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 α = 0.05 level of significance, does the nutritionist have enough evidence to reject the writer’s claim?

a. No
b. Yes
c. Cannot determine

Answer 1

Answer:

$t=\frac{78-75}{\frac{7}{\sqrt{20}}}=1.917$  

The degrees of freedom are given by:

$df= n-1 = 20-1=19$

The p value would be given by:

$p_v =2*P(t_{19}>1.917)=0.0704$  

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from 75 and the best option would be:

a. No

Step-by-step explanation:

Information given

$\bar X=78$ represent the sample mean

$s=7$ represent the sample standard deviation

$n=20$ sample size  

$\mu_o =75$ represent the value that we want to test  

$\alpha=0.05$ represent the significance level for the hypothesis test.  

t would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to check if the true mean is different from 75, the system of hypothesis would be:  

Null hypothesis:$\mu =75$  

Alternative hypothesis:$\mu \neq 75$  

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

Replacing we got:

$t=\frac{78-75}{\frac{7}{\sqrt{20}}}=1.917$  

The degrees of freedom are given by:

$df= n-1 = 20-1=19$

The p value would be given by:

$p_v =2*P(t_{19}>1.917)=0.0704$  

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from 75 and the best option would be:

a. No