a. Explanation Calculation: The given data represents the raking of 11 states based on expenditure per student and student-teacher ratio. The formula for the rank-correlation coefficient is given by: r s = 1 − 6 ∑ d 2 n ( n 2 − 1 ) . Here, ∑ d 2 is the sum of squares of the differences of the ranks of the data sets and n is the sample size . The value of the ∑ d 2 can be obtained as follows: State Expenditure per Student Student-Teacher Ratio d d 2 Arizona 9 10 ( 9 − 10 ) = − 1 1 Colorado 5 8 –3 9 Florida 4 6 –2 4 Idaho 2 11 –9 81 Iowa 6 4 2 4 Louisiana 11 3 8 64 Massachusetts 1 1 0 0 Nebraska 7 2 5 25 North Dakota 8 b. To determine Explain whether there appear to be a relationship between expenditure per student and student-teacher ratio at α = 0.05 level.

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Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 18.5, Problem 34E

To determine

Find rank-correlation between expenditure per student and student-teacher ratio.

To determine

Explain whether there appear to be a relationship between expenditure per student and student-teacher ratio at α=0.05 level.

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Chapter 18 Solutions

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Find rank- correlation between expenditure per student and student-teacher ratio.

Solution Summary: The author calculates the rank-correlation between expenditure per student and student-teacher ratio.

Find rank-correlation between expenditure per student and student-teacher ratio.

Explain whether there appear to be a relationship between expenditure per student and student-teacher ratio at α=0.05 level.

Chapter 18 Solutions