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The observed value of the chi-squared statistic is (Q12) | Under the null hypothesis, the smallest expected number of trees in any of the categories is (Q13) probability histogram of the chi-squared statistic (Q14) ? approximated well by a chi-squared curve. so, under the null hypothesis, the The appropriate chi-squared curve to use to approximate the probability histogram of the chi-squared statistic has (enter a number) (Q15) degrees of freedom. The P-value of the null hypothesis is approximately (Q16)

The observed value of the chi-squared statistic is (Q12) | Under the null hypothesis, the smallest expected number of trees in any of the categories is (Q13) probability histogram of the chi-squared statistic (Q14) ? approximated well by a chi-squared curve. so, under the null hypothesis, the The appropriate chi-squared curve to use to approximate the probability histogram of the chi-squared statistic has (enter a number) (Q15) degrees of freedom. The P-value of the null hypothesis is approximately (Q16)

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Problem 5 A logging company is bidding for logging rights on a large tract of land. The girth of a tree is the circumference of its trunk at a particular height above the ground (roughly breast height). The girth is important commercially, because it determines the width of the boards that can be milled from the tree. The owner of the property claims that the distribution of tree girth on the property is as follows: girth range <3 ft. 3-6 ft. 6-9 ft. 9-12ft >12 ft. claimed fraction of trees 10% 20% 35% 30% 5% The logging company can estimate the number of trees fairly accurately using aerial photography, but needs to do a survey on the ground to determine the distribution of girth. Ground surveys are expensive, so the logging company would like to draw its conclusions without surveying the entire tract. The company hires you as an expert statistician to test the hypothesis that the distribution of girth is what the seller claims. You select 30 locations on the property at random from a map, and send a surveyor to those 30 locations to measure all the trees within a 25 foot radius of the location. Suppose that this results in measurements of the girth of 340 trees, and that for all practical purposes, those 340 trees can be treated as a random sample with replacement from the population of trees on the property. The observed distribution of girth among the 340 trees in the sample is as follows: girth range <3 ft. 3-6 ft. 6-9 ft. 9-12ft >12 ft. observed number of trees 32 90 100 99 19 The observed value of the chi-squared statistic is (Q12) Under the null hypothesis, the smallest expected number of trees in any of the categories is (Q13) probability histogram of the chi-squared statistic (Q14) ? approximated well by a chi-squared curve. so, under the null hypothesis, the The appropriate chi-squared curve to use to approximate the probability histogram of the chi-squared statistic has (enter a number) (Q15) degrees of freedom. The P-value of the null hypothesis is approximately (Q16) |
Transcribed Image Text:Problem 5 A logging company is bidding for logging rights on a large tract of land. The girth of a tree is the circumference of its trunk at a particular height above the ground (roughly breast height). The girth is important commercially, because it determines the width of the boards that can be milled from the tree. The owner of the property claims that the distribution of tree girth on the property is as follows: girth range <3 ft. 3-6 ft. 6-9 ft. 9-12ft >12 ft. claimed fraction of trees 10% 20% 35% 30% 5% The logging company can estimate the number of trees fairly accurately using aerial photography, but needs to do a survey on the ground to determine the distribution of girth. Ground surveys are expensive, so the logging company would like to draw its conclusions without surveying the entire tract. The company hires you as an expert statistician to test the hypothesis that the distribution of girth is what the seller claims. You select 30 locations on the property at random from a map, and send a surveyor to those 30 locations to measure all the trees within a 25 foot radius of the location. Suppose that this results in measurements of the girth of 340 trees, and that for all practical purposes, those 340 trees can be treated as a random sample with replacement from the population of trees on the property. The observed distribution of girth among the 340 trees in the sample is as follows: girth range <3 ft. 3-6 ft. 6-9 ft. 9-12ft >12 ft. observed number of trees 32 90 100 99 19 The observed value of the chi-squared statistic is (Q12) Under the null hypothesis, the smallest expected number of trees in any of the categories is (Q13) probability histogram of the chi-squared statistic (Q14) ? approximated well by a chi-squared curve. so, under the null hypothesis, the The appropriate chi-squared curve to use to approximate the probability histogram of the chi-squared statistic has (enter a number) (Q15) degrees of freedom. The P-value of the null hypothesis is approximately (Q16) |
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