Question 291738: In an article appearing in Today’s Health a writer states that the average number of calories in a serving of popcorn is 75. To determine if the average number of calories in a serving of popcorn is different from 75, a nutritionist selected a random sample of 20 servings of popcorn and computed the sample mean number of calories per serving to be 78 with a sample standard deviation of 7.
1)Compute the z or t value of the sample test statistic
2)At the = .05 level of significance, does the nutritionist have enough evidence to reject the writer’s claim?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! In an article appearing in Today’s Health a writer states that the average number of calories in a serving of popcorn is 75.
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To determine if the average number of calories in a serving of popcorn is different from 75, a nutritionist selected a random sample of 20 servings of popcorn and computed the sample mean number of calories per serving to be 78 with a sample standard deviation of 7.
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Ho: u = 75
Ha: u is not equal to 75
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1)Compute the z or t value of the sample test statistic
t(78) = (78-75)/[7/sqrt(20)] = 1.9166
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p-value = 2*P(t>1.9166 with df = 19) = 0.0705
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2)At the = .05 level of significance, does the nutritionist have enough evidence to reject the writer’s claim?
Ans; Since the p-value is greater than 5%, fail to reject Ho.
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Cheers,
Stan H.
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