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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

what is permutation
Rodlett Reply
how to construct a histogram
Baalisi Reply
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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
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umar
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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
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Hassan Reply
A game in which there is no profit & no loss to any of the both player.
Milan
Differences between sample mean & population mean
mohammed Reply
***keydifferences.com/difference-between-sample-mean-and-population-mean.html
Lucien
Not difference in the formula except the notation, sample mean is denoted by x bar and population mean is denoted by mu symbol. There is formula as well as notation between difference variance and standard deviations
Akash
Likely the difference would be in the result, unless the sample is an exact representation of the population (which is unlikely.)
Ron
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Nii
Nii Avin - Data is just a simple way to refer to the numbers in the population, or in the sample used in your calculations.
Ron
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Nii
Data is the very pale android from the Star Trek Enterprise
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Emmanuel
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Handson
what is a mode?
Handson
Nii - data is whatever you are sampling. Such as the number of students in each classroom.
Ron
Handson Ndintek - the mode is the number appearing most frequently. Example: 7 9 11 7 4 6 3 7 2. 7 is the mode. In a group such as 7 9 1 4 6 3, there is no mode because no number appears more often than any other.
Ron
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qualitative and quantitative
phoenix
primary and secondary data
Peace
qualitative and quantitative
Prince
Using Cauchy Schwartz inequality,or prove that b2-b1-1=0
Md Reply
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it's a science of collection, organization, analysis and summarizing data to get useful information to make several types of conclusions.which can be used in real life.
anshika
what is the statistical probability that president Trump will remain in the white house after the election of 2020?
Terry
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Maureen
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Stephen
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Jhasaketan Reply
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average value
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Narendra
Hope this helps. There are three main types of averages. *mean -> average -> (X1+X2+X3+...+Xn) / n *mode -> the element within a set which occurs most. {3,4,5,8,12,3,4,3,3,56} mode = 3 *median -  {3,3,4,5,8,12,56} median = 5 OR {3,4,5,8,12,56} median = 6.5
Jack
conceptual approach to limits
lameck Reply
how are limits derived?
lameck
an entire section of calculus is devoted to that explanation.
Pitior
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Martin Reply
statistics :- can be defined as the branches of mathematics that deals with the summarizing, analysing,organization and interpretation of data.
Usman
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Venkat
can we find Z value on calculator with out using Z table
Maham Reply
no
Pitior
why
Maham
can another way is possible ?
Maham
Well you could make a table. And as the function you use the one used at the z table
Luca
The normal function is only one way, so you can only try using different numbers until you get the probability that you have. So that is easier if you have a table
Luca
me don't know nothing about z table and don't know how to see the z value on table can you tell me please how see the value on table
Maham
The z table is the table of the standard normal distribution
Luca
You can look it up on internet, its easier than writing down the normal distribution function (with an integral) and doing a table in the calculator
Luca
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Maham
yes use pnorm in r
Venkat
pnorm(2.3,mean=0,sd=1)
Venkat
pnorm?
Maham
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Venkat
no
Maham
its with tht u will get
Venkat
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ti83
Venkat
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Alwin
descriptive stats
Venkat
inferential stats
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outlier treatment
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boxplot
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assumption of linear regression
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Maham Reply
u can find using excel
Venkat
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Venkat
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Ibrokhim
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x:1,2,3,4,5 y:2,5,6,8,9
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regredsion equation is
Venkat
y=0.9+1.7x
Venkat
reg eq is y=0.9+1.7x
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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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