Question

In: Statistics and Probability

A survey was conducted two years ago asking college students their top motivations for using a...

A survey was conducted two years ago asking college students their top motivations for using a credit card. The percentages are shown in the table to the right. Also shown in the table is the observed frequency for these motivations from a current random sample of college students who use a credit card. Complete parts a through c below.

Response

Old Survey​ %

New Survey ​ Frequency, f

Rewards

28​%

110

Low rates

23​%

98

Cash back

21​%

108

Discounts

77%

46

Other

21%

63

a. Using

α=0.025​,

perform a​ chi-square test to determine if the probability distribution for the motivations for using a credit card changed between the two surveys.What is the null​ hypothesis, H0​?

A.The distribution of motivations is 28​% ​rewards, 23​% low​ rates,21​% cash​ back, 7%​discounts, and 21​% other.

B.

The distribution of motivations differs from the claimed or expected distribution.

C.The distribution of motivations is

110 ​rewards,98 low​ rates,108cash​ back,46 ​discounts, and 63 other.

D.

The distribution of motivations follows the normal distribution.

What is the alternate​ hypothesis, H1​?

A.

The distribution of motivations is the same as the claimed or expected distribution.

B.

The distribution of motivations differs from the claimed or expected distribution.

C.

The distribution of motivations is​ 20% rewards,​ 20% low​ rates, 20% cash​ back, 20%​ discounts, and​ 20% other.

D.

The distribution of motivations does not follow the normal distribution.

Calculate the test statistic.

χ2=

​(Round to two decimal places as​ needed.)

Determine the critical​ value,

χ2α= _____and the rejection region.

​(Round to three decimal places as​ needed.)

Choose the correct rejection region below.

A.

χ2>χ2α

B.

χ2<χ2α

C.

χ2≥χ2α

D.

χ≤χ2α

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

​p-value=____

​(Round to three decimal places as​ needed.)

Interpret the​ p-value.

The​ p-value is the probability of observing a test statistic equal to/greater than/less than

the test​ statistic, assuming

the distribution of the variable differs from the normal distribution./at least one expected frequency differs from 5./the distribution of the variable is the same as the given distribution./the distribution of the variable differs from the given distribution./the distribution of the variable is the normal distribution./the expected frequencies are all equal to 5.

c. What conclusion can be drawn about the motivation behind the use of a credit card by college students between the two​ surveys?

Reject

Do not reject

Upper H 0H0.

At the

2.52.5​%

significance​ level, there

is not

is

enough evidence to conclude that the distribution of motivations

is the same as

differs from

the claimed or expected distribution.

Click to select your answer(s).

Solutions

Expert Solution

a)NUll hypothesis HO:

A.The distribution of motivations is 28​% ​rewards, 23​% low​ rates,21​% cash​ back, 7%​discounts, and 21​% other.

alternate​ hypothesis, H1​ :

B: The distribution of motivations differs from the claimed or expected distribution

applying chi square goodness of fit test:
           relative observed Expected Chi square
Category frequency(p) Oi Ei=total*p R2i=(Oi-Ei)2/Ei
1 0.2800 110 119.00 0.6807
2 0.2300 98 97.75 0.0006
3 0.2100 108 89.25 3.9391
4 0.0700 46 29.75 8.8761
5 0.2100 63 89.25 7.7206
total 1.00 425 425 21.2170
test statistic X2= 21.22
degree of freedom =categories-1= 4
for 0.025 level and 4 df :crtiical value X2 = 11.143      from excel: chiinv(0.025,4)

e correct rejection region : C .χ2≥χ2α

The​ p-value is the probability of observing a test statistic greater than the test​ statistic, assuming the distribution of the variable is the same as the given distribution

c)

Reject HO At the 2.5 % significance​ level, there is enough evidence to conclude that the distribution of motivations   differs from the claimed or expected distribution.


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