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?

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wifilethbridge wifilethbridge · Answer 1 · 2019-09-25T07:19:32+02:00

Answer:

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 : [tex]H_0:\mu \neq 75[/tex]

Alternate hypothesis : [tex]H_a:\mu = 75[/tex]

n = 20

Sample standard deviation s = 7

[tex]\bar{x}=78[/tex]

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

So, we will use t - test .

Formula : [tex]t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}[/tex]

Substitute the values

[tex]t = \frac{78-75}{\frac{7}{\sqrt{20}}}[/tex]

[tex]t = 1.9166[/tex]

So, t calculated = 1.9166

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

Refer the t table

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

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 .