Question

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?

Answer 1

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 .