Due Date: Name: The Chi-Square Distribution. Statistics Chapter 9 - Part C Chi-Square Set: Doing a test on the mean does not tell us about how much variation there is in a process or distribution. Knowing that a machine puts an average of 6 oz. of pop in a bottle is not very helpful if sometimes it puts 0 oz. and sometimes it puts 22 oz. In this case we want to know the variance (or the standard deviation), to know if the process is working correctly. The chi-square (pronounced ki-square, as in kite) distribution allows us to make inferences about the variance of a normal population. Like the Student's t-distribution, the chi-square distribution is a family of distributions, each one identified by the degrees of freedom (n - ). The critical values for the chi-square distribution are found in Table 8 in Appendix B (Page 583). As the chi-square is not symmetrical, the values for the right hand tail and the left hand tail must be given separately. Note that the chi-square distribution is always positive. It is impossible to get a negative variance (or standard deviation). Question.) Exercise 9.05 & 9.06 Find the chi-square critical value using Table 8 in Appendix B. The chi-square is presented with the degrees of freedom, df, and the value of alpha. X^2(df,alpha) Always assume alpha is in the right hand tail, unless the question specifies that alpha is in the left hand tail. X^2(,0.0) Find the chi-square critical value using Table 8 in Appendix B. alpha = 0.9 (Alpha is in the right hand tail.) Sample size, n = 5 Find the chi-square critical value using Table 8 in Appendix B. alpha = 0.25 (Alpha is in the left hand tail, but chi-square cannot be negativ Sample size, n = 7 Find the chi-square critical value using Table 8 in Appendix B. Sample size, n = 2 alpha = 0. (This is a 2-tail test, so alpha must be divided over the two tails.) A chi-square two tail test will have two different values, because the chi-square is not symmetrical. Present the value for the left hand tail first, then the right hand tail. Page #
Question 2.) Exercise 9.07 & 9.08 What value of chi-square for 9 degrees of freedom subdivides the area under the distribution curve such that 99.5% is to the right and 0.5% is to the left? (3 points) What is the value of the 5th percentile for the chi-square distribution with 28 degrees of freedom? (3 points) The central 50% of the chi-square distribution with 5 degrees of freedom lies between what two values? Question 3.) Exercise 9.09 & 9.0 For a chi-square distribution with 2 degrees of freedom, find the area under the curve for chi-square values ranging from 20.34 to 4.4. Round your answer to 3 decimal places. For a chi-square distribution with 3 degrees of freedom, find the area under the curve between X^2(3, 0.995) and X^2(3, 0.05). Round your answer to 3 decimal places. Page #2
Question 4.) Exercise 9. When testing hypotheses about the chi-square, the Null Hypothesis and the Alternate Hypothesis must both be in terms of sigma or sigma^2 instead of mu or p. The Null Hypothesis must still contain, "=", the equals sign. Depending upon the situation, the Alternate Hypothesis can contain either "greater than", >, "less than", <, or "not equal to" <>. The value stated in the Alternate Hypothesis is always the same as the value stated in the Null Hypothesis. For each of the following situations chose the letter which represents the Alternate Hypothesis that would be used to test these claims. Remember the alternate hypothesis is always the negation of the original situation. You suspect the standard deviation has decreased from its previous value of 24. ("a") Ha: sigma > 24 ("b") Ha: sigma < 24 ("c") Ha: sigma <> 24 The new standard deviation is more than the old value of 0. ("a") Ha: sigma > 0 ("b") Ha: sigma < 0 ("c") Ha: sigma <> 0 The old standard deviation is no smaller than 0.5 oz. ("a") Ha: sigma > 0.5 oz. ("b") Ha: sigma < 0.5 oz. ("c") Ha: sigma <> 0.5 oz. The old variance is no more than 8. ("a") Ha: sigma^2 > 8 ("b") Ha: sigma^2 < 8 ("c") Ha: sigma^2 <> 8 The new variance seems to be different from the 0.025 value indicated in the specs. ("a") Ha: sigma^2 > 0.025 ("b") Ha: sigma^2 < 0.025 ("c") Ha: sigma^2 <> 0.025 f.) The new variance has decreased from 34.5. ("a") Ha: sigma^2 > 34.5 ("b") Ha: sigma^2 < 34.5 ("c") Ha: sigma^2 <> 34.5 Page #3
Question 5.) Exercise 9.2 & 9.3 & 9.4 When doing a Hypothesis test using the chi-square, we need to calculate X^2* from the sample information, and the hypothesized value of sigma. X^2* =[(n - ) * s^2 ] / sigma^2 Calculate the value of the test statistic, X^2* for the following situation. (Round your answer to 2 decimal places.) Ho: sigma^2 = 4, n = 44, s^2 = 40.99 To find the critical value of X^2, find X^2(df,alpha). For a -tail test, the sign in Ha will tell you which side of the chi-square distribution to look in. For a 2-tail test, report both values, but remember that alpha must be divided in two, before looking on the table. Determine the critical value(s) that would be used to test the following Hypothesis test using the classical approach. Ho: sigma = 0.49, vs. Ha > 0.49, n =, alpha = 0.0 (6 points) Question 6.) Exercise 9.8a A commercial farmer harvests his entire field of a vegetable crop at one time. Therefore, he would like to plant a variety of green beans that mature all at the same time. (Small standard deviation for the maturity times of individual plants.) A seed company has developed a new hybrid strain of green bean that it wants the farmer to test. The maturity times of the standard variety have an average of 54 days and a standard deviation of 2.79 days. A random sample of 24 plants of the new hybrid showed a standard deviation of 2.72 days. Does this sample show a lower standard deviation at the 0.05 level of significance? Assume that maturity times are normally distributed. State the Null and Alternate Hypotheses ("a") Ho: mu = 2.79, Ha: mu < 2.79 ("b") Ho: p = 2.79, Ha: p < 2.79 ("c") Ho: sigma = 2.79, Ha: sigma < 2.79 ("d") Ho: sigma^2 = 2.79, Ha: sigma^2 < 2.79 ("e") Ho: mu = 2.79, Ha: mu > 2.79 ("f") Ho: p = 2.79, Ha: p > 2.79 ("g") Ho: sigma = 2.79, Ha: sigma > 2.79 ("h") Ho: sigma^2 = 2.79, Ha: sigma^2 > 2.79 ("i") Ho: mu = 2.79, Ha: mu <> 2.79 ("j") Ho: p = 2.79, Ha: p <> 2.79 ("k") Ho: sigma = 2.79, Ha: sigma <> 2.79 ("l") Ho: sigma^2 = 2.79, Ha: sigma^2 <> 2.79 Determine the critical value(s). The number of decimal places depends upon the numbers in the table. Make a decision using the classical approach. ("a") Reject Ho ("b") Fail to Reject Ho ("c") Accept Ha Page #4
Question 7.) Illustration 9.5 The soft-drink bottling company wants to control the variablility in the amount of fill by not allowing the variance to exceed 0.00033. Does a sample size of 30 with a variance of 0.00056 indicate that the bottling process is out of control (with regard to variance) at the 0.025 level of significance? Question 8.) Illustration 9.6 The manufacturer claims that a photographic chemical has a shelf life that is normally distributed about a mean of 80 days with a standard deviation of 2 days. As a user of this chemical, Fast Photo is concerned that the standard deviation might be different from 2 days; otherwise, it will buy a larger quantity while the chemical is part of a special promotion. Twelve random samples were selected and tested, with a standard deviation of 6 days resulting. At the 0.05 level of significance, does the sample standard deviation present sufficient evidence to show that the standard deviation is different from 2 days? f.) Should Fast Photo buy larger quantities of the chemical when it is on sale? Page #5
Question 9.) Exercise 9.9 A car manufacturer claims that the miles per gallon for a certain model have a mean of 40.9 miles with a standard deviation equal to 3.58 miles. Use the following data, obtained from a random sample of 5 such cars, to test the hypothesis that the standard deviation differs from 3.58. Use alpha = 0.2. Assume normality. The 5 sample miles per gallon were 34, 37, 42.5, 45, 35, 3, 35, 36.5, 37, 43.5, 39, 4.5, 35, 3, 38.5. Sum of x = Sum of x^2 = Variance = St dev (Rounded to 2 decimals) = (Use the rounded standard deviation in this calculation.) f.) Question 0.) Exercise 9.50 Bright-Lite claims that its 60-watt light bulb burns with a length of life that is approximately normally distributed with a standard deviation of 79 hours. A sample of 5 bulbs had a variance of 8833. Is this sufficient evidence to reject Bright-Lite's claim in favour of the alternative; "The standard deviation is longer than 79 hours," at the 0.005 level of significance? Page #6
Statistics Chapter 9 - Part C Chi-Square Answers 6/29/2009 2:27 Question.) Question 8.) 24.7 <> 7.79 6.7 df = n - = 2 - = 4.57 9.7 alpha = 0.05 X^2(df,alpha) = X^2(, 0.975) = 3.82 X^2(df,alpha) = X^2(, 0.025) = 2.9 Question 2.) 6.84 n = 2 n - = 6.9 s^2 = 256.0 8.2 sigma^2 = 44 X^2* = 9.56 Question 3.) Diagram 0.495 0.945 RH = 3.82 LH = 2.9 X^2* = 9.56 Question 4.) b Ha: sigma < 24 Fail to Reject Ho a Ha: sigma > 0 f.) b Ha: sigma < 0.5 a Ha: sigma^2 > 8 c Ha: sigma^2 <> 0.025 f.) b Ha: sigma^2 < 34.5 Question 9.) Question 5.) 42.99 23.2 Yes, buy larger quantities because the variation is close enough to the time specified by the manufacturer. <> 56.5 2272.25 8.0238095 4.25 Question 6.) df = n - = 5 - = 4 c alpha = 0.2; alpha/2 = 0. 3. X^2(df,alpha) = X^2(4, 0.9) = 7.79 2.86 X^2(df,alpha) = X^2(4, 0.) = 2. b Fail to Reject Ho n = 5 n - = 4 s^2 = 8.0625 Question 7.) sigma^2 = 2.864 > X^2* = 9.73 df = n - = 30 - = 29 Diagram alpha = 0.025 RH = 7.79 X^2(df,alpha) = X^2(29,0.025) = 45.7 LH = 2. X^2* = 9.73 n = 30 n - = 29 s^2 = 0.00056 f.) Fail to Reject Ho sigma^2 = 0.00033 X^2* = 49.2 Question 0.) > Diagram df = n - = 5 - = 50 RH = 45.7 alpha = 0.005 X^2* = 49.2 X^2(df,alpha) = X^2(50, 0.005) = 79.5 n = 5 n - = 50 Reject Ho s^2 = 8833 sigma^2 = 624 X^2* = 70.77 Diagram RH = 79.5 X^2* = 70.77 Fail to Reject Ho Page #7