Comments on Discussion Sheet 21 and Worksheet 21 ( ) Hypothesis Tests for Population Variances and Ratios of Variances

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1 Comments on Discussion Sheet and Worksheet ( ) Hypothesis Tests for Population Variances and Ratios of Variances Discussion Sheet Hypothesis Tests for Population Variances and Ratios of Variances This discussion we look at how to use the idea of hypothesis tests for variances and ratios of variances A Two-tailed 00( - α)% Hypothesis Test for a Population Variance H 0 :σ = H a :σ ( a constant) Test statistic: χ = ( n )s (where s is the sample variance and n is the sample size) Rejection region: χ < χ α or χ > χ α (where χ α is the value such that P χ α = α and χ α is the value such that P χ α = α, both with degrees of freedom ν = n ) Note: χ is the name of the test statistic; it is not actually anything squared.). Suppose that we are told that the average stat exam score for a sample of 5 Math 4753 students is 8 with a standard deviation of 5. Perform a hypothesis test at the 95% level to see if the variance of the population is equal to 30 or not. We know that y = 8 (although it is irrelevant here), s = 5, n = 5, and α =.05 (95% significance level). We want to see if " the variance of the population is equal to 30 or not." These are the key words for a two-sided hypothesis test of whether a population variance equals a certain number, in this case = 30. This gives us the following hypotheses: H 0 :σ = 30 H a :σ 30 Next we need to find the value of our test statistic. For this two-sided test of a variance, the form of the test statistic is: χ = ( n )s where s is the sample variance and n is the sample size. This gives us as the value of our test statistic: χ = ( n )s = ( 5 )( 5) = ( 4 )( 5) = = 0 Now we need to find our rejection region. For this test the form of the rejection region is χ < χ α or χ > χ α (where χ α is the value such that P χ α = α and χ α is the value such that P χ α = α, both with degrees of freedom ν = n ). In this case, since α =.05, the rejection region is " χ < χ.05 = χ.05 = χ.975 or χ > χ α = χ.05 = χ.05 ". We know that χ.05 is the value such that P χ > χ.05 ( ) =.05 with degrees of freedom ν = n = 5 = 4. We find this in Table 8 on pages We find the column for χ.05 and the row for degrees of freedom

2 ν = n = 5 = 4. We find that χ.05,4 such that P χ > χ.975 = (page 0). We know that χ.975 is the value ( ) =.975 with degrees of freedom ν = n = 5 = 4. We find this in Table 8 on pages We find the column for χ.975 and the row for degrees of freedom ν = 4. We find that χ.975,4 =.40 (page 00). So our rejection region is χ <.40 or χ > Now we need to reach our conclusion. In this case, since "0 <.40 or 0 > " is a false statement, the value of our test statistic is not in the rejection region. This means that our conclusion is: Do not reject H 0 :σ = 30 This means that we must continue to use H 0 :σ = 30 as our working hypothesis and assume that σ = 30. We cannot reject it with these data but we do not know the probability of this assumption being true. One-tailed 00( - α)% Hypothesis Tests for a Population Variance There are two types of one-tailed hypothesis tests: variance greater than a given value and variance smaller than a given value. Variance greater than a given value: H 0 :σ = (σ 0 a constant) H 0 :σ > Test statistic: χ = ( n )s (where s is the sample variance and n is the sample size) Rejection region: χ > χ α (where χ α is the value such that P χ > χ α ν = n ) Proportion less than a given value: H 0 :σ = (σ 0 a constant) H 0 :σ < ( ) = α with degrees of freedom Test statistic: χ = ( n )s (where s is the sample variance and n is the sample size) Rejection region: χ < χ α (where χ α freedom ν = n ) is the value such that P χ > χ α ( ) = α with degrees of. Suppose that we are told that the average stat exam score for a sample of 5 Math 4753 students is 8 with a standard deviation of 5. Perform a hypothesis test at the 95% level to see if the variance of the population is greater 30. We know that y = 8 (although it is irrelevant here), s = 5, n = 5, and α =.05 (95% significance level). We want to see if "the variance of the population is greater than 30." These are the key words for a one-sided hypothesis test of whether a population variance is greater than a certain number, in this case > 30. This gives us the following hypotheses:

3 H 0 :σ = 30 H a :σ > 30 Next we need to find the value of our test statistic. For this two-sided test of a variance, the form of the test statistic is: χ = ( n )s where s is the sample variance and n is the sample size. This gives us as the value of our test statistic: χ = ( n )s = ( 5 )( 5) = ( 4 )( 5) = = 0 Notice that this is the same value as the value of the test statistic for the corresponding two-sided hypothesis test in the previous problem. Now we need to find our rejection region. For this test the form of the rejection region is χ > χ α (where χ α is the value such that P χ α = α with degrees of freedom ν = n ). In this case, since α =.05, the rejection region is χ > χ α = χ.05. We know that χ.05 is the value such that ( ) =.05 with degrees of freedom ν = n = 5 = 4. We find this in Table 8 on pages P χ > χ We find the column for χ.05 and the row for degrees of freedom ν = n = 5 = 4. We find that χ.05, 4 = (page 0). So our rejection region is χ > Now we need to reach our conclusion. In this case, since 0 > is a false statement, the value of our test statistic is not in the rejection region. This means that our conclusion is: Do not reject H 0 :σ = 30 This means that we must continue to use H 0 :σ = 30 as our working hypothesis and assume that σ = 30. We cannot reject it with these data but we do not know the probability of this assumption being true. Worksheet ( ) Hypothesis Tests for Population Variances and Ratios of Variances This is a continuation of Discussion Sheet. If you have not already completed that discussion sheet, do so now. This worksheet corresponds to the material in 9. and 9.3 of the textbook and you should read those sections. A Two-tailed 00( - α)% Hypothesis Test for the Ratio of Two Population Variances H 0 : σ σ = H a : σ σ Larger sample variance Test statistic: F = Smaller sample variance Rejection region: F > F α (with numerator degrees of freedom ν num = n num and ν denom = n denom ) 3

4 . Suppose that we are told that the average stat exam score for a sample of 5 Math 4753 students other than engineers is 8 with a standard deviation of 5 while a sample of 5 engineering students have an average exam score of 83 but a standard deviation of 8. Perform a hypothesis test at the 90% level to see if the variance of the two populations is the same or not. We know that y = 8(irrelevant here), s = 5, and n = 5 for the sample from Population (Students other than engineers) while y = 83 (irrelevant here), s = 8, and n = 5 for the sample from Population (engineering students). We also know that α =.0 (90% significance level). We want to see " if the variance of the two populations is the same or not." These are the key words for a two-sided test of the equality of two variances, which we test by testing their ratio. Thus, we have as our hypotheses: H 0 : σ σ = (which implies σ = σ ) H a : σ σ (which implies σ σ ) Next we must find the value of our test statistic. The form of this statistic is Larger sample variance F = Smaller sample variance for this test. Since s = 5 we have s = ( 5) = 5 and, since s = 8, we have s = ( 8) = 64. So we have s = 5 < 64 = s. This makes our test statistic Larger sample variance F = Smaller sample variance = s s = 64 5 =.56 Next we must find our rejection region. For this test it is of the form F > F α (with numerator degrees of freedom ν num = n num and ν denom = n denom ). In this case, Population gave us our numerator and Population gave us our denominator. So we are looking for F α = F.0 = F.05 with numerator degrees of freedom ν num = n num = n = 5 = 4 and denominator degrees of freedom ν denom = n denom = n = 5 = 4. So we are looking for F.05 (4, 4). We find values of F in either Table 9, 0, or (pages 0-09). We are looking for F.05 (4, 4), a F.05 value, so we need Table 0. The numerator degrees of freedom, ν num = 4, determines the column while the denominator degrees of freedom, ν denom = 4 determines the row. Looking at the table we see that there is a column for ν num = and another for ν num = 5. There is a row for ν denom = 4. So our value is between F.05 (, 4) =.8 and F.05 (6, 4) =.. To be conservative, we want to make it as hard to reject the null hypothesis as possible so that we are sure that our probability of being correct is at least.95. So we use the higher value, F.05 (, 4) =.8, since F.05 (4, 4) < F.05 (, 4) =.8. This makes our rejection region F >.8 Now we need to draw our conclusion. Since.56 >.8 is a true statement, the value of our test statistic is in the rejection region and we can Reject H 0 : σ σ = and accept our alternative hypothesis H a : σ. This means that from now on, given these data, we can assume that σ σ σ and thus that σ σ with a probability of.90 of being correct. Thus the answer to our question is that it is not likely that the variances are the same. 4

5 One-tailed 00( - α)% Hypothesis Tests for the Ratio of Two Variances For the first variance greater than the second: H 0 : σ σ = H a : σ σ > σ ( > σ ) Test statistic: F = s s Rejection region: F > F α (where ν num = n and ν denom = n ) For the first proportion less than the second: H 0 : σ σ = H a : σ σ < σ ( < σ ) Test statistic: F = s s Rejection region: F > F α (where ν denom = n and ν num = n ). Suppose that we are told that the average stat exam score for a sample of 5 Math 4753 students other than engineers is 8 with a standard deviation of 5 while a sample of 5 engineering students have an average exam score of 83 but a standard deviation of 8. Perform a hypothesis test at the 95% level to see if the variance of the engineers is really greater than the variance of the other students. We know that y = 8(irrelevant here), s = 5, and n = 5 for the sample from Population (Students other than engineers) while y = 83 (irrelevant here), s = 8, and n = 5 for the sample from Population (engineering students). We also know that α =.05 (95% significance level). We want to see "if the variance of the engineers is greater than the variance of the other students." These are the key words for a one-sided test of the two variances, which we test by testing their ratio. We want to see if the variance for Population is greater than that for Population. This would make their ratio less than. Thus, we have as our hypotheses: H 0 : σ σ = (which implies σ = σ ) H a : σ σ < (which implies σ < σ ) Next we must find the value of our test statistic. The form of this statistic for this set of hypotheses is F = s s Since s = 5 we have s = ( 5) = 5 and, since s = 8, we have s = ( 8) = 64. This makes our test statistic F = s s = 64 5 =.56 5

6 Notice that this would not necessarily be the same as the test statistic for the two-sided test. If we were testing H a : σ σ > the test statistic would be F = s which would have a different value. s Next we must find our rejection region. For this test it is of the form F > F α (with numerator degrees of freedom ν num = n and ν denom = n ). So we are looking for F α = F.05 with numerator degrees of freedom ν num = n = 5 = 4 and denominator degrees of freedom ν denom = n = 5 = 4. So we are looking for F.05 (4, 4). We find values of F in either Table 9, 0, or (pages 0-09). We are looking for F.05 (4, 4), a F.05 value, so we need Table 0. The numerator degrees of freedom, ν num = 4, determines the column while the denominator degrees of freedom, ν denom = 4 determines the row. Looking at the table we see that there is a column for ν num = and another for ν num = 5. There is a row for ν denom = 4. So our value is between F.05 (, 4) =.8 and F.05 (6, 4) =.. To be conservative, we want to make it as hard to reject the null hypothesis as possible so that we are sure that our probability of being correct is at least.95. So we use the higher value, F.05 (, 4) =.8, since F.05 (4, 4) < F.05 (, 4) =.8. This makes our rejection region F >.8 Now we need to draw our conclusion. Since.56 >.8 is a true statement, the value of our test statistic is in the rejection region and we can Reject H 0 : σ σ = and accept our alternative hypothesis H a : σ <. This means that from now on, given these data, we can assume that σ σ σ < and thus that σ < σ with a probability of.95 of being correct. Thus the answer to our question is that it is likely that the variance for non-engineering majors is less than the variance for engineering majors. Suggested Homework: 9.58, 9.6, 9.6, 9.65, 9.66, 9.67 Solutions to be Posted: 9.58,