Problem 5. A logging company is bidding for logging rights on a large tract of l
ID: 3267488 • Letter: P
Question
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:
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 20 locations on the property at random from a map, and send a surveyor to those 20 locations to measure all the trees within a 25 foot radius of the location.
Suppose that this results in measurementsof the girth of 300 trees, and that for all practical purposes, those 300 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 300 trees in the sample is as follows:
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) so, under the null hypothesis, the probability histogram of the chi-squared statistic (Q14) ? A: is not B: is approximated well by a chi-squared curve.
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)
Should the null hypothesis be rejected at significance level 10%? (Q17)
girth range <3 ft. 3-6 ft. 6-9 ft. 9-12ft >12 ft. claimed fraction of trees 10% 20% 35% 30% 5%Explanation / Answer
applying chi square goodness of fit on above:
Q12) observed value of the chi-squared statistic is =0.7968
Q13) smallest expected number of trees in any of the categories is = 15
(Q14) under the null hypothesis, the probability histogram of the chi-squared statistic B: is approximated well by a chi-squared curve.
(Q15) degree of freedom =(number of categories-1) =5-1=4
(Q16) P-value of the null hypothesis is approximately =0.9389
(Q17) as p value is significantly high we can not reject null hypothesis at significance level 10%
observed Expected Chi square Probability O E=total*p =(O-E)^2/E <3 0.100 31.000 30.00 0.03 3-6 0.200 62.000 60.00 0.07 6-9 0.350 108.000 105.00 0.09 9-12 0.300 83.000 90.00 0.54 >12 0.050 16.000 15.00 0.07 1 300 300 0.7968