An associate at a large home improvement store claims that she can build a composite 12 × 20-foot deck for an average cost of $7200. We sample eight decks built by this associate and find that their average cost was $8215.22 with a standard deviation of $711.16. Does it appear from this sample that the claim of $7200 is not accurate (either too high or too low)? State the critical value (CV), test statistic (TS), and decision from the test. (Use α = 0.05.)

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance. And the true mean is significantly different from 7200.

Step-by-step explanation:

Data given and notation

[tex]\bar X=8215.22[/tex] represent the sample mean

[tex]s=711.16[/tex] represent the sample standard deviation for the sample

[tex]n=8[/tex] sample size

[tex]\mu_o =7200[/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)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is equal to 7200 or not, the system of hypothesis would be:

Null hypothesis:[tex]\mu = 7200[/tex]

Alternative hypothesis:[tex]\mu \neq 7200[/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".

Calculate the statistic

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

[tex]TS= t=\frac{8215.22-7200}{\frac{711.16}{\sqrt{8}}}=4.038[/tex]

Critical value

The degrees of freedom are given by:

[tex] df = n-1= 8-1=7[/tex]

And we need to find a critical value on the t distribution with df =7 who accumulate [tex]\alpha/2 =0.025[/tex] of the area on each tail and we got:

[tex] CV= t_{crit}= 2.365[/tex]

Decision

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance. And the true mean is significantly different from 7200.