Statistics for Business and Economics Chapter 7 Inferences Based on a Single Sample: Tests of Hypotheses

Content 1.The Elements of a Test of Hypothesis 2.Formulating Hypotheses and Setting Up the Rejection Region 3.Observed Significance Levels: p-Values 4.Test of Hypothesis about a Population Mean: Normal (z) Statistic 5.Test of Hypothesis about a Population Mean: Student’s t-Statistic 6.Large-Sample Test of Hypothesis about a Population Proportion

7.1 The Elements of a Test of Hypothesis

Hypothesis Testing Population I believe the population mean age is 50 (hypothesis). Mean  X = 20 Random sample Reject hypothesis! Not close.

What’s a Hypothesis? A statistical hypothesis is a statement about the numerical value of a population parameter. I believe the mean GPA of this class is 3.5! © T/Maker Co.

Determining the Target Parameter Parameter Key Words or PhrasesType of Data µMean; averageQuantitative pProportion; percentage; fraction; rate Qualitative  2 (Will not covered in this course) Variance; variability; spread Quantitative

Null Hypothesis The null hypothesis, denoted H 0, represents the hypothesis that will be “retained” unless the data provide convincing evidence that it is false. This usually represents the “status quo” or some claim about the population parameter that the researcher wants to test. You may think of null hypothesis as the “favored” hypothesis; we reject it in favor of the alternative hypothesis H a if and only if the evidence provided by the sample data are strong against H 0 and in favor of H a. “retain H 0 ” is commonly referred to as “do not reject”. Stated in one of the following forms H 0 :   some value) H 0 :  ≤  some value) H 0 :   some value)

Alternative Hypothesis The alternative (research) hypothesis, denoted H a, represents the hypothesis that will be accepted only if the data provide convincing evidence of its truth. This usually represents the values of a population parameter for which the researcher wants to gather evidence to support.

Alternative Hypothesis 1.Opposite of null hypothesis 2.The hypothesis that will be accepted only if the data provide convincing evidence of its truth 3.Designated H a 4.Stated in one of the following forms H a :   some value) H a :   some value) H a :   some value)

Identifying Hypotheses Example 1: If the hypothesis of a researcher is that the population mean is not 3, set-up the hypotheses to be tested. Steps: State the question statistically  ≠ 3 State the opposite statistically  = 3 State the null hypothesis statistically H 0 :   3 State the alternative hypothesis statistically H a :  ≠ 3

Identifying Hypotheses Example 2: If the hypothesis of a researcher is that the population mean is greater than 3, set-up the hypotheses to be tested. Steps: State the question statistically  > 3 State the opposite statistically  ≤ 3 State the null hypothesis statistically H 0 :  ≤  3 State the alternative hypothesis statistically H a :  > 3

Identifying Hypotheses State the question statistically:  = 12 State the opposite statistically:   12 Select the alternative hypothesis: H a :   12 State the null hypothesis: H 0 :  = 12 Example 3: Is the population average amount of TV viewing 12 hours?

Identifying Hypotheses State the question statistically:   12 State the opposite statistically:  = 12 Select the alternative hypothesis: H a :   12 State the null hypothesis: H 0 :  = 12 Example 4: Is the population average amount of TV viewing different from 12 hours?

Identifying Hypotheses State the question statistically:   20 State the opposite statistically:   20 Select the alternative hypothesis: H a :   20 State the null hypothesis: H 0 :   20 Example 5: Is the average cost per hat less than or equal to $20?

Identifying Hypotheses State the question statistically:   25 State the opposite statistically:   25 Select the alternative hypothesis: H a :   25 State the null hypothesis: H 0 :   25 Example 6: Is the average amount spent in the bookstore greater than $25?

Test Statistic The test statistic is a sample statistic, computed from information provided in the sample, that the researcher uses to decide between the null and alternative hypotheses.

Type I Error A Type I error occurs if the researcher rejects the null hypothesis in favor of the alternative hypothesis when, in fact, H 0 is true. The probability of committing a Type I error is denoted by .

Example

Rejection Region The rejection region of a statistical test is the set of possible values of the test statistic for which the researcher will reject H 0 in favor of H a.

Type II Error A Type II error occurs if the researcher retains the null hypothesis when, in fact, H 0 is false. The probability of committing a Type II error is denoted by .

Conclusions and Consequences for a Test of Hypothesis True State of Nature ConclusionH 0 TrueH a True Do not reject H 0 (Assume H 0 True) Correct decisionType II error (probability  ) Reject H 0 (Assume H a True) Type I error (probability  ) Correct decision

Elements of a Test of Hypothesis 1.Null hypothesis (H 0 ): A theory about the specific values of one or more population parameters. The theory generally represents the status quo, which we adopt until it is proven false. 2.Alternative (research) hypothesis (H a ): A theory that contradicts the null hypothesis. The theory generally represents that which we will adopt only when sufficient evidence exists to establish its truth.

Elements of a Test of Hypothesis 3.Test statistic: A sample statistic used to decide whether to reject the null hypothesis. 4.Rejection region: The numerical values of the test statistic for which the null hypothesis will be rejected. The rejection region is chosen so that the probability is  that it will contain the test statistic when the null hypothesis is true, thereby leading to a Type I error. The value of  is usually chosen to be small (e.g.,.01,.05, or.10) and is referred to as the level of significance of the test.

Elements of a Test of Hypothesis 5.Assumptions: Clear statement(s) of any assumptions made about the population(s) being sampled. 6.Experiment and calculation of test statistic: Performance of the sampling experiment and determination of the numerical value of the test statistic.

Elements of a Test of Hypothesis 7.Conclusion: a.If the numerical value of the test statistic falls in the rejection region, we reject the null hypothesis and conclude that the alternative hypothesis is true. b.We know that the hypothesis-testing process will lead to this conclusion incorrectly (Type I error) only 100  % of the time when H 0 is true.

Elements of a Test of Hypothesis 7.Conclusion: b.If the test statistic does not fall in the rejection region, we do not reject H 0. Thus, we reserve judgment about which hypothesis is true. c.We do not conclude that the null hypothesis is true because we do not (in general) know the probability  that our test procedure will lead to an incorrect acceptance of H 0 (Type II error).

7.2 Setting Up the Rejection Region

Steps for Selecting the Null and Alternative Hypotheses 1.The rejection region is determined by alternative hypothesis. a. One-tailed, upper-tailed (e.g., H a : µ > 2,400) b. One-tailed, lower-tailed (e.g., H a : µ < 2,400) c. Two-tailed (e.g., H a : µ ≠ 2,400)

One-Tailed Test A one-tailed test of hypothesis is one in which the alternative hypothesis is directional and includes the symbol “.”

Two-Tailed Test A two-tailed test of hypothesis is one in which the alternative hypothesis does not specify departure from H 0 in a particular direction and is written with the symbol “ ≠.”

Rejection Region (One-Tail Test-lower tailed) HoHo Value Critical Sample Statistic Rejection Region Fail to Reject Region Sampling Distribution 1 –  Level of Confidence

Rejection Regions (One-Tailed Test-Upper-tailed) HoHo Value Critical Sample Statistic Rejection Region Fail to Reject Region Sampling Distribution 1 –  Level of Confidence

Rejection Regions (Two-Tailed Test) HoHo Value Critical Sample Statistic Rejection Region Rejection Region Fail to Reject Region Sampling Distribution 1 –  Level of Confidence

Rejection Regions (Two-Tailed level  =0.05 Test) HoHo Value Critical  Sample Statistic Rejection Region Rejection Region Fail to Reject Region Sampling Distribution 1 –  Level of Confidence 

Rejection Regions Alternative Hypotheses Lower- Tailed Upper- Tailed Two-Tailed  =.10 z < –1.282z > 1.282z  =.05 z < –1.645z > 1.645z 1.96  =.01 z < –2.326z > 2.326z 2.575

7.3 Observed Significance Levels: p-Values

p-Value Probability of obtaining a test statistic more extreme (  or  than actual sample value, given H 0 is true Can be thought of as a measure of the “credibility” of the null hypothesisH 0.  is the nominal level of significance. This value is assumed by an analyst. p-value is also probability for making type-I error. But, p-value is called “observed level of significance”. It is used to make rejection decision If p-value  , do not reject H 0 If p-value < , reject H 0

The p-value shows our confidence to reject null hypothesis. If this value is smaller than , then the probability that we will reject null hypothesis when it is true is even smaller than the maximum tolerated error probability, . So we can conclude that null hypothesis is wrong and can be rejected in favor of alternative hypothesis. The smaller the p-value is, the more confident we are with our decision to reject H 0.

Steps for Calculating the p-Value for a Test of Hypothesis 1.Determine the value of the test statistic z corresponding to the result of the sampling experiment.

Steps for Calculating the p-Value for a Test of Hypothesis 2a. If the test is one-tailed, the p-value is equal to the tail area beyond z in the same direction as the alternative hypothesis. Thus, if the alternative hypothesis is of the form >, the p-value is the area to the right of, or above, the observed z-value. Conversely, if the alternative is of the form <, the p-value is the area to the left of, or below, the observed z-value.

Steps for Calculating the p-Value for a Test of Hypothesis 2b.If the test is two-tailed, the p-value is equal to twice the tail area beyond the observed z-value in the direction of the sign of z – that is, if z is positive, the p-value is twice the area to the right of, or above, the observed z-value. Conversely, if z is negative, the p-value is twice the area to the left of, or below, the observed z-value.

Reporting Test Results as p-Values: How to Decide Whether to Reject H 0 1.Choose the maximum value of  that you are willing to tolerate. 2.If the observed significance level (p-value) of the test is less than the chosen value of , reject the null hypothesis. Otherwise, do not reject the null hypothesis. 3.Decision is always the same as found with the level , critical value, rejection region approach. 4.Typical values for  are 0.01, 0.05, 0.10.

Two-Tailed z Test p-Value Example Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes showed x = The company has specified  to be 15 grams. Find the p-value. How does it compare to  =.05? 368 gm.

Two-Tailed z Test p-Value Solution z z value of sample statistic (observed) 

Two-Tailed Z Test p-Value Solution 1/2 p-Value z value of sample statistic (observed) p-Value is P(z  –1.50 or z  1.50) z 01.50–1.50 From z table: lookup .5000 –  

Two-Tailed z Test p-Value Solution 1/2 p-Value /2 p-Value.0668 p-Value is P(z  –1.50 or z  1.50) =.1336 z 01.50–1.50

Two-Tailed z Test p-Value Solution 01.50–1.50 z Reject H 0 1/2 p-Value = /2  =.025 p-Value =.1336   =.05 Do not reject H 0. Test statistic is in ‘Do not reject’ region

One-Tailed z Test p-Value Example Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed x = The company has specified  to be 15 grams. Find the p- value. How does it compare to  =.05? 368 gm.

One-Tailed z Test p-Value Solution z  z value of sample statistic

One-Tailed z Test p-Value Solution Use alternative hypothesis to find direction p-Value is P(z  1.50) z value of sample statistic   p-Value z From z table: lookup .5000 – 

One-Tailed z Test p-Value Solution p-Value.0668 z value of sample statistic From z table: lookup 1.50 Use alternative hypothesis to find direction.5000 –   p-Value is P(z  1.50) =.0668   z

 =.05 One-Tailed z Test p-Value Solution z Reject H 0 p-Value =.0668 (p-Value =.0668)  (  =.05). Do not reject H 0. Test statistic is in ‘Do not reject’ region

p-Value Thinking Challenge You’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is less than 32 mpg. Similar models have a standard deviation of 3.8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30.7 mpg. What is the p- value? How does it compare to  =.01?

Use alternative hypothesis to find direction  p-Value Solution* z 0–2.65 z value of sample statistic  From z table: lookup  p-Value –  p-Value is P(z  -2.65) =.004. p-Value < (  =.01). Reject H 0.

7.4 Test of Hypotheses about a Population Mean: Normal (z) Statistic

Large-Sample Test of Hypothesis about µ One-Tailed TestTwo-Tailed Test H 0 : µ = µ 0 H a : µ < µ 0 H a : µ ≠ µ 0 (or H a : µ > µ 0 ) Test Statistic:Test Statistic:

Large-Sample Test of Hypothesis about µ One-Tailed Test Rejection region: z < –z  (or z > z   when H a : µ > µ 0 ) where z  is chosen so that P(z > z  ) =  You may also use p-value to give your decision. If p-value  , do not reject H 0 If p-value < , reject H 0

Large-Sample Test of Hypothesis about µ Two-Tailed Test Rejection region: |z| > z  where z  is chosen so that P(|z| > z  ) =  /2 Note: µ 0 is the symbol for the numerical value assigned to µ under the null hypothesis.

Conditions Required for a Valid Large- Sample Hypothesis Test for µ 1.A random sample is selected from the target population. 2.The sample size n is large (i.e., n ≥ 30). (Due to the Central Limit Theorem, this condition guarantees that the test statistic will be approximately normal regardless of the shape of the underlying probability distribution of the population.)

Possible Conclusions for a Test of Hypothesis 1.If the calculated test statistic falls in the rejection region, reject H 0 and conclude that the alternative hypothesis H a is true. State that you are rejecting H 0 at the  level of significance. Remember that the confidence is in the testing process, not the particular result of a single test.

Possible Conclusions for a Test of Hypothesis 2.If the test statistic does not fall in the rejection region, conclude that the sampling experiment does not provide sufficient evidence to reject H 0 at the  level of significance. [Generally, we will not “accept” the null hypothesis unless the probability  of a Type II error has been calculated.]

Two-Tailed z Test Example Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes had x = The company has specified  to be 25 grams. Test at the.05 level of significance. 368 gm.

Two-Tailed z Test Solution H 0 : H a :   n  Critical Value(s): Test Statistic: Decision: Conclusion:  = 368   z 01.96– Reject H Do not reject at  =.05 No evidence average is not 368

Two-Tailed z Test Thinking Challenge You’re a Q/C inspector. You want to find out if a new machine is making electrical cords to customer specification: average breaking strength of 70 lb. with  = 3.5 lb. You take a sample of 36 cords & compute a sample mean of 69.7 lb. At the.05 level of significance, is there evidence that the machine is not meeting the average breaking strength?

Two-Tailed z Test Solution* H 0 : H a :  = n = Critical Value(s): Test Statistic: Decision: Conclusion:  = 70   z 01.96– Reject H Do not reject at  =.05 No evidence average is not 70

One-Tailed z Test Example Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed x = The company has specified  to be 25 grams. Test at the.05 level of significance. 368 gm.

One-Tailed z Test Solution H 0 : H a :  = n = Critical Value(s): Test Statistic: Decision: Conclusion:  = 368  > z Reject Do not reject at  =.05 No evidence average is more than 368

One-Tailed z Test Thinking Challenge You’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is at least 32 mpg. Similar models have a standard deviation of 3.8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30.7 mpg. At the.01 level of significance, is there evidence that the miles per gallon is less than 32?

One-Tailed z Test Solution* H 0 : H a :  = n = Critical Value(s): Test Statistic: Decision: Conclusion:  = 32  < z Reject Reject at  =.01 There is evidence average is less than 32

7.5 Test of Hypothesis about a Population Mean: Student’s t-Statistic

Small-Sample Test of Hypothesis about µ One-Tailed Test H 0 : µ = µ 0 H a : µ µ 0 ) Test statistic: Rejection region: t < –t  (or t > t  when H a : µ > µ 0 ) where t  and t  are based on (n – 1) degrees of freedom

Small-Sample Test of Hypothesis about µ Two-Tailed Test H 0 : µ = µ 0 H a : µ ≠ µ 0 Test statistic: Rejection region: |t| > t 

Conditions Required for a Valid Small- Sample Hypothesis Test for µ 1.A random sample is selected from the target population. 2.The population from which the sample is selected has a distribution that is approximately normal.

Two-Tailed t Test Example Does an average box of cereal contain 368 grams of cereal? A random sample of 36 boxes had a mean of and a standard deviation of 12 grams. Test at the.05 level of significance. 368 gm.

Two-Tailed t Test Solution H 0 : H a :  = df = Critical Value(s): Test Statistic: Decision: Conclusion:  = 368   – 1 = 35 t Reject H Reject at  =.05 There is evidence population average is not 368

Two-Tailed t Test Thinking Challenge You work for the FTC. A manufacturer of detergent claims that the mean weight of detergent is 3.25 lb. You take a random sample of 64 containers. You calculate the sample average to be lb. with a standard deviation of.117 lb. At the.01 level of significance, is the manufacturer correct? 3.25 lb.

Two-Tailed t Test Solution* H 0 : H a :   df  Critical Value(s): Test Statistic: Decision: Conclusion:  = 3.25   – 1 = 63 t Reject H Do not reject at  =.01 There is no evidence average is not 3.25

One-Tailed t Test Example Is the average capacity of batteries less than 140 ampere-hours? A random sample of 20 batteries had a mean of and a standard deviation of Assume a normal distribution. Test at the.05 level of significance.

One-Tailed t Test Solution H 0 : H a :  = df = Critical Value(s): Test Statistic: Decision: Conclusion:  = 140  < – 1 = 19 t Reject H 0 Reject at  =.05 There is evidence population average is less than 140

One-Tailed t Test Thinking Challenge You’re a marketing analyst for Wal- Mart. Wal-Mart had teddy bears on sale last week. The weekly sales ($ 00) of bears sold in 10 stores was: At the.05 level of significance, is there evidence that the average bear sales per store is more than 5 ($ 00)?

One-Tailed t Test Solution* H 0 : H a :  = df = Critical Value(s): Test Statistic: Decision: Conclusion:  = 5  > – 1 = 9 t Reject H 0 Do not reject at  =.05 There is no evidence average is more than 5

7.6 Large-Sample Test of Hypothesis about a Population Proportion

Large-Sample Test of Hypothesis about p One-Tailed Test H 0 : p = p 0 H a : p p 0 ) Test statistic: Rejection region: z z  when H a : p > p 0 ) Note: p 0 is the symbol for the numerical value of p assigned in the null hypothesis

Large-Sample Test of Hypothesis about p Two-Tailed Test H 0 : p = p 0 H a : p ≠ p 0 Test statistic: Rejection region: |z| < z  Note: p 0 is the symbol for the numerical value of p assigned in the null hypothesis

Conditions Required for a Valid Large- Sample Hypothesis Test for p 1.A random sample is selected from a binomial population. 2.The sample size n is large. (This condition will be satisfied if both np 0 ≥ 15 and nq 0 ≥ 15.)

One-Proportion z Test Example The present packaging system produces 10% defective cereal boxes. Using a new system, a random sample of 200 boxes had  11 defects. Does the new system produce fewer defects? Test at the.05 level of significance.

One-Proportion z Test Solution H 0 : H a :  = n = Critical Value(s): Test Statistic: Decision: Conclusion: p =.10 p < z Reject H 0 Reject at  =.05 There is evidence new system < 10% defective

One-Proportion z Test Thinking Challenge You’re an accounting manager. A year-end audit showed 4% of transactions had errors. You implement new procedures. A random sample of 500 transactions had 25 errors. Has the proportion of incorrect transactions changed at the.05 level of significance?

One-Proportion z Test Solution* H 0 : H a :  = n = Critical Value(s): Test Statistic: Decision: Conclusion: p =.04 p  z Reject H Do not reject at  =.05 There is evidence proportion is not 4%