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?

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

Step-by-step explanation:

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

Null hypothesis :

Alternate hypothesis :  

n = 20

Sample standard deviation s = 7

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

So, we will use t - test .

Formula :

Substitute the values

So, t calculated = 1.9166

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

Refer the t table

So,

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 .

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step-by-step explanation:

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