Aim: Review Session 4 Hypothesis Testing

Final 5/9 and 5/11

Recap: Sample Proportion

• When a random variable has two possible outcomes, we can use the sample proportion as a summary

• X is the interpreted random variable used• n = known sample size

Xp

n

Example: Sample Proportion

• In a random sample of 150 seniors, 45 report taking transportation to school. What is the sample proportion?

45.30

150p 105

.70150

p

1. Tells you the proportion that take transportation

2. Tells you the proportion that doesn’t take transportation.

Sample Size

• A sample size of 30 or more is considered significantly large enough to conduct statistical analysis

• Sample size of 30 or more z test• Sample size less than 30 t test

Two-Tailed Vs. One-Tailed Test

• Null hypothesis: the actual information given• Alternative hypothesis: the claim/what you

are testing

Z Test

• FOR SAMPLE GREATER THAN OR EQUAL TO 30

Z Test: Table E• You recently received a job with a company that manufactures an

automobile antitheft device. To conduct an advertising campaign for your product, you need to make a claim about the number of automobile thefts per year. Since the population of various cities in the United States varies, you decide to use rates per 10,000 people. (The rates are based on the number of people living in the cities.) Your boss said that last year the theft rate per 10,000 people was 44 vehicles. You want to see if it has changed. The following are rates per 10,000 people for 36 randomly selected locations in the United States.

50 42 125 62 134 73 39 69 23 94 73

24 51 55 26 66 41 67 15 53 56 91

20 78 70 25 62 115 17 36 58 56 33

75 20 16

Procedure

1. Hypothesis2. Find test value

3. Find critical value (P-value) Table E4. Make a decision5. Draw a conclusion/Summary

Xz

n

Decision Rule when using P-values

• If P-value ≤ α, reject the null hypothesis.• If P-value > α do not reject the null hypothesis.

T Test

• SAMPLE SIZE LESS THAN 30

T test: Table F• A tobacco company claims that its best-selling cigarettes

contain at most 40mg of nicotine. This claim is tested at the 1% significance level by using the result of 15 randomly selected cigarettes. The mean was 42.6mg and the standard deviation was 3.7mg. Evidence suggests that nicotine is normally distributed. Information from a computer output of the hypothesis test is listed.

Sample mean = 42.6 Significant level = 0.01 Sample Standard Deviation = 3.7 P-Value = 0.008

Procedure

1. Hypothesis 2. Decide if two-tailed test and find d.f.3. Find critical value from table F4. Find test value

5. Make a decision6. Draw a conclusion/Summary

Xt

s

n

Decision for t test

• When test value is greater than critical value, we reject the null hypothesis.

Solution to Examples

• Solutions to both examples are scanned in a pdf and can be found on the course website titled: “hypothesis testing 2.solutions.pdf”

What you need tomorrow?

• Bring your own z and t table….I will not provide you one and you may not share with the person next to you

• Calculator if you choose• You must have memorized the formula to get

the test values