}{{\sigma }^2}$ Test of a single variance statistic where:
n : sample size
s : sample standard deviation
σ : population standard deviation

df = n – 1 Degrees of freedom

Test of a single variance

• Use the test to determine variation.
• The degrees of freedom is the number of samples – 1.
• The test statistic is $\frac{(n–1)\cdot s^2}{{\sigma }^2}$ , where n = the total number of data, s ² = sample variance, and σ ² = population variance.
• The test may be left-, right-, or two-tailed.

Use the following information to answer the next three exercises: An archer’s standard deviation for his hits is six (data is measured in distance from the center of the target). An observer claims the standard deviation is less.

Use the following information to answer the next three exercises: The standard deviation of heights for students in a school is 0.81. A random sample of 50 students is taken, and the standard deviation of heights of the sample is 0.96. A researcher in charge of the study believes the standard deviation of heights for the school is greater than 0.81.

Use the following information to answer the next four exercises: The average waiting time in a doctor’s office varies. The standard deviation of waiting times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the variance of waiting times is greater than originally thought.