11.6 Test of a single variance

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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:

$\frac{\left(n-1\right){s}^{2}}{{\sigma }^{2}}$

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

Try it

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

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:

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.

Try it

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.

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

${\chi }^{2}=$ $\frac{\left(n-1\right)\cdot {s}^{2}}{{\sigma }^{2}}$ 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 $\frac{\left(n–1\right)\cdot {s}^{2}}{{\sigma }^{2}}$ , 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.

What type of test should be used?

a test of a single variance

State the null and alternative hypotheses.

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

a left-tailed test

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?

State the null and alternative hypotheses.

H 0 : σ 2 = 0.81 2 ;

H a : σ 2 >0.81 2

df = ________

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

What is the test statistic?

What is the p -value?

0.0542

What can you conclude at the 5% significance level?

what is standard deviation?
It is the measure of the variation of certain values from the Mean (Center) of a frequency distribution of sample values for a particular Variable.
Dominic
Yeah....the simplest one
IRFAN
what is the number of x
10
Elicia
Javed Arif
Jawed
how will you know if a group of data set is a sample or population
population is the whole set and the sample is the subset of population.
umair
if the data set is drawn out of a larger set it is a sample and if it is itself the whole complete set it can be treated as population.
Bhavika
hello everyone if I have the data set which contains measurements of each part during 10 years, may I say that it's the population or it's still a sample because it doesn't contain my measurements in the future? thanks
Alexander
Pls I hv a problem on t test is there anyone who can help?
Peggy
Dominic
Bhavika is right
Dominic
what is the problem peggy?
Bhavika
hi
Sandeep
Hello
hi
Bhavika
hii Bhavika
Dar
Hi eny population has a special definition. if that data set had all of characteristics of definition, that is population. otherwise that is a sample
Hoshyar
three coins are tossed. find the probability of no head
three coins are tossed consecutively or what ?
umair
umair
or .125 is the probability of getting no head when 3 coins are tossed
umair
🤣🤣🤣
Simone
what is two tailed test
if the diameter will be greater than 3 cm then the bullet will not fit in the barrel of the gun so you are bothered for both the sides.
umair
in this test you are worried on both the ends
umair
lets say you are designing a bullet for thw gun od diameter equals 3cm.if the diameter of the bullet is less than 3 cm then you wont be able to shoot it
umair
In order to apply weddles rule for numerical integration what is minimum number of ordinates
excuse me?
Gabriel
why?
didn't understand the question though.
Gabriel
which question? ?
We have rules of numerical integration like Trapezoidal rule, Simpson's 1/3 and 3/8 rules, Boole's rule and Weddle rule for n =1,2,3,4 and 6 but for n=5?
John
geometric mean of two numbers 4 and 16 is:
10
umair
really
iphone
quartile deviation of 8 8 8 is:
iphone
sorry 8 is the geometric mean of 4,16
umair
quartile deviation of 8 8 8 is
iphone
can you please expalin the whole question ?
umair
mcq
iphone
h
iphone
can you please post the picture of that ?
umair
how
iphone
hello
John
10 now
John
how to find out the value
can you be more specific ?
umair
yes
KrishnaReddy
what is the difference between inferential and descriptive statistics
descriptive statistics gives you the result on the the data like you can calculate various things like variance,mean,median etc. however, inferential stats is involved in prediction of future trends using the previous stored data.
umair
if you need more help i am up for the help.
umair
Thanks a lot
Anjali
Inferential Statistics involves drawing conclusions on a population based on analysis of a sample. Descriptive statistics summarises or describes your current data as numerical calculations or graphs.
fred
my pleasure😊. Helping others offers me satisfaction 😊
umair
for poisson distribution mean............variance.
both are equal to mu
Faizan
mean=variance
Faizan
what is a variable
something that changes
Festus
why we only calculate 4 moment of mean? asked in papers.
why we only 4 moment of mean ? asked in BA exam
Faizan
Hello, can you please share the possible questions that are likely to be examined under the topic: regression and correlation analysis.
Refiloe
for normal distribution mean is 2 & variance is 4 find mu 4?
repeat quastion again
Yusuf
find mu 4. it can be wrong but want to prove how.
Faizan
for a normal distribution if mu 4 is 12 then find mu 3?
Question hi wrong ha
Tahir
ye BA mcqs me aya he teen he. 2dafa aya he
Faizan
if X is normally distributed. (n,b). then its mean deviation is?
Faizan
The answer is zero, because all odd ordered central moments of a normal distribution are Zero.
nikita
which question is zero
Faizan
sorry it is (5,16) in place of (n,b)
Faizan
I got. thanks. it is zero.
Faizan
a random variable having binomial distribution is?
Bokaho