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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:

( n - 1 ) s 2 σ 2

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.

Math instructors are not only interested in how their students do on exams, on average, but how the exam scores vary. To many instructors, the variance (or standard deviation) may be more important than the average.

Suppose a math instructor believes that the standard deviation for his final exam is five points. One of his best students thinks otherwise. The student claims that the standard deviation is more than five points. If the student were to conduct a hypothesis test, what would the null and alternative hypotheses be?

Even though we are given the population standard deviation, we can set up the test using the population variance as follows.

  • H 0 : σ 2 = 5 2
  • H a : σ 2 >5 2
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A SCUBA instructor wants to record the collective depths each of his students dives during their checkout. He is interested in how the depths vary, even though everyone should have been at the same depth. He believes the standard deviation is three feet. His assistant thinks the standard deviation is less than three feet. If the instructor were to conduct a test, what would the null and alternative hypotheses be?

H 0 : σ 2 = 3 2

H a : σ 2 <3 2

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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: χ 24 2 , where:

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

Calculate the test statistic:

χ 2 = ( n     1 ) s 2 σ 2 = ( 25     1 ) ( 3.5 ) 2 7.2 2 = 5.67

where n = 25, s = 3.5, and σ = 7.2.

Graph:

This is a nonsymmetrical chi-square curve with values of 0 and 5.67 labeled on the horizontal axis. The point 5.67 lies to the left of the peak of the curve. A vertical upward line extends from 5.67 to the curve and the region to the left of this line is shaded. The shaded area is equal to the p-value.

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

χ 2 = ( n 1 ) s 2 σ 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 ( n 1 ) s 2 σ 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.

What type of test should be used?

a test of a single variance

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State the null and alternative hypotheses.

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Is this a right-tailed, left-tailed, or two-tailed test?

a left-tailed test

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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.

What type of test should be used?

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State the null and alternative hypotheses.

H 0 : σ 2 = 0.81 2 ;

H a : σ 2 >0.81 2

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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.

What type of test should be used?

a test of a single variance

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What is the test statistic?

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What is the p -value?

0.0542

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What can you conclude at the 5% significance level?

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Questions & Answers

7.The following data give thenumber of car thefts that occurred in a city in the past 12 days. 63711438726915 Calculate therange, variance, and standard deviation.
Mitu Reply
express the confidence interval 81.4% ~8.5% in interval form
Xx Reply
a bad contain 3 red and 5 black balls another 4 red and 7 black balls, A ball is drawn from a bag selected at random, Find the probability that A is red?
Shazain Reply
The information is given as, 30% of customers shopping at SHOPNO will switch to DAILY SHOPPING every month on the other hand 40% of customers shopping at DAILY SHOPPING will switch to other every month. What is the probability that customers will switch from A to B for next two months?
sharmin Reply
Calculate correlation coefficient, where SP(xy) = 144; SS(x) = 739; SS(y) = 58. (2 Points)
Ashfat Reply
The information are given from a randomly selected sample of age of COVID-19 patients who have already survived. These information are collected from 200 persons. The summarized information are as, n= 20; ∑x = 490; s^2 = 40. Calculate 95% confident interval of mean age.
Ashfat
The mode of the density of power of signal is 3.5. Find the probability that the density of a random signal will be more than 2.5.
Ashfat
The average time needed to repair a mobile phone set is 2 hours. If a customer is in queue for half an hour, what is the probability that his set will be repaired within 1.6 hours?
Ashfat
A quality control specialist took a random sample of n = 10 pieces of gum and measured their thickness and found the mean 9 and variance 0.04. Do you think that the mean thickness of the spearmint gum it produces is 8.4
nazrul Reply
3. The following are the number of mails received in different days by different organizations: Days (x) : 23, 35, 38, 50, 34, 60, 41, 32, 53, 67. Number of mails (y) : 18, 40, 52, 45, 32, 55, 50, 48, 26, 25. i) Fit a regression line of y on x and test the significance of regression. ii) Estimate y
Atowar Reply
The number of problem creating computers of two laboratories are as follows: Number of computers: 48, 6, 10, 12, 30, 11, 49, 17, 10, 14, 38, 25, 15, 19, 40, 12. Number of computers: 12, 10, 26, 11, 42, 11, 13, 12, 18, 5, 14, 38. Are the two laboratories similar in respect of problem creating compute
Tamim Reply
Is the severity of the drug problem in high school the same for boys and girls? 85 boys and 70 girls were questioned and 34 of the boys and 14 of the girls admitted to having tried some sort of drug. What can be concluded at the 0.05 level?
Ashfat Reply
null rejected
Pratik
a quality control specialist took a random sample of n=10 pieces of gum and measured their thickness and found the mean 7.6 and standered deviation 0.10. Do you think that the mean thickness of the spearmint gum it produces is 7.5?
Shanto Reply
99. A one sample, one-tail t-test is conducted and the test statistic value is calculated to be 2.56. The degrees of freedom for the test are 10. Which of the following conclusions for the test would be correct? a
Niaz Reply
A one sample, one-tail t-test is conducted and the test statistic value is calculated to be 2.56. The degrees of freedom for the test are 10. Which of the following conclusions for the test would be correct?
Niaz
what is null Hypothesis
Niaz
what is null Hypothesis
Niaz
when median is greater than mode?
Hafiza Reply
hello
Amaano
is this app useful
Worthy
little bit 😭
G-
oh
Worthy
when tail is positive
Jungjoon
define hypothesis
Worthy
I'm struggling to type it's on my laptop...statistics
Yoliswa
types of averages .mean median mode quarantiles MCQ question
Rupa Reply
what a consider data?
JAGESH Reply
Out of 25 students, 15 are male. Is the overall proportion of male students 0.7 in AIUB? (4 Points)
Omer Reply
15/25=0.6 or 60% standard calculation
Andrea
A quality control specialist took a random sample of n = 10 pieces of gum and measured their thickness and found the mean 7.6 and variance 0.01. Do you think that the mean thickness of the spearmint gum it produces is 7.5? (4 Points)
Omer
10 gums mean = 7.6 variance= 0.01 standard deviation= ? what us the data set?
Andrea
0.6
Rubina

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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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