Ch. 9 examples

9.1/9/2: Test on a single population mean Intro: So far our work has been on estimation, or confidence intervals Now the focus is on Hypothesis Testing

Types of Hypotheses The null hypothesis H0 is one which expresses the current state of nature of belief about a population The alternative (or research) hypothesis (H 1 or H a ) is one which reflects the researcher’s belief. (It will always disagree with the null hypothesis). Note: the alternative hypothesis can be one or two tailed.

Level of significance Generally, this value is .05, .10, or .01

Step 1 and 2: H 0 and H 1 We have a 7 step process to follow (template available on the webpage) The first two steps cover defining the two hypotheses

Should you use a 2 tail, or a right, or left tail test? 2-tailed ex H0: µ= __ H1: µ ≠ __ Left tail ex H0: µ = ___ H1: µ < ___ Right tail ex H0: µ = __ H1: µ > __ Test whether the average in the bag of numbers is or isn’t 100. Test if a drug had any effect on heartrate. Test if a tutor helped the class do better on the next test. Test if a drug improved elevated cholesterol.

Summary of Hypothesis test steps Null hypothesis H0, alternative hypothesis H1, and preset α Test statistic and sampling distribution P-value and/or critical value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results 2-tailed ex H0: µ= 100 H1: µ ≠ 100 α = 0.05 Left tail ex H0: µ = 200 H1: µ < 200 Right tail ex H0: µ = 50 H1: µ > 50

Type I and Type II error

Probabilities associated with error

Example #1- numbers in a bag Recall that I claimed that my bag of numbers had a mean µ = 100 and a standard deviation =21.9. Test this hypothesis if your sample size n= 20 and your sample mean x-bar was 90.

Ex #1- Hypothesis Test for numbers in a bag α = 0.05 Z = = P-value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results

Ex #2– new sample mean for numbers in a bag If the sample mean is 95, redo the test: H0: µ = 100 H1: µ ≠ 100 α = 0.05 Z = = P-value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results

Ex #3: Left tail test- cholesterol A group has a mean cholesterol of 220. The data is normally distributed with σ= 15 After a new drug is used, test the claim that it lowers cholesterol. Data: n=30, sample mean= 214.

Ex #3- cholesterol- test H0: µ 220 (fill in the correct hypotheses here) H1: µ 220 α = 0.05 Z = = P-value and/or critical value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results

Ex #4- right tail- tutor Scores in a MATH117 class have been normally distributed, with a mean of 60 all semester. The teacher believes that a tutor would help. After a few weeks with the tutor, a sample of 35 students’ scores is taken. The sample mean is now 62. Assume a population standard deviation of 5. Has the tutor had a positive effect?

Ex #4: tutor Z = = P-value and/or critical value 4. Test conclusion H0: µ 60 (fill in the correct hypotheses here) H1: µ 60 α = 0.05 Z = = P-value and/or critical value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results

9.2– t tests Just like with confidence intervals, if we do not know the population standard deviation, we substitute it with s (the sample standard deviation) and Run a t test instead of a z test

Ex #5– t test – placement scores The placement director states that the average placement score is 75. Based on the following data, test this claim. Data: 42 88 99 51 57 78 92 46 57

Ex #5 t test – placement scores H0: µ 75 fill in the correct hypothesis here H1: µ 75 α = 0.05 t = = P-value and/or critical value 4. Test conclusion If p-value ≤ α, we reject H0 and say the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results

Ex #6- placement scores The head of the tutoring department claims that the average placement score is below 80. Based on the following data, test this claim. Data: 42 88 99 51 57 78 92 46 57

Ex #6– t example H0: µ 80 (fill in the correct hypotheses here) α = 0.05 t = = P-value and/or critical value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results

Ex #7- salaries– t A national study shows that nurses earn $40,000. A career director claims that salaries in her town are higher than the national average. A sample provides the following data: 41,000 42,500 39,000 39,999 43,000 43,550 44,200

Ex #7- salaries H0: µ 40000 (fill in the correct hypotheses here) α = 0.05 t = = P-value and/or critical value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results

Traditional Critical Value Approach Redo Example #1 Recall that I claimed that my bag of numbers had a mean µ = 100 and a standard deviation =21.9. Test this hypothesis if your sample size n= 20 and your sample mean x-bar was 90.

Ex#1 redone with CV Z = = CV 4. Test conclusion α = 0.05 Z = = CV 4. Test conclusion If p-value ≤ α, then test value is in RR, and we reject H0 and say that the data are significant at level α If p-value > α, then test value is not in RR, and we do not reject H0 5. Interpretation of test results

Ex #3 redone with CV A group has a mean cholesterol of 220. The data is normally distributed with σ= 15 After a new drug is used, test the claim that it lowers cholesterol. Data: n=30, sample mean= 214.

Ex#3- 5 steps- done with CV H1: µ 220 (fill in) α = 0.05 Z = = CV 4. Test conclusion If p-value ≤ α, then test value is in RR, and we reject H0 and say that the data are significant at level α If p-value > α, then test value is not in RR, and we do not reject H0 5. Interpretation of test results

9.3 Testing Proportion p Recall confidence intervals for p: ± z

Hypothesis tests for proportions Null hypothesis H0, alternative hypothesis H1, and preset α 2. Test statistic and sampling distribution P-value and/or critical value z= = 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results 2-tailed ex H0: p= .5 H1: p ≠ .5 α = 0.05 Left tail ex H0: p = .7 H1: p < .7 Right tail ex H0: p = .2 H1: p > .2

Ex #8- proportion who like job The HR director at a large corporation estimates that 75% of employees enjoy their jobs. From a sample of 200 people, 142 answer that they do. Test the HR director’s claim.

Ex #8 H0: p=.75 (fill in hypothesis) H1: p α = Null hypothesis H0, alternative hypothesis H1, and preset α H0: p=.75 (fill in hypothesis) H1: p α = Test statistic and sampling distribution Z = = 3. P-value and/or critical value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results

Ex #9 Previous studies show that 29% of eligible voters vote in the mid-terms. News pundits estimate that turnout will be lower than usual. A random sample of 800 adults reveals that 200 planned to vote in the mid-term elections. At the 1% level, test the news pundits’ predictions.

Ex #9 H0: p (fill in hypothesis) H1: p α = Null hypothesis H0, alternative hypothesis H1, and preset α H0: p (fill in hypothesis) H1: p α = Test statistic and sampling distribution Z = = 3. P-value and/or critical value 4. Test conclusion If p-value ≤ α, we reject H0 and say that the data are significant at level α If p-value > α, we do not reject H0 5. Interpretation of test results