Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). The test statistic in a left-tailed test is z = -1.83.
Using Excel or a statistical calculator, the p-value that corresponds to a z-value of -1.83 in a left-tailed test is 0.0336.
(You can use Excel to calculate this by entering “=Normsdist(-1.83)” in a cell. Don’t include the quotation marks though.)
Since the p-value of 0.0336 is less than the level of significance of 0.05, the decision would be to reject the null hypothesis.
Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). With H1: p < 3/5, the test statistic is z = -1.68.
Based on the alternative hypothesis (H1), this is a left-tailed test.
The corresponding p-value is 0.0465.
Since the p-value of 0.0465 is less than the level of significance of 0.05, the decision is to reject the null hypothesis.
Formulate the indicated conclusion in nontechnical terms. Be sure to address the original claim. The owner of a football team claims that the average attendance at games is over 694, and he is therefore justified in moving the team to a city with a larger stadium. Assuming that a hypothesis test of the claim has been conducted and that the conclusion is failure to reject the null hypothesis, state the conclusion in nontechnical terms.
The null hypothesis would be that average attendance is less than or equal to 694.
The alternative hypothesis would be that average attendance is over 694.
Since the conclusion is “failure to reject the null hypothesis”, there is insufficient evidence to support the claim that attendance is now over 694. Based on this test,
the owner does not have sufficient justification to move the team to a city with a
larger stadium.
Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type II error for the test. A consumer advocacy group claims that the mean mileage for the Carter Motor Company's new sedan is less than 32 miles per gallon. Identify the type I error for the test.
In this case the null hypothesis would be that the mean mileage is 32 or more miles per gallon.
The alternative hypothesis would be that the mean mileage is less than 32 miles per gallon.
A Type I error occurs when the null hypothesis is rejected, although it is actually true. In this scenario that would mean that it is concluded that the new sedan’s mileage is less than 32 miles per gallon, when the mileage is actually greater than or equal to 32 miles per gallon.