Answer:
a) Null and alternative hypothesis
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
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:
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:
Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size are:
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):
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.
Answer:
Step-by-step explanation:
Since; the density function diagrams were not included in the question; we will be unable to determine the best which depicts this problem.
However;
Let use X to represent the time required for the delivery.
Then X~N(3.8 ,0.8)
i.e
E(x) = 3.8
s.d(x) = 0.8
NOW; P(x>4) = P(X-3.8/0.8 > 4-3.8/0.8)
= P(Z > 0.25)
= 1-P(Z < 0.25)
=1 - Φ (0.25)
= 1 - 0.5987 ( from Normal table Φ (0.25) = 0.5987 )
= 0.4013
Thus; the probability a single delivery would take more than 4 hours is 0.4013
What is the z value corresponding to the interval boundary?
The z value is calculated as:
z = 0.25