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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

I don't get the example
Hadekunle Reply
ways of collecting data at least 10 and explain
Ridwan Reply
Example of discrete variable
Bada Reply
sales made monthly.
Gbenga
I am new here, can I get someone to guide up?
alayo
dies outcome is 1, 2, 3, 4, 5, 6 nothing come outside of it. it is an example of discrete variable
jainesh
continue variable is any value value between 0 to 1 it could be 4digit values eg 0.1, 0.21, 0.13, 0.623, 0.32
jainesh
How to answer quantitative data
Alhassan Reply
hi
Kachalla
what's up here ... am new here
Kachalla
sorry question a bit unclear...do you mean how do you analyze quantitative data? If yes, it depends on the specific question(s) you set in the beginning as well as on the data you collected. So the method of data analysis will be dependent on the data collecter and questions asked.
Bheka
how to solve for degree of freedom
saliou
Quantitative data is the data in numeric form. For eg: Income of persons asked is 10,000. This data is quantitative data on the other hand data collected for either make or female is qualitative data.
Rohan
*male
Rohan
Degree of freedom is the unconditionality. For example if you have total number of observations n, and you have to calculate variance, obviously you will need mean for that. Here mean is a condition, without which you cannot calculate variance. Therefore degree of freedom for variance will be n-1.
Rohan
data that is best presented in categories like haircolor, food taste (good, bad, fair, terrible) constitutes qualitative data
Bheka
vegetation types (grasslands, forests etc) qualitative data
Bheka
I don't understand how you solved it can you teach me
Caleb Reply
solve what?
Ambo
mean
Vanarith
What is the end points of a confidence interval called?
ZIMKHITHA Reply
lower and upper endpoints
Bheka
Class members write down the average time (in hours, to the nearest half-hour) they sleep per night.
William Reply
how we make a classes of this(170.3,173.9,171.3,182.3,177.3,178.3,174.175.3)
Sarbaz
6.5
phoenix
11
Shakir
7.5
Ron
why is always lower class bundry used
Caleb
Assume you are in a class where quizzes are 20% of your grade, homework is 20%, exam _1 is 15%,exam _2 is 15%, and the final exam is 20%.Suppose you are in the fifth week and you just found out that you scored a 58/63 on the fist exam. You also know that you received 6/9,8/10,9/9 on the first
Diamatu Reply
quizzes as well as a 9/11,10/10,and 4.5/7 on the first three homework assignment. what is your current grade in the course?
Diamatu
the answer is 2.6
Abdul
if putting y=3x examine that correlation coefficient between x and y=3x is 1.
Aadrsh Reply
what is permutation
Rodlett Reply
how to construct a histogram
Baalisi Reply
You have to plot the class midpoint and the frequency
Wydny
ok so you use those two to draw the histogram right.
Amford
yes
Wydny
ok can i be a friend so you can be teaching me small small
Amford
how do you calculate cost effectiveness?
George
Hi everyone, this is a very good statistical group and am glad to be part of it. I'm just not sure how did I end up here cos this discussion just popes on my screen so if I wanna ask something in the future, how will I find you?
Bheka
To make a histogram, follow these steps: On the vertical axis, place frequencies. Label this axis "Frequency". On the horizontal axis, place the lower value of each interval. ... Draw a bar extending from the lower value of each interval to the lower value of the next interval.
Divya
I really appreciate that
umar Reply
I want to test linear regression data such as maintenance fees vs house size. Can I use R square, F test to test the relationship? Is the good condition of R square greater than 0.5
Mok Reply
yes of course must have use f test and also use t test individually multple coefficients
rishi
Alright
umar
hi frnd I'm akeem by name, I wanna study economics and statistics wat ar d thing I must do to b a great economist
akeem
Is R square cannot analysis linear regression of X vs Y relationship?
Mok
To be an economist you have to be professional in maths
umar
hi frnds
Shehu
what is random sampling what is sample error
Nistha Reply
@Nistha Kashyap Random sampling is the selection of random items (or random numbers) from the group. A sample error occurs when the selected samples do not truely represent the whole group. The can happen when most or all of the selected samples are taken from only one section of the group;
Ron
Thus the sample is not truely random.
Ron
What is zero sum game?
Hassan Reply
A game in which there is no profit & no loss to any of the both player.
Milan

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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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