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Sampling Distribution and Test Proportion

Use the following information for problems 1-3. Decide whether the normal sampling distribution can be used. If it can be used, test the claim about the population proportion p at the given level of significance using the given sample statistics.

1. Claim: p > 0:30;
= 0:05. Sample statistics: p^ = 0:35, n = 50.

2. Claim: p = 0:80;
= 0:10. Sample statistics: p^ = 0:78, n = 19.

3. Claim: p < 0:22;
= 0:10. Sample statistics: p^ = 0:17, n = 200.

4. Do You Eat Breakfast?
A research center estimates that at least 35% of U.S. adults eat breakfast every day. In a random sample of 300 U.S. adults, 38% say they eat breakfast every day. At alpha = 0:05, conclude a hypothesis test for the population proportion, p. (Note: Please see note above which explains requirements for a hypothesis test.)

5. Genetically Modified Foods
An environmentalist claims that more than 60% of British consumers are concerned about the use of genetic modification in food production and want to avoid genetically modified foods. You want to test this claim. You find that in a random sample of 100 British consumers, 68% say that they are concerned about the use of genetic
modification in food production and want to avoid genetically modified foods. At alpha = 0:10, can you support the environmentalist's claim?

Use the following information for problems 6-8. Find the critical values for the indicated test for a population variance, sample size n, and the level of significance

6. Right-tailed test, n = 18,
= 0:025

7. Left-tailed test, n = 13,
= 0:10

8. Two-tailed test, n = 22,
= 0:05

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Use the following information for problems 1-3. Decide whether the normal sampling distribution can be used. If it can be used, test the claim about the population proportion p at the given level of significance using the given sample statistics.
1. Claim: p > 0:30;
= 0:05. Sample statistics: p^ = 0.35, n = 50.
Since np=0.35*50=17.5>5, n(1-p)=50*(1-0.35)=32.5>5,
We could use the normal distribution.
Ho: p<=0.30
Ha: P>0.30
This is a one tailed t test.
At 0.05 significance level, the degree of freedom is 50-1=49,
The critical t value is 1.68.
Test value t=(0.35-0.30)/sqrt(0.30*(1-0.30)/50)=0.772
Since 0.772<1.68, we could not reject the null hypothesis.
Based on the test, we could not conclude that p>0.30.
2. Claim: p = 0:80;
= 0:10. Sample statistics: p^ = 0:78, n = 19.
Since ...

Solution Summary

The expert examines sampling distribution and test proportions.

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