Ch 07 Solutions

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Chapter 7

Hypothesis testing

7.1 Learning activity A7.1

Question:

Think about each of the following statements. Then give the nulland alternative hypotheses and say whether they will need one- ortwo-tailed tests.

a) The general mean level of family income in a population isknown to be 10,000 ulam a year. You take a random sample in anurban area U and find the mean family income is 6,000 ulam a yearin that area. Do the families in the chosen area have a lower incomethan the population as a whole?

b) You are looking at data from two schools on the heights andweights of children by age. Are the mean weights for girls aged1011 the same in the two schools?

c) You are looking at reading scores for children before and after anew teaching programme. Have their scores improved?

Solution:

Look at the wording of a) to c) carefully. You can get clues as towhether you are dealing with a one-tailed test (where H1 will use a), or a two-tailed test (where H1 will involve =) according tothe use of words like:

increase higher greater diminished

which all imply aone-tailed test, or the use of words like:

equal changed different from

which all imply atwo-tailed test.

a) Here you want to know whether the mean of incomes in a chosenarea (U) islessthan the general population mean ( =10,000).

So you need a one-tailedtest.

H0 : U =10,000 (i.e.Uis equal to)H1 : U


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Statistics 1 Solutions to learning activities

H0 : A =B (i.e. the means are the same)H1 : A=B (i.e. the means are different)

c) We look at the reading score means for children before, xB, andafter, xA, a teaching programme with populations means B andArespectively.

H0 : A =B (i.e. B is no different fromA)H1 : A > B (i.e. the mean score after the programme is greaterthan before).


Question:

Complete the following chart:

Result from your test Real situation

H0 true H0 false

H0 true Correct Type II error

Probability(1 )called theconfidence

intervalof the test

H0 false

Probability(1 )calledthepowerof the test

Solution:

Your completed table should look like this:

Result from your test Real situationH0 true H0 false

H0 true Correct Type II error

Probability(1 ) Probability =called theconfidence

intervalof the test

H0 false Type I error Correct

Probability called the Probability(1 )calledsignificance level thepowerof the test

of the test

2


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CHAPTER 7. HYPOTHESIS TESTING


Question:

The manufacturer of a patient medicine claimed that it was 90%effective in relieving an allergy for a period of 8 hours. In a sampleof 200 people suffering from the allergy, the medicine providedrelief for 160 people.

Determine whether the manufacturers claim is legitimate. (Becareful. Your parameter here will be .) Is your test one- ortwo-tailed?

Solution:

Here the manufacturer is claiming that 90% of the population willbe relieved over 8 hours. That is =0.9.

A sample ofn =200, and 160 gained relief. That isp = 160200

=0.8.

For the manufacturers claim to be accepted, I think it is fair toassume that we are asking whetherp is less than or not (we wouldaccept the claim otherwise). So we have a one-tailed test with:

H0 : =0.9H1 : >0.9.

We use the population value to work out the standard error, and

so calculate:

z =0.8 0.9

0.90.1200

=0.1

0.0212= 4.71698.

This goes beyond the tables (thez given is ()4.417, so this result ishighly significant thep-value is nearly zero). Looked at another

way, you could look at Table 10 (Percentage points of thet-distribution) and take the bottom line (where =

).

Note that the 5% value is 1.645 for a one-tailed test and the 1%and the 0.1% values are 2.326 and 3.090 respectively. Thisconfirms that the result is highly significant. So werejectH0 and themanufacturers claim is not met. The population given relief fromthe sample taken is significantly less than the 90% he claims.


Question:

A sample of seven is taken at random from a large batch of

3


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(nominally 12 volt) batteries. These are tested and their truevoltages are shown below:

12.9 11.6 13.5 13.9 12.1 11.9 13.0

a) Test if the mean voltage of the whole batch is 12 volts.

b) Test if the mean batch voltage is less than 12.

Solution:

In part a) you are asked to do a two-sided test; in part b) it is aone-sided test. Which is more appropriate will depend on thepurpose of the experiment, and your suspicions before you conductit.

If you suspectedbeforecollecting the data that the mean voltagewas less than 12 volts, the one-sided test could be appropriate.

If you had no prior reason to believe that the mean was lessthan 12 volts you would do a two-sided test.

General rule: decide on whether it is a one- or two-sided testbeforecalculating the test statistic.

a) We are to test H0 : =12 v. H1 : =12. The key points here arethatn is small and that2 is unknown. We can use the t-testand this is valid provided the data are normally distributed. Thetest statistic value is

t =x 12s/

7=

12.7 120.858/

7=2.16.

This is compared to a Studentst distribution on 6 degrees offreedom. The critical value corresponding to a 5% test is 2.447.Hence we cannot reject the null hypothesis at the 5% level. (Wecan reject at the 10% level, but the convention on this course isto regard such evidence merely as casting doubt on H0, ratherthan justifying rejection as such.)

b) We are to test H0 : =12 v. H1 :


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CHAPTER 7. HYPOTHESIS TESTING

Solution:

Significant at the 5% level means there is a less than 5% chance ofgetting data as extreme as those observed if the null hypothesis was

true. This implies that the data are incompatible with the nullhypothesis, which we reject.

Significant at 10% but not at 5% is often interpreted as meaningthere is some doubt about the null hypothesis, but not enough toreject it.

Here we are testing a proportion: H0 : =0.25v. H1 : >0.25.Note that this is a one-sided test we have reason to believe thatthe sales campaign hasincreasedsales, and we believe this beforecollecting any data. As this is a test for proportions andn is large,

we compute the test statistic value

z = p (1 )/n =

100

300 0.250.000625

=3.33.

Compare this to the critical values of a normal distribution and youwill see that it is significant at the 1% level, say. That is, there is verystrong evidence that more than 25% of homemakers use SNOLITE.It appears that the campaign has been successful.


Question:

If you live in California, the decision to purchase earthquakeinsurance is a critical one. An article in the Annals of the Associationof American Geographers (June 1992) investigated many factorsthat California residents consider when purchasing earthquakeinsurance. The survey revealed that only 133 of 337 randomlyselected residences in Los Angeles County were protected byearthquake insurance.

a) What are the appropriate null and alternative hypotheses to testthe research hypothesis that less than 40% of the residents of

Los Angeles County were protected by earthquake insurance?b) Do the data provide sufficient evidence to support the research

hypothesis? (Use =0.10.)

c) Calculate and interpret thep-value for the test.

Solution:

a) Appropriate hypotheses are H0 : =0.4v. H1 :


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Compare this to the critical vales of a normal distribution andyou will see that it is not significant at even the 10% level.

c) To compute thep-value for this lower-tailed test, use

P(Z

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