# Question: Use the original weights of pre 1964 quarters and post 1964 quarters

Use the original weights of pre-1964 quarters and post-1964 quarters listed in Data Set 21 in Appendix B. Instead of using the A test, use the following procedure for a “count five” test of equal variation. What do you conclude?

a. For the first sample, find the absolute deviation of each value. The absolute deviation of a sample value x is |x – x̄|. Sort the absolute deviation values. Do the same for the second sample.

b. Let cx be the count of the number of absolute deviation values in the first sample that are greater than the largest absolute deviation value in the other sample. Also, let c2 be the count of the number of absolute deviation values in the second sample that are greater than the largest absolute deviation value in the other sample. (One of these counts will always be zero.)

c. If the sample sizes are equal (n1 = n2)> use a critical value of 5. If n1 ≠ n2, calculate the critical value shown below.

d. If c1 ≥ critical value, then conclude that σ21> σ22 . If c2 ≥ critical value, then conclude that σ21> σ22. Otherwise, fail to reject the null hypothesis of σ21 = σ22.

a. For the first sample, find the absolute deviation of each value. The absolute deviation of a sample value x is |x – x̄|. Sort the absolute deviation values. Do the same for the second sample.

b. Let cx be the count of the number of absolute deviation values in the first sample that are greater than the largest absolute deviation value in the other sample. Also, let c2 be the count of the number of absolute deviation values in the second sample that are greater than the largest absolute deviation value in the other sample. (One of these counts will always be zero.)

c. If the sample sizes are equal (n1 = n2)> use a critical value of 5. If n1 ≠ n2, calculate the critical value shown below.

d. If c1 ≥ critical value, then conclude that σ21> σ22 . If c2 ≥ critical value, then conclude that σ21> σ22. Otherwise, fail to reject the null hypothesis of σ21 = σ22.

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