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Fundamentals of Hypothesis Testing

Identify the Population Assume the population mean TV sets is 3. (Null Hypothesis) REJECT Compute the Sample Mean to be 2.0 Take a Sample Null Hypothesis Hypothesis Testing Process Do a statistical test and conclude

1. State H 0 State H 1 2. Choose  3. Collect data 4. Compute test statistic (and/or the p value) 5. Make the decision General Steps in Hypothesis Testing

Step 1 Define the null and alternative hypotheses

A hypothesis is a claim about the population parameter.  Examples of a parameter are population mean or proportion The mean number of TV sets per household is 3.0! © 1984-1994 T/Maker Co. What is a Hypothesis?

States the assumption to be tested e.g. The mean number of TV sets is 3 (H 0 :  3) not The null hypothesis is always about a population parameter (H 0 :  3), not about a sample statistic (H 0 : X  3) The Null Hypothesis, H 0

TRUEBegins with the assumption that the null hypothesis is TRUE (Similar to the notion of innocent until proven guilty) Always contains the “ = ” sign. There may be enough evidence to reject the Null Hypothesis. Otherwise it is not rejected. (continued)

Is the opposite of the null hypothesis e.g. The mean number of TV sets is not 3 (H 1 :   3) Never contains an “=” sign. The Alternative Hypothesis, H 1

One and Two Tail Hypotheses H 0 :   3 H 1 :  < 3 H 0 :   3 H 1 :  > 3 H 0 :   3 H 1 :   3

Step 2 Set the level of significance 

H 0 : Innocent The Truth Verdict InnocentGuilty Decision H 0 TrueH 0 False Innocent CorrectError Do Not Reject H 0 1 -  Type II Error (  ) Guilty Error Correct Reject H 0 Type I Error (  ) Power (1 -  ) Result Possibilities Jury Trial Hypothesis Test

The  (level of significance) is selected by the researcher at the start of the research. Typical values are 0.01, 0.05, and 0.10 The  defines unlikely values of sample statistic if null hypothesis is true. This is called rejection region. Level of Significance, 

Level of Significance,  and the Rejection Region H 0 :   3 H 1 :  < 3 H 0 :   3 H 1 :  > 3 H 0 :   3 H 1 :   3    /2 Critical Value(s)

Type I error Is when you reject the null hypothesis when it is true. Probability of type I error is  and is set by the researcher. Errors in Making Decisions

Type II Error Is failing to Reject a False Null Hypothesis Probability of Type II Error Is  (beta) Power(1-  ) The Power of The Test Is (1-  )

  Reduce probability of one error and the other one goes up holding everything else unchanged.  &  Have an Inverse Relationship

  increases when difference between hypothesized parameter & its true value decreases.  increases when you are willing to take a bigger chance of a type I error (  decreases). Factors Affecting Type II Error   

 increases when population standard deviation   increases  increases when sample size n decreases Factors Affecting Type II Error     n (continued)

How to Choose between Type I and Type II Errors Choice depends on the costs of the errors. Choose smaller type I error when the cost of rejecting the hypothesis is high.  At a criminal trial, the presumption is innocence. A type I error is convicting an innocent person.

Step 3 Gather the data

Step 4 Conduct a statistical test to measure the strength of the evidence

Sample Mean  = 3 Sampling Distribution of we get a sample mean of this value... the population mean is 2 If H 0 is true Weighting the evidence

Step 5 Reach a conclusion

Sample Mean  = 3 Sampling Distribution of It is unlikely that we would get a sample mean of this value...... if in fact this were the population mean.... Therefore, we reject the null hypothesis that  = 3. 2 If H 0 is true Rejecting H 0

An Example of the 5 Steps For the two tail test on the mean where the population standard deviation  is known.

Assumptions Population is normally distributed. If not normal, use large samples. Z test statistic: Two-Tail Z Test for the Mean (  Known)

Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes showed X = 372.5. The company has specified  to be 15 grams. Test at the  0.05 level 368 gm. Example: Two Tail Test H 0 :  368 H 1 :  368

  = 0.05 n = 25 Critical value: ±1.96 Test Statistic: Decision: Conclusion: Do Not Reject at  =.05 Cannot prove that the population mean is other than 368 Z 0 1.96.025 Reject Example Solution: Two Tail -1.96.025 H 0 :  368 H 1 :  368

p Value Solution (p Value = 0.1336)  (  = 0.05) Do Not Reject. 0 1.50 Z Reject  = 0.05 1.96 p Value = 2 x 0.0668 Look up 1.5 in the z table (.9332) and subtract from 1.00 to get.0668. Double if two tail test. Reject

The Z Test and Confidence Interval You will find both give the same conclusion but the test is easier to use and provides more information.

Connection to Confidence Intervals For X = 372.5 oz,  = 15 and n = 25, The 95% Confidence Interval is (equation 8.1): 372.5 - (1.96) 15/ to 372.5 + (1.96) 15/ or 366.62    378.38 If this interval contains the hypothesized mean (368), we do not reject the null hypothesis. _

An Example of the 5 Steps For the one tail test on the mean where the population standard deviation  is known.

Z 0  Reject H 0 Z 0 0  H 0 :  0 H 1 :  <  0 H 0 :  0 H 1 :  >  0 Z Must Be Significantly Below  to reject H 0 Z Must Be Significantly Above  to reject H 0 One Tail Tests

Assumptions not changed: Population is normally distributed. If not normal, use large samples. Z test statistic not changed: One-Tail Z Test for the Mean (  Known)

Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed X = 372.5. The company has specified  to be 15 grams. Test at the  0.05 level 368 gm. Example: One Tail Test H 0 :  368 H 1 :  > 368

Z.04.06 1.6.9495.9505.9515 1.7.9591.9599.9608 1.8.9671.9678.9686.9738.9750 Z0  Z = 1 1.645. 05 1.9.9744 Standardized Cumulative Normal Distribution Table (Portion) What is Z given  = 0.05?  =.05 Finding Critical Values: One Tail Tests Critical Value = 1.645.95

  = 0.5 n = 25 Critical value: 1.645 Test Statistic: Decision: Conclusion: Do Not Reject at  =.05 Cannot prove that the population mean is more than 368 Z 0 1.645.05 Reject Example Solution: One Tail H 0 :  368 H 1 :  > 368

Z0 1.50 p Value =. 0668 Z Value of Sample Statistic From Z Table: Lookup 1.50 to Obtain.9332 Use the alternative hypothesis to find the direction of the rejection region. 1.000 -.9332.0668 p Value is P(Z  1.50) = 0.0668 p Value Solution

0 1.50 Z Reject (p Value = 0.0668)  (  = 0.05) Do Not Reject. p Value = 0.0668  = 0.05 Test Statistic 1.50 Is In the Do Not Reject Region p Value Solution 1.645

An Example of the 5 Steps For the test on the mean where the population standard deviation  is not known.

Assumption: The population is normally distributed or only slightly skewed & a large sample taken. t test with n-1 degrees of freedom t Test for the Mean (  Unknown)

Example: One Tail t-Test Does an average box of cereal contain more than 368 grams of cereal? A random sample of 36 boxes showed X = 372.5, and  s  15. Test at the  0.01 level. 368 gm. H 0 :  368 H 1 :  368  is not given

  = 0.01 n = 36, df = 35 Critical value: 2.4377 Test Statistic: Decision: Conclusion: Do Not Reject at  =.01 Cannot prove that the population mean is more than 368 t 35 0 2.4377.01 Reject Example Solution: One Tail H 0 :  368 H 1 :  368 1.80

The t test and p-value You will be unable to calculate the p-value for the t test as the book does not provide the necessary table.

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