<< Chapter < Page Chapter >> Page >

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
Got questions? Get instant answers now!
Got questions? Get instant answers now!

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

Got questions? Get instant answers now!

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.

Got questions? Get instant answers now!
Got questions? Get instant answers now!

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.

Got questions? Get instant answers now!

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

Got questions? Get instant answers now!

State the null and alternative hypotheses.

Got questions? Get instant answers now!

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

a left-tailed test

Got questions? Get instant answers now!


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?

Got questions? Get instant answers now!

State the null and alternative hypotheses.

H 0 : σ 2 = 0.81 2 ;

H a : σ 2 >0.81 2

Got questions? Get instant answers now!


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

Got questions? Get instant answers now!

What is the test statistic?

Got questions? Get instant answers now!

What is the p -value?

0.0542

Got questions? Get instant answers now!

What can you conclude at the 5% significance level?

Got questions? Get instant answers now!

Questions & Answers

What Is The Confidence Interval
ala Reply
sample mean 25, sample standard deviation 20, sample size 200, calculate the confidence interval using the given values and the original confidence level of 90%.
Cady Reply
Can you help me in mathematical statistics problems?
bint-e-taj Reply
yes
Kc
Pls who can help me to teach me statistics
nasir
i need tutor for statistics plz
Rana
ok
Ekene
the power of the test is
Ejaz Reply
please can anyone help me solve these questions below? I need help please.
MMSI
a)An investor wants to eliminate seven of the investments in her portfolio by selling 4 stocks and 3 bonds. In how many can these be sold if among 25 securities in the portfolio,13 are stocks and the rest bonds?
MMSI
a)If a random variable has the standard normal distribution,what are the probabilities that it will take on a value: i)Less than 1.64 ii)Greater than-0.47
MMSI
b)A random variable has a normal distribution with a mean of 60 and standard deviation 5.2.What are the probabilities that the random variable will take on a value: i)Less than 65.2 ii)Between 48 and 72?
MMSI
b)If the probability that an individual suffers a bad reaction from injection of a given serum is 0.001,use the Poisson law to calculate the probability that out of 2000 individuals i)Exactly 3 individuals will suffer a bad reaction. ii)More than 2 individuals will suffer a bad reaction.
MMSI
b)The breakfast menu serve data popular 5-star Hotel in Accra consists of the following items: Juice-Mango,Grape,Apple. Toast-Whitewheat,Whole wheat. Egg:Fried,Hard-boiled,Scrambled. Beverage:Coffee,Tea,Cocoa.
MMSI
Continuation of the last question.Assist the Hotel manager to determine the number of possible breakfast combinations that can be served, one from each category
MMSI
please I need help.
MMSI
3x2x3
Vince
Are you answering the last question?
MMSI
please you guys should help me I need it so badly
MMSI
bias came in sampling due to
Muzammil Reply
sampling error
Vikram
what is the difference between population and sample
Inam
Sample is the group of individual who participate in your study. Sample is a subset of population. Population is the broader group of people to whom you intend to generalize the results of your study.
Ekene
how do you find z if you only know the area of .0808
Cady Reply
construct a frequency distribution
Sana
How to take a random sample of 30 observations
Hamna Reply
you can use the random function to generate 30 numbers or observation
smita
How we can calculate chi-square if observed x٫y٫z/frequency 40,30,20 Total/90
Insha Reply
calculate chi-square if observed x,y,z frequency 40,30,20total 90
Insha
find t value,if boysN1, ،32,M1,87.43 S1square,39.40.GirlsN2,34,M2,82.58S2square,40.80 Determine whether the results are significant or insignificant
Insha
The heights of a random sample of 100 entering HRM Freshman of a certain college is 157 cm with a standard deviation of 8cm. test the data against the claim that the overall height of all entering HRM students is 160 cm. previous studies showed that
Crispen Reply
complete the question.. as data given N = 100,mean= 157 cm, std dev = 8 cm..
smita
Z=x-mu/ std dev
smita
the power of the test is
Ejaz
find the mean of 25,26,23,25,45,45,58,58,50,25
Asmat Reply
add all n divide by 10 i.e 38
smita
38
hhaa
amit
1 . The “average increase” for all NASDAQ stocks is the:
Jamshaid Reply
STATISTICS IN PRACTICE: This is a group assignment that seeks to reveal students understanding of statistics in general and it’s practical usefulness. The following are the guidelines; 1.      Each group has to identify a natural process or activity and gather data about/from the process. 2.     
Kofi Reply
The diameter of an electric cable,say, X is assumed to be continoues random variable with p.d.f f(x)=6x(1-x); ≤x≤1 a)check that f(X) is p.d.f b) determine a number b such that p(Xb)
Syed Reply
A manufacturer estimate 3% of his output is defective. Find the probability that in a sample of 10 items (a) less than two will be defective (b) more than two will be defective.
ISAIAH Reply
A manufacturer estimates that 3% of his output of a small item is defective. Find the probabilities that in a sample of 10 items (a) less than two and (b) more than two items will be defective.
ISAIAH
use binomial distribution with parameter n=10, p= 0.03, q=0.97
Shivprasad
the standard deviation of a symmetrical distribution is 7.8 . what must be the value of forth moment about the mean in order that distribution be a) leptokurtic b) mesokurtic c) platy kyrtic intrept the obtain value of a b and c
Tushar Reply

Get the best Introductory statistics course in your pocket!





Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Introductory statistics' conversation and receive update notifications?

Ask