10
Stat 2263
Chapter 7 Hypothesis Testing with One Sample
Section 7-1 Overview
· Researchers are interested in answering many types of questions
Ø Does a new medication lower a patient’s blood pressure significantly?
Ø Does a new diet soda have a higher preference rating?
· In statistics, a hypothesis is a claim or statement about a property of a population
· The null hypothesis, denoted H0 is usually a tentative assumption about a population parameter and, in our case, will be a statement about the value of a population parameter, like p, m or s. It will contain a condition of equality.
· The alternative hypothesis, denoted HA is the opposite of what is stated in the null hypothesis.
· Our objective is to determine whether to reject H0 (there is sufficient evidence to infer that HA is true) or fail to reject H0 (there is insufficient evidence to infer that HA is true). We will be making inferences on the population based on the sample results.
Section 7-2 Basics of Hypothesis Testing
Procedure for Test of Hypothesis
· Null and Alternative Hypothesis
There are three types of null/alternative hypotheses we can use:
left- tailed right-tailed two tailed
Note the condition of equality in each of the null hypotheses. In your text, they solely use = symbol in the null hypothesis. This is theoretically incorrect, but is adopted by many texts. On tests and assignments I will accept either.
Eg. The manager of the Danvers-Hilton Resort Hotel stated that the average guest bill for a weekend is $600 or less. A member of the hotel’s accounting staff noticed that guests’ bills have been increasing of late. The accountant will use a sample of weekend guest bills to test the manager’s claim. What type of test is this? Right tailed
Eg. State the null and alternative hypothesis for each conjecture.
a. A researcher thinks that if expectant mothers use vitamin pills the birthweight of the babies will increase. The average birthweights of the population is 8.6 pounds.
b. An engineer hypothesizes that the mean number of defects can be decreased in a manufacturing process of compact discs by using robots instead of humans for certain tasks. The mean number of defective discs per 1000 is 18.
c. A psychologist feels that playing soft music during a test will change the results of the test. In the past, the mean of the scores is 73.
· Level of Significance (a) If unspecified use a=0.05 It is the probability of rejecting the null hypothesis when the null hypothesis is actually true or the probability of a Type I error.
· Test Statistic: or or
· Rejection or Critical Region: This is the set of all values of the test statistic that will lead us to reject the null hypothesis. The set up of the Critical Region depends on the test of hypothesis
· Calculations & P-Value – The p value or probability value is the probability of getting a value of the test statistic as extreme or more extreme than the one calculated.
· Conclusion – reject or fail to reject the null hypothesis. We decide this using two different methods: either based on the Rejection Region or based on the p-value. If the p-value is less than the level of significance we reject the null hypothesis.
This format is to be used at all times hypothesis testing is used.
P-Value Approach
· The p-value of a test of hypothesis is the probability of observing a test statistic at least as extreme as the one computed, given H0 is true. It measures the support (or lack of support) provided by the sample for H0
· If the p-value £ a, the value of the test statistic is in the rejection region
· Reject H0 if p-value £ a
· Most computer packages calculate p-value and not the rejection region. In all of our tests and assignments I will expect both the p-value approach and the rejection region, If a is unspecified, assume it to be 0.05.
There are two possible errors to be made.
Ø Type I error or rejecting H0 when H0 is true. The probability of Type I error is denoted a. This is type of error is controlled by the experimenter who specifies a maximum allowable error, usually less than 10 %.
Ø Type II error or accepting H0 when it is false. The probability of Type II error is denoted b.
Null Hypothesis / True / FalseReject / a
Significance level
P(Type I error) / 1-b
Power of the test
Fail to Reject / 1-a
Confidence Level / b
P(Type II error)
Exercises pages 335 – 336
Section 7-3 Testing a Claim About a Population Proportion
Ø Psychology Today reported that 56 % of the households in the United States have Internet access (May 21, 2001). The parameter 56 % is called a proportion.
Ø Proportions can be obtained from samples or populations. The population parameter is denoted p, while the sample proportion is denoted .
Ø , where x is the number of successes in the sample and n is the sample size. In Chapter 9, we learned that was approximately normally distributed, with mean p and standard deviation , , as long as and . Therefore, , as long as and .
There are three types of null/alternative hypotheses we can use:
left- tailed right-tailed two tailed
Procedure for Test of Hypothesis
· Null and Alternative Hypothesis (one of the above sets)
· Level of Significance (a) If unspecified use a=0.05
· Test Statistic:
· Rejection Region:
Two-tailed : Reject H0 if Z < or if Z >
Right-tailed: Reject H0 if Z >
Left-tailed: Reject H0 if Z < -
· Calculations & P-Value
· Conclusion
Example
A telephone company representative estimates that 40% of its customers have call-waiting service. To test this hypothesis, she selected a sample of 100 customers and found that 37 of them had call waiting. At a=0.01, is there enough evidence to reject the claim?
Note that and
a=0.01
Test Statistic:
Rejection Region: Reject H0 if Z < -2.575 or if Z > 2.575
Calculations:
Conclusion: Fail to reject the null hypothesis at a=0.01. There is insufficient evidence to infer that the true proportion of customers with call-waiting service is significantly different from 40 %
Example
A statistician read that at least 77% of the population oppose replacing $5 bills with $5 coins. To see if this claim is valid, the statistician selected a sample of 80 people and found that 55 were opposed to replacing the $5 bills. At a=0.01, test the claim that at least 77 % of the population are opposed to the change.
Note that and
a=0.01
Test Statistic:
Rejection Region: Reject H0 if Z < -2.33
Calculations:
,
Conclusion: Fail to reject the null hypothesis at a=0.01. There is insufficient evidence to infer that the alternative hypothesis is true. The claim may be correct.
Try Exercises on pages 344 – 347
Section 7-4 Testing a Claim About the Population Mean m, when the Population Standard Deviation, s, is Known
Assumptions:
1. That the population that the sample is drawn from is normally distributed or n>30
2. Sample is a simple random sample
3. The population standard deviation, ơ, is known.
Note: If s is unknown we will use the t distribution , in Sec 7-5.
There are three types of null/alternative hypotheses we can use:
left- tailed right-tailed two tailed
Procedure for Test of Hypothesis
· Null and Alternative Hypothesis (one of the above sets)
· Level of Significance (a) If unspecified use a=0.05
· Test Statistic:
· Rejection Region:
Two-tailed : Reject H0 if Z < or if Z >
Right-tailed: Reject H0 if Z >
Left-tailed: Reject H0 if Z < -
· Calculations & P-Value
· Conclusion
Example:
The Medical Rehabilitation Foundation reports that the average cost of rehabilitation for stroke victims is $ 24 672 (September 1995). To see if the average cost of rehabilitation at a particular hospital, a researcher selected a random sample of 35 stroke victims at the hospital and found that the average cost of their rehabilitation is
$25 226. The standard deviation of the population is $3251. At , an it be concluded that the average cost of stroke rehabilitation at this particular hospital is different from $24 672?
Z=1.01 There is insufficient sample evidence to support the claim that the average cost of stroke rehab is different from 24 672.
Example:
A researcher wishes to test the claim that the average age of lifeguards in Ocean City exceeds 24 years. She selects a sample of 36 guards and finds the sample mean to be 24.7. The population standard deviation is 2 years. Is there evidence to support this claim at 0.01 level of significance?
Z=2.10, p-value= .0179, Fail to reject null hypothesis
Exercises pages 351 – 352
Section 7-5 Testing a Claim About the Population Mean m, when the Population Standard Deviation, s, is Not Known
Assumptions:
1. That the population that the sample is drawn from is normally distributed or n>30
2. Sample is a simple random sample
3. The population standard deviation, ơ, is unknown.
There are three types of null/alternative hypotheses we can use:
left- tailed right-tailed two tailed
Procedure for Test of Hypothesis
· Null and Alternative Hypothesis (one of the above sets)
· Level of Significance (a) If unspecified use a=0.05
· Test Statistic:
· Rejection Region:
Two-tailed : Reject H0 if t < or if t >
Right-tailed: Reject H0 if t >
Left-tailed: Reject H0 if t < -
· Calculations & P-Value
· Conclusion
Example:
A job placement director claims that the average salary for a nursing assistant is $24 000. A sample of 10 nursing assistants has a mean of $23 450 and a standard deviation of $400. Assuming salaries are normally distributed, is there sufficient evidence to reject the director’s claim at 0.05 level of significance?
a=0.05
Test Statistic:
Rejection Region: Reject H0 if t < -2.262 or t > 2.262
Calculations:
p-value > 2(0.10)
Conclusion: Reject H0, there is sufficient evidence to infer that m is not equal to 50
Example:
A physician claims that a runner’s maximal volume oxygen is greater than the average of all adults. A sample of 15 joggers has a mean of 40.6 ml/kg and a standard deviation of 6 ml/kg. If the average of all adults is 36.7 ml/kg, is there enough evidence to support the physician’s claim at 0.05 level of significance? Assume normality.
t=2.517, .01<p value<.025
Exercises page 360 – 363
Section 7-6 Testing a Claim About a Standard Deviation or Variance of a Population
Assumptions for the use of distribution:
Ø The sample must be randomly selected from the population
Ø The population must be normally distributed for the variable under study
Procedure for Hypothesis Testing About :
1. Null and Alternative Hypothesis
Two-tailed Right-tailed Left-tailed
Note: For tests about the population standard deviation, s, everything about this procedure remains the same.
2. Level of Significance a
3. Test Statistic
4. Rejection Region
a) For 2-tailed test: Reject H0 if < or if >
b) For a right tailed test: Reject H0 if >
c) For a left-tailed test: Reject H0 if <
5. Calculations & a bound on p-value
6. Conclusion
Example:
A researcher knows from past studies that the standard deviation of the time it takes to examine a patient is 16.8 minutes. A random sample of 24 patients is selected and examination time recorded. The standard deviation was 12.5 minutes. At .05 level of significance can it be concluded that the standard deviation has changed? Assume examination time is normally distributed.
Chi squared = 12.733, p value is between .05 and .10, insufficient evidence to conclude that std dev has changed.
Example:
A hospital administrator believes that the standard deviation of the number of people using outpatient surgery per day is greater than eight. A random sample of 15 days is selected and the data is shown below. At 0.10 level of significance is there enough evidence to support the administrator’s claim? Assume normality.
25 30 5 15 18
42 16 9 10 12
12 38 8 14 27
Test Statistic:
Rejection Region: Reject H0 if χ2 > 21.064
Calculations:
.01< p-value <.025
Conclusion:
Reject the null hypothesis. There is sufficient evidence to
infer that the variance has exceeds 64.
Exercises pages 368 - 369