PE6

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Ramir Oliver Thomas Edison State University STA -2010: Principles of Statistics Practice Exercise 6 December 10, 2023

Practice Exercise 6 Practice Exercise 6 1. Does the scatter plot appear linear? Strong or weak? Positive or negative? The scatter plot appears to be linear and negative. The data has more than a few outliers and therefore tends to lean more weak than strong. 2. Use the following information to answer the next problem. A random sample of 10 professional athletes produced the following data where x is the number of endorsements the player has and y is the amount of money made (in millions of dollars).

Practice Exercise 6 3. The table below shows the life expectancy for an individual born in the United States in certain years. a. Decide which variable should be the independent variable and which should be the dependent variable. Independent variable should be the Year of Birth Dependent variable is Life Expectancy b. Draw a scatter plot of the ordered pairs. c. Calculate the least squares line. Put the equation in the form of: ŷ = a + bx ŷ = -377.24 + 0.2275 x d. Find the correlation coefficient. Is it significant? r = 0.961 1920 1930 1940 1950 1960 1970 1980 1990 2000 2010 2020 40 45 50 55 60 65 70 75 80 85 Life Expectancy Year of Birth L ife Expectancy (years) Year of Birth Life Expectancy 1930 59.7 1940 62.9 1950 70.2 1965 69.7 1973 71.4 1982 74.5 1987 75 1992 75.7 2010 78.7

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Practice Exercise 6 Null Hypothesis: H 0 : ρ = 0 Null Hypothesis H 0 : The population correlation coefficient IS NOT significantly different from zero. There IS NOT a significant linear relationship (correlation) between x and y in the population. Alternate Hypothesis: H a : ρ ≠ 0 Alternate Hypothesis Ha: The population correlation coefficient IS significantly DIFFERENT FROM zero. There IS A SIGNIFICANT LINEAR RELATIONSHIP (correlation) between x and y in the population. Per table of critical values at a=0.05 and using df=9 we get a critical value of 0.666 Since r>critical value (0.961>0.666) correlation coefficient IS significant. e. Find the estimated life expectancy for an individual born in 1950 and for one born in 1982. ŷ = -377.24 + 0.2275 x -> -377.24 + 0.2275*(1950) = 66.385 ŷ = -377.24 + 0.2275 x -> -377.24 + 0.2275*(1982) = 73.665 f. Why aren’t the answers to part e the same as the values in the table that correspond to those years? The best fit line ensures that the sum of the squared errors is minimized (small as possible). Any data point off the best fit line is considered an “error”. Use the two points in part e to plot the least squares line on your graph from part b.

Practice Exercise 6