8.3 The Z-Test for the Population Mean:
A one-tailed test indicates that the null hypothesis should be rejected when the
test value is in the critical region on one side of the mean. A one-tailed test is either a right-tailed test or left-tailed test, depending on the direction of the inequality of the alternative hypothesis.
Example
Critical and noncritical regions on the Standard Normal Distribution N(0,1)
(here )
Example
Finding the Critical Value for =0.01 (Right-Tailed Test) using the standard normal distribution N(0,1).
Solution
The critical value is for =0.01 is 2.33.
Let’s Do It! Critical Values.
a. Finding the Critical Value for =0.05 (Left-Tailed Test) using The standard normal distribution N(0,1).
b. Finding the Critical Value for =0.05 (Two-Tailed Test) using The standard normal distribution N(0,1).
Case 1: Testing the Mean Using a Large Sample n30
Example
A researcher reports that the average salary of assistant professors is more than $42,000. A sample of 30 assistant professors has a mean salary of $43,260. Test the claim that assistant professors earn more than $42,000 a year. The standard deviation of the sample is $5230. Use =0.05.
a) Write the hypothesis of the researcher.
b) What type of test? (Circle one)
One-sided to the right One-sided to the left Two-Sided
c) Find the critical value for =0.05 on the Z-distribution.
d) Is the observed average an extreme under the null hypothesis?
e) Summarize your result?
Solution
a) (Or )
ß This is the researcher’s claim
b) This test is a one-Sided to the right (b/c the direction of the extreme under the null is to the right. Values on far right are unlikely under the null and more likely under the alternative)
c) Critical value C.V. = InvNorm (0.95, 0, 1) = 1.65
d) To determine if the observed sample average $43,260 is extreme under the null, first standardize under the null. The standardized value is called The Test Statistic.
Test Statistic=
== 1.32
Notice how the standardized value of the observed sample average is outside the critical region. This implies that $43,260 is NOT extreme under the null hypothesis.
Therefore we don’t reject he null hypothesis.
e) Result: There is not enough evidence to support the researchers claim that the average salary is more than $42,000.
Let’s Do it! Weight Loss
A diet pill promises a weight loss of 10lbs in the first week (the fine print on the bottle indicates that the reported amount is an average weight loss). A researcher wishes to test the effectiveness of the pill. 36 subjects we selected randomly and given the pill. The amount of their weight lost in the first week is given below:
3 2 1 1 0 5 5 5 7 2 2 0
4 8 8 1 0 0 2 2 2 4 6 2
2 1 1 0 0 0 3 2 1 1 0 2
a) Write the hypothesis.
b) What type of test? (Circle one)
One-sided to the right One-sided to the left Two-Sided
c) Find the critical value for =0.10 on the Z-distribution.
d) Is the observed average an extreme under the null hypothesis?
e) Summarize your result?
Let’s Do it! Teen Spending
According to Teenage Research Unlimited, in 2007, teens spent an average of $103 per week (over the course of the year that equals $5356). A researcher believes that in the year of 2008, this average has changed. He collected information from 125 randomly selected teenagers. His sample resulted in an average spending of $118 per week with a standard deviation of $122.
a) Write the hypothesis.
b) What type of test? (Circle one)
One-sided to the right One-sided to the left Two-Sided
c) Find the critical value for =0.10 on the Z-distribution.
d) Is the observed average an extreme under the null hypothesis?
e) State your conclusion.
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8.4 The t-Distribution and Testing Hypothesis about the mean for small Sample
Properties of the t-distribution
§ The t-distribution has a symmetric bell-shaped density centered at 0, similar to the N(0,1) distribution.
§ The t-distribution is “flatter” and has “heavier tails” than the N(0,1) distribution.
§ As the sample size increases, the t-distribution approaches the N(0,1) distribution.
Example
Find the critical t value for =0.05 with d.f. = 16 for a right-tailed t test.
Solution
Find the 0.05 column in the top row and 16 in the left-hand column. Where the row and column meet, the appropriate critical value is found; it is +1.746
Let’s Do It!
Find the critical t value for 0.01 with d.f. = 22 for a left-tailed test.
Find the critical values for 0.10 with d.f. =18 for a two-tailed t test.
Example Hospital Infections
A medical investigation claims that the average number of infections per week at a hospital in southwestern Pennsylvania is 16.3. A random sample of 10 weeks had a mean number of 17.7 infections. The sample standard deviation is 1.8. Is there enough evidence to reject the investigator’s claim at 0.05?
Solution
Let’s Do It!
An educator claims that the average salary of substitute teachers in school districts in Allegheny County, Pennsylvania, is less than $60 per day. A random sample of eight school districts is selected, and the daily salaries (in dollars) are shown. Is there enough evidence to support the educator’s claim at 0.10?
Let’s Do It! State and Local Taxes
The U.S. average for state and local taxes for a family of four is $4172. A random sample of 20 families in a northeastern state indicates that they paid an annual amount of $4560 with a standard deviation of $1590. At 0.05, is there sufficient evidence to conclude that they pay more than the national average of $4172?
Let’s Do It! Doctor Visits
A report by the Gallup Poll stated that on average a woman visits her physician 5.8 times a year. A researcher randomly selects 20 women and obtained these data. At 0.05 can it be concluded that the average is different from 5.8 visits per year?
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