In: Statistics and Probability
A study published in 2008 in the American Journal of Health Promotion (Volume 22, Issue 6) by researchers at the University of Minnesota (U of M) found that 124 out of 1,923 U of M females had over $6,000 in credit card debt while 61 out of 1,236 males had over $6,000 in credit card debt.
a. Construct a 95% confidence interval for the difference between the proportions of female and male U of M students who have more than $6000 in credit card debt (pf−pm). Round your sample proportions and margin of error to four decimal places
Answer: ( , )
b. Test, at 5% level, if there is evidence that the proportion of female students at U of M with more than $6000 credit card debt is greater than the proportion of males at U of M with more than $6000 credit card debt.
H0:pf=pm
Ha:pf _ pm
where pf: proportion of female U of M students with more than
$6000 credit card debt,
pm: proportion of male U of M students with more than $6000 credit
card debt
Pooled proportion (for standard error) p̂= ? (round to 4 decimal places)
Test statistic z= ? (round to 3 decimal places)
p-value= ? (round to 3 decimal places)
a)
sample #1 ----->
first sample size, n1= 1923
number of successes, sample 1 = x1=
124
proportion success of sample 1 , p̂1=
x1/n1= 0.0644826
sample #2 ----->
second sample size, n2 = 1236
number of successes, sample 2 = x2 =
61
proportion success of sample 1 , p̂ 2= x2/n2 =
0.049353
level of significance, α = 0.05
Z critical value = Z α/2 =
1.960 [excel function: =normsinv(α/2)
Std error , SE = SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 *
(1-p̂2)/n2) = 0.00833
margin of error , E = Z*SE = 1.960
* 0.0083 = 0.01632
confidence interval is
lower limit = (p̂1 - p̂2) - E = 0.015
- 0.0163 = -0.0011896
upper limit = (p̂1 - p̂2) + E = 0.015
+ 0.0163 = 0.0314493
so, confidence interval is (
-0.0012 < p1 - p2 <
0.0314 )
b)
Ho: p1 = p2
Ha: p1 > p2
sample #1 ----->
first sample size, n1=
1923
number of successes, sample 1 = x1=
124
proportion success of sample 1 , p̂1=
x1/n1= 0.0644826
sample #2 ----->
second sample size, n2 =
1236
number of successes, sample 2 = x2 =
61
proportion success of sample 1 , p̂ 2= x2/n2 =
0.049353
difference in sample proportions, p̂1 - p̂2 =
0.0645 - 0.0494 =
0.0151
pooled proportion , p =
(x1+x2)/(n1+n2)= 0.0586
std error ,SE = =SQRT(p*(1-p)*(1/n1+
1/n2)= 0.00856
Z-statistic = (p̂1 - p̂2)/SE = (
0.015 / 0.0086 ) =
1.767
p-value = 0.039
[excel function =NORMSDIST(-z)]
decision : p-value<α,Reject null hypothesis