Answer:
a) Null and alternative hypothesis
![H_0: \mu=32.48\\\\H_a:\mu< 32.48](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu%3D32.48%5C%5C%5C%5CH_a%3A%5Cmu%3C%2032.48)
b) Test statistic t=-1.565
P-value = 0.0612
NOTE: the sample size is n=65.
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
![H_0: \mu=32.48\\\\H_a:\mu< 32.48](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu%3D32.48%5C%5C%5C%5CH_a%3A%5Cmu%3C%2032.48)
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:
![H_0: \mu=32.48\\\\H_a:\mu< 32.48](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu%3D32.48%5C%5C%5C%5CH_a%3A%5Cmu%3C%2032.48)
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:
![s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12}{\sqrt{65}}=1.4884](https://tex.z-dn.net/?f=s_M%3D%5Cdfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%3D%5Cdfrac%7B12%7D%7B%5Csqrt%7B65%7D%7D%3D1.4884)
Then, we can calculate the t-statistic as:
![t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{30.15-32.48}{1.4884}=\dfrac{-2.33}{1.4884}=-1.565](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7BM-%5Cmu%7D%7Bs%2F%5Csqrt%7Bn%7D%7D%3D%5Cdfrac%7B30.15-32.48%7D%7B1.4884%7D%3D%5Cdfrac%7B-2.33%7D%7B1.4884%7D%3D-1.565)
The degrees of freedom for this sample size are:
![df=n-1=65-1=64](https://tex.z-dn.net/?f=df%3Dn-1%3D65-1%3D64)
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):
![\text{P-value}=P(t](https://tex.z-dn.net/?f=%5Ctext%7BP-value%7D%3DP%28t%3C-1.565%29%3D0.0612)
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.
<u>Critical value approach</u>
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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.