diane charges $10 for a house call plus $35 an hour for electrical work. if she made $150 on one job,how many hours did she work
2 answers:
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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.