According to the Vivino website, suppose the mean price for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is $32.48. A New England-based lifestyle magazine wants to determine if red wines of the same quality are less expensive in Providence, and it has collected prices for 65 randomly selected red wines of similar quality from wine stores throughout Providence. The mean and standard deviation for this sample are $30.15 and $12, respectively.
(a) Develop appropriate hypotheses for a test to determine whether the sample data support the conclusion that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48. (Enter != for ≠ as needed.)
H0:
Ha:
(b) Using the sample from the 60 bottles, what is the test statistic? (Round your answer to three decimal places.)
Using the sample from the 60 bottles, what is the p-value? (Round your answer to four decimal places.)
p-value =
(c) At α = 0.05, what is your conclusion?
Do not reject H0. We can conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.
Reject H0. We can conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.
Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.
Reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.
(d) Repeat the preceding hypothesis test using the critical value approach.
State the null and alternative hypotheses. (Enter != for ≠ as needed.)
H0:
Ha:
Find the value of the test statistic. (Round your answer to three decimal places.)
State the critical values for the rejection rule. Use
α = 0.05.
(Round your answers to three decimal places. If the test is one-tailed, enter NONE for the unused tail.)
test statistic ≤
test statistic ≥
State your conclusion.
Do not reject H0. We can conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.
Reject H0. We can conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.
Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.
Reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

c) Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

d) Null and alternative hypothesis

[tex]H_0: \mu=32.48\\\\H_a:\mu< 32.48[/tex]

Test statistic t=-1.565

Critical value tc=-1.669

t>tc --> Do not reject H0

Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Then, the null and alternative hypothesis are:

[tex]H_0: \mu=32.48\\\\H_a:\mu< 32.48[/tex]

The significance level is 0.05.

The sample has a size n=65.

The sample mean is M=30.15.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12.

The estimated standard error of the mean is computed using the formula:

[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12}{\sqrt{65}}=1.4884[/tex]

Then, we can calculate the t-statistic as:

[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{30.15-32.48}{1.4884}=\dfrac{-2.33}{1.4884}=-1.565[/tex]

The degrees of freedom for this sample size are:

[tex]df=n-1=65-1=64[/tex]

This test is a left-tailed test, with 64 degrees of freedom and t=-1.565, so the P-value for this test is calculated as (using a t-table):

[tex]\text{P-value}=P(t<-1.565)=0.0612[/tex]

As the P-value (0.0612) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Critical value approach

At a significance level of 0.05, for a left-tailed test, with 64 degrees of freedom, the critical value is t=-1.669.

As the test statistic is greater than the critical value, it falls in the acceptance region.

The null hypothesis failed to be rejected.