Hypothesis Testing Questions

Practice Questions

1. The director of manufacturing at a clothing factory needs to determine whether a new machine is producing a particular type of cloth according to the manufacturer's specifications, which indicate that the cloth should have a mean breaking strength of 32 kilograms and a standard deviation of 2.72 kilograms. A sample of 49 pieces of cloth reveals a sample mean breaking strength of 31.3 kg. ( solution )

a) State the null and alternative hypotheses.

b) Is there evidence that the machine is not meeting the manufacturer's specifications for average breaking strength? (Use a 5% level of significance.)

c) Compute the p-value and interpret its meaning.

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2. According to a newspaper article, "More than three-quarters of Canadians believed the quality of the environment declined over the last decade…" This statement was based on a poll in which 1161 of the 1504 Canadians who responded agreed. Is there sufficient evidence to conclude that over 75% of the population agreed with the statement? Test at the 5% level of significance. ( solution )

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3. A machine that fills bottles of salad dressing at a food plant is working properly when it fills each bottle with 250 millilitres of liquid dressing mix. The manufacturer must be sure that the bottles contain on average no less than 250 millilitres. The standard deviation of the process is 4.5 millilitres. A sample of 50 bottles is selected periodically, and the filling line is stopped if there is evidence that the mean fill has dropped to less than 250 millilitres. Suppose the average fill is 249.5 millilitres from a sample of 50 bottles. ( solution )

At the 5% level of significance, using critical values, test if there is evidence to show that the mean fill amount is less than 250 millilitres.

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4. A trucking firm suspects the claim that the average lifetime of certain tires is at least 28,000 miles. To check the claim, the firm tests 40 such tires on its trucks and gets a mean lifetime of 27,563 miles with a standard deviation of 1,348 miles. Perform the test at the 0.01 level of significance. ( solution )

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5. A law student wants to test the claim that convicted embezzlers spend an average of 12.3 months in jail. He tests a random sample of 35 such cases and gets an average of 11.5 months with s = 3.8 months. Test at the 0.05 level of significance. ( solution )

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6. A diesel engine manufacturer claims that his engines have a mean thermal efficiency of at least 32.3%. Tests performed on 40 of his engines yielded a mean thermal efficiency of 31.8% with a standard deviation of 2.2%. Test his claim at the 0.05 level of significance. Find the tail probability or p-value associated with these test results. ( solution )

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7. A food study reports that 1 pound of goodfish yields an average of 2.41 ounces of FPC (fish-protein concentrate) used to enrich food. Is this average supported by a study
of 30 samples of goodfish which show an average of 2.44 ounces of FPC/lb. and a standard deviation of 0.07 ounces if we use the: ( solution )

a) 0.05 level of significance;

b) 0.01 level of significance?

c) Find the tail probability corresponding to the 2 values of z in a) and b) to decide whether Ho can be rejected at the 0.02 level of significance.

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8. An automobile rustproofing company claims that their method protects cars
for an average of 55 months. This hypothesis is tested against the alternative
that the protection lasts for more than 55 months. A random sample of 200 cars
produces an average protection time of 56 months with a standard deviation of 15 months.
Test the claim at the 5% level of significance and draw a conclusion. ( solution )

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9. A deodorant manufacturer claims that the mean drying time of their product is at most
15 minutes. A sample of 16 cans yielded a mean drying time of 18 min. with s = 6 minutes.

a) Test the claim at the 5% significance level and draw a conclusion.

b) Find the p -value and decide at what level of significance we would not reject Ho.

( solution )

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10. The yield of alfalfa, in tons per acre, from 6 test plots of organic soil is : ( solution )

Test at the 5% level of significance whether this supports the claim that the average yield for this kind of alfalfa in this type of organic soil is 1.5 tons per acre.

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11. The owner of a gasoline station wants to study gasoline purchasing habits by motorists at his station. A random sample of 60 motorists in a certain week gave these results:

· Mean or Average volume purchased: = 42.8 litres, s = 6.7 litres

· 11 motorists purchased premium-grade gasoline

a) At the 5% level of significance, does evidence show that the mean purchase is different from 40 litres? ( solution )

b) At the 5% level of significance, is there enough evidence to assume that fewer than 20% of the motorists purchase premium-grade gasoline? ( solution )

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12. A television critic claims that 80% of all viewers find the noise level of a certain commercial objectionable. If a random sample of 320 viewers includes 245 who find the noise level objectionable, do these results support his claim at the 5% level of significance? ( solution )

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Solutions

1. a) Solution

b) Solution: Since we are given a standard deviation, we use the z-test:

Level of significance = 0.05.

Criteria: if |z | > 1.96, reject Ho.

Now we calculate z

Since – 1.80 > – 1.96, we cannot reject Ho.

c) Solution:

Since z = – 1.80, the p-value = 2(P (z) > 1.80) = 2(0.5000 – 0.4641) = 2(0.0359) = 0.0718

The p-value is 0.0718. This means that if the null hypothesis is true, the probability of obtaining a sample mean more extreme than 31.3 kg is 7.18%.

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2. Solution:

Since this question deals with proportions, we must use the test statistic for proportions. The formula for this statistic is:

upper tail test at 5%

We need to know for the statistic, so we calculate it:

Test Criteria: If z > 1.645 reject Ho.

We can now calculate the statistic:

Since 2 > 1.645, we must reject Ho.

Testing with the p-value:

P(z > 2) = (0.5000 – 0.4772) = 0.0228

Since 0.0228 < 0.05, there is evidence to reject the null hypothesis and to conclude that over 75% of the population believed that the quality of the environment declined over the last decade.

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3. Solution:

Hypothesis Test on Mean
Value of interest: µ , mean fill of bottles
Hypotheses: Ho:	Ha: l < 250
Level of significance: = 5 %	Criteria: if z < – 1.645, reject Ho
Test Statistic:	Decision: Since – 0.756 is not less than – 1.645, we can't reject Ho
Conclusion: the test indicates there is not enough evidence to prove the bottles are being filled with less than 250 mL of salad dressing.

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4. Solution:

Lower Tail Test on a Mean
Ho: l m 28,000 miles	Ha: l < 28,000 miles
Level of significance: = 1 %	Criteria: if z < – 2.33, reject Ho
Statistic:
Decision: we cannot reject the null hypothesis Ho. Conclusion: we conclude that the average lifetime of certain tires is at least 28,000 miles.

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5. Solution:

2-tail test on mean length of jail sentence for embezzlers.
Ho:µ = 12.3 months	Ha: months
Level of significance: = 5 %	Criteria: if | z | > 1.96, reject Ho
Statistic:
Decision: we cannot reject the null hypothesis Ho. Conclusion: the claim that convicted embezzlers spend an average of 12.3 months in jail cannot be refuted by the results of this test.

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6. Solution:

Lower Tail Test on a Mean
Ho: l m 32.3 % efficiency rating	Ha: l < 32.3 % efficiency rating
Level of significance: = 5 %	Criteria: if z < – 1.645, reject Ho
Statistic:
Decision: we cannot reject the null hypothesis Ho Conclusion: his engines have a mean thermal efficiency of at least 32.3% the p-value = the probability of being in the tail created by the statistic z = – 1.44 the p-value = 0.5000 – 0.4251 = 0.0749 p > 5%, so we don't heave Ho.

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7. Solution:

a) 0.05 level of significance; .......... b) 0.01 level of significance?

2-Tail Test on a Mean
Ho:µ = 2.41 ounces	Ha: ounces
Level of significance: a) = 5 %, b) = 1 %	Criteria: a) if | z | > 1.96, reject Ho b) if | z | > 2.575, reject Ho
Statistic :
Decision: a) since 2.35 > 1.96, we must reject the null hypothesis Ho. b) since 2.35 < 2.575, we cannot reject Ho.

c) Solution:

p-value = 2(0.5000 – 0.4906) = 0.0188
since 0.0188 < 0.02, we would reject Ho at the 2% level of significance.

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8. Solution:

value of interest: mean protection time from rustproofing.
Ho: l = 55	Ha: l > 55
Level of significance: = 5 %	Criteria: if z > 1.645, reject Ho
Statistic:
Decision: we cannot reject the null hypothesis Ho. Conclusion: it seems that the rustproofing protection does last for an average of 55 months.

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9) Solution:

hypothesis test on mean drying time (small sample)
Ho: l [ 15	Ha: l > 15
Level of significance: = 5 %, df = 15	Criteria: if t > 1.753, reject Ho
Statistic:
Decision: we must reject the null hypothesis Ho Conclusion: the test shows the maximum drying time is more than 15 minutes.

b) Solution:

From the table, we find that for 15 degrees of freedom, a t-value of 2.131 creates an upper tail of 2.5%. Since the test t-value = 2, we would not reject Ho at the 3% level of significance.

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10. Solution:

hypothesis test on mean yield of alfalfa (small sample)
	s by formula = 0.434
Ho: µ = 1.5	Ha:
Level of significance: = 5 %, df = 5	Criteria: if | t | > 2.571, reject Ho
Statistic:
Decision: we cannot reject the null hypothesis Ho Conclusion: the test shows that the average yield for this kind of alfalfa in this type of organic soil is 1.5 tons per acre.

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11. a) Solution:

hypothesis test on mean volume of gasoline purchased
	s = 6.7 litres
Ho: µ = 40 litres	Ha:
Level of significance: = 5 %,	Criteria: if | z | > 1.96, reject Ho
Statistic:
Decision: since 3.24 > 1.96, we must reject the null hypothesis Ho Conclusion: the test shows that the mean gasoline purchase is different from 40 litres.

b) Solution:

Ho: p = 0.20	Ha: p < 0.20 (1-tail test)
Level of significance:	Criteria: if z < – 1.645, reject Ho
Statistic:
Decision: – 0.329 > – 1.645, so we cannot reject the null hypothesis Conclusion: we can't assume that fewer than 20% of the motorists buy premium-grade gas.

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12. Solution:

Ho: p = 0.80	Ha: (2-tail test)
Level of significance:	Criteria: if | z | > 1.96, reject Ho
Statistic:
Decision: – 1.55 is > – 1.96, so we cannot reject the null hypothesis Conclusion: the critic's claim is probably true and 80% of the viewers find the noise in the ad hard to take.

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