## A diet guide claims that you will consume 120 calories from a serving of vanilla yogurt. A random sample of 22 brands of vanilla yogur

A diet guide claims that you will consume 120 calories from a serving of vanilla yogurt. A random

sample of 22 brands of vanilla yogurt produced a mean of 118.3 calories with standard deviation 3.7

calories.

Suppose that mu is the true mean number of calories for a serving of vanilla yogurt. We want to test if

the diet guide’s claim is accurate. Use significance level 0.05.

1. What are the null and alternative hypotheses?

2. What is the value of the test statistic?

3. What is the rejection region?

4. Do we reject the null hypothesis?

5. Based on this hypothesis test, do we conclude the diet guide is accurate or inaccurate?

6. Should the p-value for this test be greater than or less than the significance level?

## Answers ( )

Answer:1) Null hypothesis:[tex]\mu = 120[/tex]

Alternative hypothesis:[tex]\mu \neq 120[/tex]

2) [tex]t=\frac{118.3-120}{\frac{3.7}{\sqrt{22}}}=-2.155[/tex]

3) [tex] t_{cric}=\pm 2.08[/tex]

And the rejection zone of the null hypothesis would be [tex] |t_{calc}|>2.08[/tex]

4) Since our statistic calculated is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance

5) Since we reject the null hypothesis of accuracy we have enough evidence to conclude at 5% of significance that the procedure is not accurate

6) Since we reject the null hypothesi we need to expect that the p value would be lower than the significance level provided and that means:

[tex] p_v <\alpha[/tex]

Step-by-step explanation:Data given and notation[tex]\bar X=118.3[/tex] represent the sample mean

[tex]s=3.7[/tex] represent the sample standard deviation

[tex]n=22[/tex] sample size

[tex]\mu_o =120[/tex] represent the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

Part 1: State the null and alternative hypotheses.We need to conduct a hypothesis in order to check if the true mean is equal to 120 because that means accurate, the system of hypothesis would be:

Null hypothesis:[tex]\mu = 120[/tex]

Alternative hypothesis:[tex]\mu \neq 120[/tex]

If we analyze the size for the sample is < 30 and we don’t know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)

t-test:“Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value”.Part 2: Calculate the statistic

We can replace in formula (1) the info given like this:

[tex]t=\frac{118.3-120}{\frac{3.7}{\sqrt{22}}}=-2.155[/tex]

Part 3: Rejection regionWe calculate the degrees of freedom given by:

[tex] df = n-1= 22-1 =21[/tex]

We need to find a critical value who accumulates [tex]\alpha/2 = 0.025[/tex] of the area in the tails of the t distribution with 21 degrees of freedom and we got:

[tex] t_{cric}=\pm 2.08[/tex]

And the rejection zone of the null hypothesis would be [tex] |t_{calc}|>2.08[/tex]

Part 4Since our statistic calculated is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance

Part 5Since we reject the null hypothesis of accuracy we have enough evidence to conclude at 5% of significance that the procedure is not accurate

Part 6Since we reject the null hypothesi we need to expect that the p value would be lower than the significance level provided and that means:

[tex] p_v <\alpha[/tex]