HYPOTHESIS TESTING. Hypothesis Testing

Transcription

1 MBA 605 Business Analytics Don Conant, PhD. HYPOTHESIS TESTING Hypothesis testing involves making inferences about the nature of the population on the basis of observations of a sample drawn from the population. This involves determining the magnitude of the difference between an observed value of the statistic ( ) and the hypothesized value of the parameter (µ) and then deciding whether the magnitude of this difference justifies rejection of the hypothesis. 1

2 Hypothesis testing consists of four steps. State the null hypothesis (H 0 ). Set the criterion for rejecting H 0. Compute the test statistic Decide about H 0. We can test the hypothesis that the population of freshman psychology students has a mean quantitative SAT score (µ) of 455. State the null hypothesis (H 0 ). In a mean difference test H 0 is stated in terms of no difference between the means. H 0 : µ = 455 or H 0 : µ = 0 We test H 0 against the alternative hypothesis (H a ) H a includes the possible outcomes not covered by H 0. H a : µ 455 or H a : µ Set the criterion for rejecting H 0. Now we determine how different the sample statistic ( ) from the hypothesized population parameter (µ). Suppose we randomly select 144 freshman psychology students from the population and find a mean ( ) of 535. Is an of 535 sufficiently different from a µ of 455? Three points to consider in answering this question. Errors in hypothesis testing Level of significance Region of rejection 2

3 Errors in hypothesis testing. There are four possible outcomes when testing hypotheses. A true hypothesis is rejected (Type I error) A true hypothesis is not rejected (Correct) A false hypothesis is not rejected (Type II error) A false hypothesis is rejected (Correct) The possibility of making an error cannot be eliminated completely. Which type of error is more serious? Example H 0 : New drug benefits = Old drug benefits Type I error = Patients incur additional cost for the new drug even though it is not more beneficial. Type II error = Patients do not incur additional cost for the new drug nor do they receive its additional benefits. So, which is more serious? How much more cost? How much more benefit? It is often a value judgment so minimize both errors. 3

4 Levels of significance The level of significance or alpha (α) is defined as the probability of making a type 1 error when testing H 0. The most often used levels of significance are.05 and.01. The research knows that the decision to reject H 0 may be incorrect 5% or 1% of the time respectively. The researcher determines the level and accepts the risk. Once we determine the α for the test we can calculate the region of rejection. Region of rejection things we need to know. H 0 contains an = sign, therefore a value can be too great or to small thus a two-tailed test. The α value is.05. Because the test is two-tailed the α value in each tail is is α/2 =.025 The number of standard deviations an α of.025 is on the standard normal distribution. NORM.S.INV(.025) = NORM.S.INV (.975) =

5 µ = 455, σ = 100, = 535 The standard error of the mean Now we can calculate the upper and lower bounds of the region of rejection (the critical values) Compute the statistic. Using z-scores we determine how different is from µ, or the number of standard deviation units the observed sample is from the hypothesized value The test statistic indicates the observed sample mean (535) is 9.60 standard deviations above the hypothesized value for the population parameter (455). 5

6 Decide about H 0 We have determined that acceptable values for z with an α of.05 are z 1.96, our critical value (±1.96) The observed value of the test statistic (+9.60) exceeds the critical value (±1.96). Therefore the probability is less than.05 (p <.05) that 535 will have occurred by chance if H 0 is true (µ = 455). Thus, reject H 0. The mean SAT level for the population of freshman psychology students does not appear to be equal to 455. NOTE: Common probability levels in social science research are p <.05, p <.01, p <.001. Suppose we had found that the sample mean () for 144 students was not 535 but 465. Calculate the z value. Remember 8.33 and µ = 455 Calculate the probability (p) for our new z value. 1-NORM.S.DIST(z,True) Decide about H 0. What is the value of p? 6

7 Compute the statistic Decide about H 0 We have determined that acceptable values for z with an α of.05 are z 1.96, our critical value (±1.96) The observed value of the test statistic (+1.20) is within the critical range (±1.96). We fail to reject H 0 Probability (p) If we fail to reject H 0, the probability of hypothesized parameter (µ) of 455 is.115. p=.115 Directional Null Hypothesis Suppose that the researcher has information about the variable under investigation so that the direction of the results is anticipated or is only interested in one direction of the results. In such a case H 0 would indicate direction. H 0 : µ 455 H a : µ > 455 NORM.S.INV(.95) =

8 Unknown σ Up to this point we have used the normal standard distribution. This distribution can only be used when the population standard deviation (σ) is known as with SAT scores. However σ is rarely known. When σ is unknown the researcher estimates it by using the standard deviation of the sample (s). Similarly, the standard error of the mean is estimated. t Distributions This adjustment effects the distribution used for testing especially when the sample is small (< 120). In such cases the standard normal distribution is inappropriate. A t distribution must be used. The t distribution changes as the sample size changes and is effected by degrees of freedom (a restriction of sampling). 8

9 t Distributions Degrees of freedom The concept of degrees of freedom is fundamentally mathematical. In the example of calculating the sample variance, because the mean is known a restriction is placed on the sample size (n). Imagine the population mean (25) is being estimated from a sample of five values. The first four of those values are free to assume any value (21, 23, 24, 30). However, the fifth value must result in the known mean of 25. Thus the fifth value must be 27. The calculated mean results in a loss of 1 degree of freedom. t Distributions We use the t distributions in hypothesis testing the same way we used the normal distribution. Unlike the normal distribution table each row represents a different t distribution based on the degrees of freedom. 9

10 t Distributions As with the standard normal distribution, Excel has functions the provide the cumulative area under the curve and the number of standard deviations for a given area. t.dist(x, degrees of freedom, cumulative) x = number of standard deviations df = n 1 (the sample size less 1) Yields the area under the curve t.inv(probability, degrees of freedom) Probability = area under the curve df = n 1 (the sample size less 1) Yields the number of standard deviations F Distributions In mean difference testing (z tests and t tests) when two independent samples are compared it is important to whether or not the variances of the samples are equal. The F test is often used to determine equality of variance. The result of the test is checked against an F distribution. Unlike the normal and t distributions, the F distributions are not symmetrical. F distributions start at 0 and go to (rather than to ). The shapes of the F distributions vary drastically, especially when degrees of freedom are small. 10

11 F Distributions The formula for the F test is It is a ratio using the variances (s 2 ) of the two data sets. Rather than use a table, the result is found in Excel. F.DIST(x, df 1, df 2, cumulative) x = the ratio of the variances (F). df = the sample size of each data set less 1. cumulative = is set to True. 11