Question

A fast-food chain claims one large order of its fries weighs 170 grams. Joe thinks he is getting less than what the restaurant advertises. He weighs the next 12 random orders of fries before he eats them and finds the sample mean is 165.9 grams and the standard deviation is 11.98 grams. What conclusion can be drawn at α = 0.10? (2 points)

Answer 1

There is sufficient evidence to prove the fast-food chain advertisement is false

Explanation:

Answer 2

`t=(165.9-170)/((11.98)/(√(12)))=-1.186`

`p_v =P(t_((11))<-1.186)=0.1303`

If we compare the p value and the significance level given
`\alpha=0.1` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is not significantly lower than 170 at 10% of signficance.

Explanation:

Data given and notation

`\bar X=165.9` represent the sample mean

`s=11.98` represent the sample standard deviation

`n=12` sample size

`\mu_o =170` represent the value that we want to test

`\alpha=0.1` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is lower than 170, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 170`

Alternative hypothesis:
`\mu < 170`

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:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (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".

Calculate the statistic

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

`t=(165.9-170)/((11.98)/(√(12)))=-1.186`

P-value

The first step is calculate the degrees of freedom, on this case:

`df=n-1=12-1=11`

Since is a one side lower test the p value would be:

`p_v =P(t_((11))<-1.186)=0.1303`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.1` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is not significantly lower than 170 at 10% of signficance.