The accompanying table shows the market share for automotive manufacturers in 2013. Also shown is the frequency of car purchases by manufacturer from a random sample of 210 customers in March of 2018. Use these data to complete parts a and b.

Manufacturer Market Share 2013 (%) Frequency 2018 A 19.6% 37 B 16.4% 32 C 13.9% 31 D 12.1% 28 E 10.3% 20 Other 27.7% 62

The test statistic is

enter your response here.

(Round to two decimal places as needed.)

b. Determine the p-value for the chi-square test statistic 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).

equals CHISQ.DIST left parenthesis x comma deg_freedom comma cumulative right parenthesis=CHISQ.DIST(x, deg_freedom, cumulative)

equals T.DIST.RT left parenthesis x comma deg_freedom right parenthesis=T.DIST.RT(x, deg_freedom)

equals T.DIST. 2 Upper T left parenthesis x comma deg_freedom right parenthesis=T.DIST.2T(x, deg_freedom)

equals CHISQ.DIST.RT left parenthesis x comma deg_freedom right parenthesis=CHISQ.DIST.RT(x, deg_freedom)

equals NORM.S.DIST left parenthesis z comma cumulative right parenthesis=NORM.S.DIST(z, cumulative)

Determine the p-value.

p-value=enter your response here

(Round to three decimal places as needed.)

Interpret the p-value.

The p-value is the probability of observing a value for the test statistic

equal to

less than

greater than

the calculated test statistic, assuming

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

at least one expected frequency differs from 5.

the distribution of the variable is the normal distribution.

the distribution of the variable differs from the given distribution.

the distribution of the variable differs from the normal distribution.

the expected frequencies are all equal to 5.

Draw a conclusion.

Because the p-value is

greater than

less than

equal to

=0.05,

reject

fail to reject

H0.

At the

5%

significance level, there

is not

is

enough evidence to conclude that the distribution of education levels

differs from

is the same as

the

uniform distribution.

normal distribution.

claimed or expected distribution.