Question

In​ 1991, a study was conducted to determine the education levels of people who declare bankruptcy. The percentages are shown in the accompanying table. Also shown in the table is the observed frequency for these education levels from a random sample of individuals who files for bankruptcy in 2017. Use these data to complete parts a through c.

Education level Probability 1991​ (%) Observed Frequency 2017

No high school diploma 21.5%    12

High school diploma    30.3%    38

Some college 35.5%    42

College diploma 9.9%    18   

Advanced degree 2.8​%    5

Total ​100% 115

a. Using alpha α=0.05​, perform a​ chi-square test to determine if the probability distribution for the education levels of individuals who files for bankruptcy changed between 1991 and 2017.

What is the null​ hypothesis, H0​?

A. The distribution of education levels is 12 no high school​ diploma, 38 high school​ diploma, 42 some​ college,18 college​ diploma, and 5 advanced degree.

B. The distribution of education levels follows the normal distribution.

C. The distribution of education levels is 21.5​% no high school​ diploma, 30.3​% high school​ diploma, 35.5​% some​ college, 9.9​% college​ diploma, and 2.8​% advanced degree.

D. The distribution of education levels differs from the claimed or expected distribution.

What is the alternative​ hypothesis, H1​?

A. The distribution of education levels differs from the claimed or expected distribution.

B. The distribution of education levels is​ 20% no high school​ diploma, 20% high school​ diploma, 20% some​ college, 20% college​ diploma, and​ 20% advanced degree.

C. The distribution of education levels does not follow the normal distribution.

D. The distribution of education levels is the same as the claimed or expected distribution.

The test​ statistic, combining the College diploma and Advanced degree​ rows, is . ​(Round to two decimal places as​ needed.)

b. Determine the​ p-value using Excel and interpret its meaning.

Identify a function that can be used in Excel to directly calculate the​ p-value (with no other calculations needed other than calculating the arguments of the function​ itself). (=CHISQ.DIST(x, deg_freedom, cumulative)/=NORM.S.DIST(z, cumulative)/ =CHISQ.DIST.RT(x, deg_freedom)/ =T.DIST.2T(x, deg_freedom)/=T.DIST.RT(x, deg_freedom) pick one

Determine the​ p-value, combining the College diploma and Advanced degree rows.

​p-value=nothing ​(Round to three decimal places as​ needed.)

Interpret the​ p-value.

The​ p-value is the probability of observing a test statistic (greater than/less than/equal to) the test​ statistic, assuming (the distribution of the variable is the normal distribution/the expected frequencies are all equal to 5/the distribution of the variable differs from the given distribution/at least one expected frequency differs from 5/the distribution of the variable differs from the normal distribution/the distribution of the variable is the same as the given distribution) pick one.

c. What conclusion can be drawn about the type of individual who filed for bankruptcy in 2017 vs. the type of individual who filed for bankruptcy in​ 1991?

(Reject/ Do not reject) H0. At the 5​% significance​ level, there (is not/is) enough evidence to conclude that the distribution of education levels (is the same as/differs from) the( normal distribution/claimed or expected distribution/uniform distribution.) pick one Click to select your answer(s).

Answer 1

What is the null​ hypothesis, H0​?

The distribution of education levels is 21.5​% no high school​ diploma, 30.3​% high school​ diploma, 35.5​% some​ college, 9.9​% college​ diploma, and 2.8​% advanced degree.

What is the alternative​ hypothesis, H1​?

The distribution of education levels differs from the claimed or expected distribution.

Test statistics:

= 11.96

df = n-1

= 4

p-value = 0.0198

greater than

the distribution of the variable is the same as the given distribution

c. What conclusion can be drawn about the type of individual who filed for bankruptcy in 2017 vs. the type of individual who filed for bankruptcy in​ 1991?

Reject

is

differs from

claimed or expected distribution

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