Chapter Nine Hypothesis Testing

Hypothesis testing is used to make decisions concerning the value of a parameter. 9.1 Introduction to Statistical Tests

Null Hypothesis: H0 a working hypothesis about the population parameter in question

The value specified in the null hypothesis is often: a historical value a claim a production specification

Alternate Hypothesis: H1 any hypothesis that differs from the null hypothesis

An alternate hypothesis is constructed in such a way that it is the one to be accepted when the null hypothesis must be rejected.

A manufacturer claims that their light bulbs burn for an average of 1000 hours. We have reason to believe that the bulbs do not last that long. Determine the null and alternate hypotheses.

A manufacturer claims that their light bulbs burn for an average of 1000 hours. ... The null hypothesis (the claim) is that the true average life is 1000 hours. H0: m = 1000

… A manufacturer claims that their light bulbs burn for an average of 1000 hours. We have reason to believe that the bulbs do not last that long. ... If we reject the manufacturer’s claim, we must accept the alternate hypothesis that the light bulbs do not last as long as 1000 hours. H1: m < 1000

Types of Statistical Tests Left-tailed: H1 states that the parameter is less than the value claimed in H0. Right-tailed: H1 states that the parameter is greater than the value claimed in H0. Two-tailed: H1 states that the parameter is different from (  ) the value claimed in H0. Pg. 362 – yellow box

Given the Null Hypothesis H0:  = k If you believe that  is less than k, Use the left-tailed test: H1:  < k Pg. 363 – Table 9-1

Given the Null Hypothesis H0:  = k If you believe that  is more than k, Use the right-tailed test: H1:  > k Pg. 363 – Table 9-1

Given the Null Hypothesis H0:  = k If you believe that  is different from k, Use the two-tailed test: H1:   k Pg. 363 – Table 9-1

General Procedure for Hypothesis Testing Formulate the null and alternate hypotheses. Take a simple random sample. Compute a test statistic corresponding to the parameter in H0. Assess the compatibility of the test statistic with H0.

Hypothesis Testing about the Mean of a Normal Distribution with a Known Standard Deviation  Pg. 363 – yellow box

P-value of a Statistical Test Assuming H0 is true, the probability that the test statistic (computed from sample data) will take on values as extreme as or more than the observed test statistic is called the P-value of the test The smaller the P-value computed from sample data, the stronger the evidence against H0. Pg. 365 – yellow box

P-values for Testing a Mean Using the Standard Normal Distribution

P-value for a Left-tailed Test P-value = probability of getting a test statistic less than Pg. 366 – Table with all 3 cases

P-value for a Right-tailed Test P-value = probability of getting a test statistic greater than

P-value for a Two-tailed Test P-value = probability of getting a test statistic lower than or higher than

Types of Errors in Hypothesis Testing Type I Type II

Type I Error rejecting a null hypothesis which is, in fact, true

Type II Error not rejecting a null hypothesis which is, in fact, false

Type I and Type II Errors Pg. 367 – Table 9-2

Level of Significance, Alpha (a) the probability of rejecting a true hypothesis Alpha is the probability of a type I error Pg. 367 – yellow box

Type II Error Beta = β = probability of a type II error (failing to reject a false hypothesis) In hypothesis testing α and β values should be chosen as small as possible. Usually α is chosen first. 157

Power of the Test = 1 – β The probability of rejecting H0 when it is in fact false = 1 – b. The power of the test increases as the level of significance (a) increases. Using a larger value of alpha increases the power of the test but also increases the probability of rejecting a true hypothesis. Pg. 367 – paragraph before numbering

Probabilities Associated with a Statistical Test Pg. 368 – Table 9-3

Hypotheses and Types of Errors A fast food restaurant indicated that the average age of its job applicants is fifteen years. We suspect that the true age is lower than 15. We wish to test the claim with a level of significance of a = 0.01. Determine the Null and Alternate hypotheses and describe Type I and Type II errors.

… average age of its job applicants is fifteen years … average age of its job applicants is fifteen years. We suspect that the true age is lower than 15. H0: m = 15 H1: m < 15

H0: m = 15 H1: m < 15 a = 0.01 A type I error would occur if we rejected the claim that the mean age was 15, when in fact the mean age was 15 (or higher). The probability of committing such an error is as much as 1%.

H0: m = 15 H1: m < 15 a = 0.01 A type II error would occur if we failed to reject the claim that the mean age was 15, when in fact the mean age was lower than 15. The probability of committing such an error is called beta.

Concluding a Hypothesis Test Using the P-value and Level of Significance α If P-value < α reject the null hypothesis and say that the data are statistically significant at the level α. If P-value > α, do not reject the null hypothesis. Pg. 369 – green box procedure

Basic Components of a Statistical Test Null hypothesis, alternate hypothesis and level of significance Test statistic and sampling distribution P-value Test conclusion Interpretation of the test results

Null Hypothesis, Alternate Hypothesis and Level of Significance If the sample data evidence against H0 is strong enough, we reject H0 and adopt H1. The level of significance, α, is the probability of rejecting H0 when it is in fact true.

Test Statistic and Sampling Distribution Mathematical tools to measure compatibility of sample data and the null hypothesis

P-value The probability of obtaining a test statistic from the sampling distribution that is as extreme as or more extreme than the sample test statistic computed from the data under the assumption that H0 is true

Test Conclusion If P-value < α reject the null hypothesis and say that the data are statistically significant at the level α. If P-value > α, do not reject the null hypothesis.

Interpretation of Test Results Give a simple explanation of conclusion in the context of the application.

Reject or ... When the sample evidence is not strong enough to justify rejection of the null hypothesis, we fail to reject the null hypothesis. Use of the term “accept the null hypothesis” should be avoided. When the null hypothesis cannot be rejected, a confidence interval is frequently used to give a range of possible values for the parameter.

Fail to Reject H0 There is not enough evidence to reject H0. The null hypothesis is retained but not proved.

Reject H0 There is enough evidence to reject H0. Choose the alternate hypothesis with the understanding that it has not been proven.

9.2 Testing the Mean  When  is Known Let x be the appropriate random variable. Obtain a simple random sample (of size n) of x values and compute the sample mean x. State the null and alternate hypotheses and set the level of significance α. If x has a normal distribution, any sample size will work. If we cannot assume a normal distribution, use n > 30.

Testing the Mean  When  is Known Use the test statistic:

Testing the Mean  When  is Known Use the standard normal distribution and the type of test (one-tailed or two-tailed) to find the P-value corresponding to the test statistic. If the P-value < α, then reject H0. If the P-value > α, then do not reject H0. State your conclusion.

Testing the Mean  When  is Known: Example Your college claims that the mean age of its students is 28 years. You wish to check the validity of this statistic with a level of significance of a = 0.05. Assume the standard deviation is 4.3 years. A random sample of 49 students has a mean age of 26 years.

Hypothesis Test Example Perform a ________-tailed test. two Level of significance = α = 0.05

Sample Test Statistic

For a two-tailed test: P-value = 2P(z < 3.26) = 2(0.0006) = 0.0012 Sample Results For a two-tailed test: P-value = 2P(z < 3.26) = 2(0.0006) = 0.0012 154

P-value and Conclusion α = 0.05. Since the P-value < α , we reject the null hypothesis. We conclude that the true average age of students is not 28.

Testing the Mean  When  is Unknown Let x be the appropriate random variable. Obtain a simple random sample (of size n) of x values and compute the sample mean x. State the null and alternate hypotheses and set the level of significance α. If x has a mound shaped symmetric distribution, any sample size will work. If we cannot assume this, use n > 30.

Testing the Mean  When  is Unknown Use the test statistic:

Testing the Mean  When  is Unknown Use the Student’s t distribution and the type of test (one-tailed or two-tailed) to find (or estimate) the P-value corresponding to the test statistic. If the P-value < α, then reject H0. If the P-value > α, then do not reject H0. State your conclusion.

Using Table 4 to Estimate P-values Use one-tailed areas as endpoints of the interval containing the P-value for one-tailed tests.

P-value for One-tailed Tests

Using Table 4 to Estimate P-values Use two-tailed areas as endpoints of the interval containing the P-value for one-tailed tests.

P-value for Two-tailed Tests

Testing the Mean  When  is Unknown: Example The Parks Department claims that the mean weight of fish in a lake is 2.1 kg. We believe that the true average weight is lower than 2.1 kg. Assume that the weights are mound-shaped and symmetric and a sample of five fish caught in the lake weighed an average of 1.99 kg with a standard deviation of 0.09 kg.

Determine the P-value when testing the claim that the mean weight of fish caught in a lake is 2.1 kg (against the alternate that the weight is lower). A sample of five fish weighed an average of 1.99 kg with a standard deviation of 0.09 kg.

Test the Claim Using α = 10% Null Hypothesis: H0:  = 2.1 kg Alternate Hypothesis: H1:  < 2.1 kg α = 0.10

We will complete a left-tailed test with:

The Test Statistic t

Using Table 4 with t = 2.73 and d.f. = 4 Sample t =  2.73

The t value is between two values in the chart The t value is between two values in the chart.Therefore the P-value will be in a corresponding interval. Sample t =  2.73

Since we are performing a one-tailed test, we use the “one-tail area” line of the chart. Sample t =  2.73

Since we are performing a one-tailed test, we use the “one-tail area” line of the chart. Sample t =  2.73

0.025 < P-value < 0.050 … Sample t = 2.73

0.025 < P-value < 0.050 Since the range of P-values is less than a (10%), we reject the null hypothesis.

Interpret the results: At level of significance 10% we rejected the null hypothesis that the mean weight of fish in the lake was 2.1 kg. Based on our sample data, we conclude that the true mean weight is actually lower than 2.1 kg.

Critical Region (Traditional) Method for Hypothesis Testing An alternate technique to the P-value method Logically equivalent to the P-value method

Critical Region Procedure for Testing  When  is Known Let x be the appropriate random variable. Obtain a simple random sample (of size n) of x values and compute the sample mean x. State the null and alternate hypotheses and set the level of confidence α. If x has a normal distribution, any sample size will work. If we cannot assume a normal distribution, use n > 30.

Critical Region Method for Testing the Mean  When  is Known Use the test statistic:

Critical Region Method for Testing the Mean  When  is Known Using the level of significance α and the alternate hypothesis, show the critical region and critical values on a graph of the sampling distribution. Conclude the test. If the test statistic is in the critical region, then reject H0. If not, do not reject H0. State your conclusion.

Most Common Levels of Significance α = 0.05 and α = 0.01

Critical Region(s) The values of x for which we will reject the null hypothesis. The critical values are the boundaries of the critical region(s).

Concluding Tests Using the Critical Region Method Compare the sample test statistics to the critical value(s) For a left-tailed test: If the sample test statistic is < critical value, reject H0. If the sample test statistic is > critical value, fail to reject H0.

Critical Region for H0:  = k Left-tailed Test

Concluding Tests Using the Critical Region Method Compare the sample test statistics to the critical value(s) For a right-tailed test: If the sample test statistic is > critical value, reject H0. If the sample test statistic is < critical value, fail to reject H0.

Critical Region for H0:  = k Right-tailed Test

Concluding Tests Using the Critical Region Method Compare the sample test statistics to the critical value(s) For a two-tailed test: If the sample test statistic lies beyond the critical values, reject H0. If the sample test statistic lies between the critical values, fail to reject H0.

Critical Region for H0:  = k Two-tailed Test

Critical Values z0 for α = 0.05 and α = 0.01: Left-tailed Test

Critical Values z0 for α = 0.05 and α = 0.01: Right-tailed Test

Critical Values z0 for α = 0.05 and α = 0.01: Two-tailed Test

Testing the Mean  When  is Known: Example Your college claims that the mean age of its students is 28 years. You wish to check the validity of this statistic with a level of significance of a = 0.05. Assume the standard deviation is 4.3 years. A random sample of 49 students has a mean age of 26 years.

Hypothesis Test Example H0: m = 28 H1: m ¹ 28 Perform a ________-tailed test. two Level of significance = α = 0.05

Sample Test Statistic

Sample Results 154

Critical Region for a Two-tailed Test with α = 0.05

Our z =  3.26 falls within the critical region.

Since the test statistic is in the critical region we… Reject the Null Hypothesis.

Conclusion We conclude that the true average age of students is not 28.

9.3 Testing a Proportion p We will test claims that a given percentage of the population fits a certain description.

Let r be the binomial random variable, the number of successes out of n independent trials.

For large samples (np > 5 and nq > 5):

Three Types of Tests of Hypotheses for Tests of Proportions Left-tailed tests Right-tailed tests Two-tailed tests

Left-Tailed Test H0: p = k H1: p < k

Right-Tailed Test H0: p = k H1: p > k

Two-Tailed Test H0: p = k H1: p  k

Testing a Proportion p Consider a binomial experiment with n trials. Let p represent the population probability of success. Let q = 1  p represent the population probability of failure. Let r be a random variable that represents the number of successes out of the n binomial trials.

Testing a Proportion p State the null and alternate hypotheses and set the level of significance α. The number of trials should be sufficiently large so that both np > 5 and nq > 5. (Use p from the null hypothesis.)

Testing a Proportion p

Testing a Proportion p Use the standard normal distribution and the type of test (one-tailed or two-tailed) to find the P-value corresponding to the test statistic. If the P-value < α, then reject H0. If the P-value > α, then do not reject H0. State your conclusion.

In the past, college officials observed that 40% of students took advantage of early registration. This semester, of 4830 students, 2077 took advantage of early registration. Use a 5% level of significance to test the claim that a higher percentage of students now participates in early registration.

In the past, college officials observed that 40% of students took advantage of early registration. ... H0: m = 0.40

Use a right-tailed test. … test the claim that a higher percentage of students now participates in early registration. H1: m > 0.40 Use a right-tailed test.

… This semester, of 4830 students, 2077 took advantage of early registration. ...

The corresponding z value:

Use Table 3 to Determine the P-value Associated with z = 4.26, The P-value is approximately 0.000

Since α = 0.05 and P < α REJECT the null hypothesis.

Conclusion Since we have rejected the hypothesis that 40% of students participate in early registration, we conclude that: A percentage higher than 40% of students now participates in early registration.

Test the hypothesis involving a proportion: H0: p = 0.70 H1: p ¹ 0.70 Use a = 0.01 Suppose that in 120 trials there were 80 successes.

We find that:

Use Table 3 to find the P-value associated with z = –0.72. P(z £ –0.72) = 0.2358 Since the test is a two-tailed test, double the area in the left tail to find P. P = 2(0.2358) = 0.4716

When the P-value is greater than the level of significance, we do not reject the null hypothesis. FAIL to reject the null hypothesis.