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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. At the a = .05 level of significance, does the nutritionist have enough evidence to reject the writer’s claim?

User Rabbitt
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Answer:

The nutritionist have enough evidence to reject the writer’s claim .

Explanation:

Claim : The average number of calories in a serving of popcorn is different from 75

Null hypothesis :
H_0:\mu \\eq 75

Alternate hypothesis :
H_a:\mu = 75

n = 20

Sample standard deviation s = 7


\bar{x}=78

Since n < 30 and population standard deviation is not given .

So, we will use t - test .

Formula :
t = (x-\mu)/((s)/(√(n)))

Substitute the values


t = (78-75)/((7)/(√(20)))


t = 1.9166

So, t calculated = 1.9166

Degree of freedom = n-1 = 20-1 = 19

Refer the t table

So,
t_{((\alpha)/(2),df)}=t_{((0.05)/(2),19)}=2.093

Since t critical > t calculated

So, we are fail to reject the null hypothesis .

So, the nutritionist do not have enough evidence to reject the writer’s claim .

User Kevin Wittek
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