Answered step by step
Verified Expert Solution
Link Copied!

Question

1 Approved Answer

The data set calledExercise versus Life Expectancy A simple random sample of 10 counties from Mississippi with an exercise rate of 13.5%, 10 counties from

image text in transcribedimage text in transcribed

The data set called"Exercise versus Life Expectancy" A simple random sample of 10 counties from Mississippi with an exercise rate of 13.5%, 10 counties from Oklahoma with an exercise rate of 19%, and 10 counties from Colorado with a 32.5% exercise rate were selected. The male life expectancies for the 30 counties were recorded. Use the data to test theclaimthat the average life expectancy differs for at least one of the states (exercise rates). Use a 5% significance rate.

I am using the ANOVA to test if the average life expectancy differs for at least one of the states (exercise rates). The P value (.000) is lower than the significance value (.05), therefore we reject the null hypothesis, there is sufficient evidence to support the claim that the average life expectancy differs for at least one of the states (exercise rates). It was appropriate to use the Bonferroni's test because the standard deviations are equal, and it shows us that Colorado state has significantly different averages in life expectancies (exercise rate) to Mississippi state (P- value - 0.000) and Oklahoma state (P-value - 0.000) (Table B and C).

See the attachment, Graph, boxplot, Anova attached

image text in transcribedimage text in transcribedimage text in transcribed
Appendix Graph A: Side-by side Boxplots showing the distribution of life expectancyes for the each of the states (exercise rates) 85.0- 80.0- Life Expectancy (Years) 75.0- 70.0- 65.0- Mississippi Oklahoma Colorado State D DD XTable A: One-Way ANOVA chart for comparing the average difference in life expectancyes between states (exercise rates) Graph B: Means Plot for comparing the average difference in life expectancyes between states (exercise rates) ANOVA Life Expectancy (Years) Sum of Mean Squares df Square Sig. 78.0- Between 233.091 2 116.545 26.346 000 Groups Within Groups 119.439 27 4.424 Total 352.530 29 76.0- Mean of Life Expectancy (Years) 74.0- Table B: The "Descriptives" for comparing the average difference in life expectancyes between states (exercise rates) 72.0- Descriptives Life Expectancy (Years) 95% Confidence Interval for N Mean Deviation Std. Error Lower Bound Upper Bound Minimum Maximum 1.7698 .5597 69.424 71.956 67.1 73.5 70.0- Mississippi 10 70.690 Oklahoma 10 72.130 1.9861 .6281 70.709 73.551 69.2 75.7 Mississippi Oklahoma Colorado Colorado 10 77.190 2.4888 .7870 75.410 78.970 73.4 81.8 Total 30 73.337 3.4866 6366 72.035 74.63 67.1 81.8 StateTable C: Bonferroni's multiple comparisons test Multiple Comparisons Dependent Variable: Life Expectancy (Years) Bonferroni Mean Difference (1- 95% Confidence Interval (1) State () State J) Std. Error Sig. Lower Bound Upper Bound Mississippi Oklahoma -1.4400 9406 412 3.841 .961 Colorado 6.5000 9406 000 8.90 -4.099 Oklahoma Mississippi 1.4400 .9406 412 -.961 3.841 Colorado -5.0600 9406 .000 -7.461 -2.659 Colorado Mississippi 6.5000 9406 .000 4.099 8.901 Oklahoma 5.0600 9406 .000 2.659 7.461 *. The mean difference is significant at the 0.05 level. K D DD X

Step by Step Solution

There are 3 Steps involved in it

Step: 1

blur-text-image

Get Instant Access to Expert-Tailored Solutions

See step-by-step solutions with expert insights and AI powered tools for academic success

Step: 2

blur-text-image

Step: 3

blur-text-image

Ace Your Homework with AI

Get the answers you need in no time with our AI-driven, step-by-step assistance

Get Started

Recommended Textbook for

Partial Differential Equations For Scientists And Engineers

Authors: Stanley J Farlow

1st Edition

0486134733, 9780486134734

More Books

Students also viewed these Mathematics questions

Question

Summarize the ABCDE method for overcoming irrational beliefs.

Answered: 1 week ago