7 . Sampling distribution of the difference between two means A statistician is interested in the effectiveness of a weight-loss supplement.

Question

7 . Sampling distribution of the difference between two means A statistician is interested in the effectiveness of a weight-loss supplement. She randomly selects two independent samples. Individuals in the first sample of size n1 = 24 take the weight-loss supplement. Individuals in the second sample of size n2 = 21 take a placebo. Individuals in both samples follow identical exercise and diet programs. At the end of the study, the statistician measures the weight loss (in percent) of each participant. Suppose that the weight loss (in percent) for individuals taking the supplement is normally distributed with μ 1 = 13.00 and σ 21 = 38.56, and that the weight loss (in percent) for individuals taking a placebo is normally distributed with μ 2 = 2.00 and σ 22 = 9.73. The difference between the two sample means follows anormal distribution with a mean of11.00 and standard deviation equal to1.4387 . Use the Distributions tool to help answer the question that follows. What is the probability that the mean weight loss (in percent) for the first sample exceeds the mean weight loss (in percent) for the second sample by at least 12%?

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Brielle 4 weeks 2021-09-14T14:47:26+00:00 1 Answer 0

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    2021-09-14T14:49:24+00:00

    Answer:

    0.229

    Step-by-step explanation:

    Given that the difference between the two sample means follows anormal distribution with a mean of11.00 and standard deviation equal to1.4387

    A statistician is interested in the effectiveness of a weight-loss supplement. She randomly selects two independent samples. Individuals in the first sample of size n1 = 24 take the weight-loss supplement. Individuals in the second sample of size n2 = 21 take a placebo. Individuals in both samples follow identical exercise and diet programs. At the end of the study, the statistician measures the weight loss (in percent) of each participant.

    We find that mean difference actual = 13-2 = 11

    Probability that difference >12 =P(Z>\frac{12-11}{1.3487} \\)

    =P(Z>0.742)=.0.229

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