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    Sampling Distribution and Test Proportion

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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.

    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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    Solution Preview

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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 ...

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    The expert examines sampling distribution and test proportions.

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