Just do 3 and 5 question only, i finished other, thank you. Chi-Squrare distribu
ID: 3176593 • Letter: J
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Just do 3 and 5 question only, i finished other, thank you. Chi-Squrare distribution question.
A certain county park was surveyed for plant species ten years ago. At that time there were 40% Smilax, 10% Scotch Broom, 30% Toyon and 20% Poison Oak. Recently, to determine if the distribution of plant species was the same, a sample of 600 plants was taken with the following results. Use a 2% level of significance. Answer questions 1 through 3. Hint: convert the expected values to whole numbers to use. Specie Smilax Scotch Broom Toyon Poison Oak 170 150 Observed in sample 200 80 a Goodness of Fit problem b) Test of Independence 1. This is a 2. The degrees of freedom are a) 5 b) 4 c) 3 d 2 3. The distribution of species >a) has changed b) has not changedExplanation / Answer
A country park was sureyed for the plant species ren years ago.at that time were 40% smilax,10% scotch broom,30% toyon and 20% poison oak
Recently the surveyed with 600 palnts results.
smilax 200
scotch broom 80
toyon 170
poison oak 150
from this problem 40% 600 means 40*600/100 =240
10% of 600 means 10*600/100 =60
30% of 600 means 30* 600/100 =180
20% of 600 means 20*600/100 =120
1. A: Godness of freedom problem.
because goodness of fit test is a way of determining whether a set of categorical data came from a claimed discrete discrete distribution or not.
The goodness-of-fit test expands the one-proportion z-test. The one-proportion z-test is used if the outcome has only two categories. The goodness-of-fit test is used if you have two or more categories.
2.A:
The degrees of freedom in a statistical calculation represent how many values involved in a calculation have the freedom to vary. The degrees of freedom can be calculated to help ensure the statistical validity of chi-square tests, t-tests and even the more advanced f-tests. These tests are commonly used to compare observed data with data that would be expected to be obtained according to a specific hypothesis.
The statistical formula to determine degrees of freedom is quite simple. It states that degrees of freedom equal the number of values in a data set minus 1, and looks like this:
df = N-1
Where N is the number of values in the data set (sample size). Take a look at the sample computation.
tje degree of freedom is 3.
3 A:
has changed.
beacuse they are not in correct perchange has they are in 10 years back.
4 A:
One test statistic that follows a chi-squared distribution exactly is the test that the variance of a normally distributed population has a given value based on a sample variance. Such tests are uncommon in practice because the true variance of the population is usually unknown
for a toal 157+126+58=341
for B total 65+82+45 =162
for c total 181+142+60=383
for D total 10+46+28 =84
final total 341+162+383+54=940
small car size 157+65+181+10 =413
intermediate 126+82+142+46 =396
large 58+45+60+28 =196
341 * 413/940 =149.82
(157-149.82)^2 /149.82 =51.55/149.82=0.344
answer is 34.67
5.car size manufactures are independent
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