Suppose that a group of researchers is planning to test a new weight loss supplement. They have selected a random sample of 45 people who are trying to lose weight and plan to measure the amount of weight lost after one month of using the supplement. Assume that the researchers know from prior experiments that the standard deviation of weight lost in one month, , is 1.7 lb.To show that the supplement is effective, they plan to use a one-sample z‑test of Ho : u= 0lb against H1 : u > 0lb , where is the mean amount of weight lost in one month. They have also determined that, for a test with a significance level of 0.05, the power of the test is 0.9347 if the mean amount of weight lost is actually 0.8 lb.What is the probability that the researchers will reject their null hypothesis if the mean amount of weight lost is 0.8 lb or more? Give your answer as a percentage, precise to two decimal places.

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Asked Nov 18, 2019
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Suppose that a group of researchers is planning to test a new weight loss supplement. They have selected a random sample of 45 people who are trying to lose weight and plan to measure the amount of weight lost after one month of using the supplement. Assume that the researchers know from prior experiments that the standard deviation of weight lost in one month, , is 1.7 lb.

To show that the supplement is effective, they plan to use a one-sample z‑test of Ho : u= 0lb against H1 : u > 0lb , where is the mean amount of weight lost in one month. They have also determined that, for a test with a significance level of 0.05, the power of the test is 0.9347 if the mean amount of weight lost is actually 0.8 lb.

What is the probability that the researchers will reject their null hypothesis if the mean amount of weight lost is 0.8 lb or more? Give your answer as a percentage, precise to two decimal places.

 

 

 

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Expert Answer

Step 1

It is given that, the population standard deviation is 1.7 and the sample size considered is 45.

Here, the standard error of mean is, σ/sqrt(n) = 1.7/sqrt(45) =0.2534

For a 0.05 level of significance, critical value of z is 1.96.

The corresponding margin of error is, z× standard error = 1.96×0.2534 =0.4967.

Step 2

The probability that the researcher rejects the null hypothesis if mean...

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P(X 0.4967)-1-P(X 0.4967) =1-0.1157 Using the EXCEL formulae: =0.8843 NORMDIST(0.4967,0.8,0.253 4,1)

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