Section 9.2 Hypothesis Testing Proportions P-Value
Section 9.2
Hypothesis Testing ProportionsP-Value
p is the percent from the hypothesisn is the sample size is the percent from the sample
Finding
In the problem they will give you a sample percent or they will give you a sample space and a part of that sample space.
If they give you the sample space and the part, just divide the small number by the large number and you have . Make sure you round to 3 decimal places.
What type of test is it?
% Left-tailed Right-tailed Two-tailed
.20 80 -.84 .84 1.28
.15 85 -1.03 1.03 1.44
.10 90 -1.28 1.28 1.645
.05 95 -1.645 1.645 1.96
.02 98 -2.05 2.05 2.33
.01 99 -2.33 2.33 2.575
Z-scores for and the tailed tests
Steps to Solve the Problem:
Step 1: Define the hypothesisStep 2: Compute the value of the z test statistic or the P-valueStep 3: Determine if the hypothesis is one-tailed or two-tailedStep 4: Compare the z-test statistic to the z for the α or compare the P-value of the z-test
statistic to the α.Step 5: Make the decision based on the decision rules. Step 6: Determine the conclusion. State the decision in terms of the claim being tested.
Step 3: One-tailed or two-tailed?Since the problem states “less than” the hypothesis is one-tailed.
Step 4: Since the confidence level is 95% we can use an α of .05 or the z-score of -1.645. (this is negative since it is left tailed)Since the test statistic is -0.88, we can compare this to -1.645, or we can find the p-value of .1894 and compare it to .05. In both situations the test statistic is greater than the confidence level.
Step 5: Since the test statistic is greater than the
confidence level, we will fail to reject the null.
Meaning there is not sufficient evidence to support the claim.
The z-score for the .10 significance level is 1.645. Since 2.23 is greater it falls in the rejection region.
The P-value for 2.33 is 0.0129. Since this value is less than .10 it falls in the rejection region.
NOTE: Since this is two-tailed, for comparing the p-values you must divide the significance level by 2.