Consumers with at least one credit card have a mean of 4.84 credit cards. [Sourc

Question

Consumers with at least one credit card have a mean of 4.84 credit cards. [Source: Sumit Agarwal, John C. Driscoll, Xavier Gabaix, and David Laibson, "Learning in the Credit Card Market, " Working Paper 13822, National Bureau of Economic Research (NBER), February 2008.] You want to test the hypothesis that the mean number of credit cards held by high-income consumers (consumers with annual incomes over $100,000) is different from the reported mean of 4.84. A random sample of 220 high-income consumers provides a sample mean number of credit cards of x = 5.40. Use a significance level of alpha = .05 for the test. Use a confidence interval estimate approach to conduct the hypothesis test. To use a confidence interval estimate approach to conduct the hypothesis test, you construct the ____ confidence interval estimate of the population mean. Use the Distributions tool to help answer the questions that follow. Assume that the population standard deviation is known and equal to 3.56, the standard deviation from the NBER study. Use the tool to compute the confidence interval estimate. The confidence interval estimate is ___ to ___. Since the confidence interval estimate ___ __, the null hypothesis (that the mean number of credit cards help by high-income consumers is 4.84) is ___.

Explanation / Answer

a.
Given that,
population mean(u)=4.84
standard deviation, =3.56
sample mean, x =5.4
number (n)=220
null, Ho: =4.84
alternate, H1: !=4.84
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 5.4-4.84/(3.56/sqrt(220)
zo = 2.33319
| zo | = 2.33319
critical value
the value of |z | at los 5% is 1.96
we got |zo| =2.33319 & | z | = 1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 2.33319 ) = 0.01964
hence value of p0.05 > 0.01964, here we reject Ho
ANSWERS
---------------
null, Ho: =4.84
alternate, H1: !=4.84
test statistic: 2.33319
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.01964
mean number of cousmers is diffrent from reported mean

b.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=5.4
Standard deviation( sd )=3.56
Sample Size(n)=220
Confidence Interval = [ 5.4 ± Z a/2 ( 3.56/ Sqrt ( 220) ) ]
= [ 5.4 - 1.96 * (0.24) , 5.4 + 1.96 * (0.24) ]
= [ 4.9296,5.8704 ]
mean number of cousmers is diffrent from reported mean

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