Question

Directions for questions 4 & 5: We selected a random sample of 100 StatCrunchU students, 67 females and 33 males, and analyzed their responses to the question, "What is the total amount (in dollars) of credit card debt you have accrued to date?" With more than 30 in each random and independent sample, conditions are met for modeling the distribution of differences in sample means using a T-model. Therefore, we will proceed with finding a confidence interval to estimate the gender difference in credit card debt for StatCrunchU students. Summary statistics for CC Debt: Group by: Gender Gender Mean Std. dev. n Female 2577.75 1916.29 67 Male 3809.42 2379.47 33 Use StatCrunch to find the 95% confidence interval estimating the difference µ1 – µ2, where µ1 is the mean credit card debt for all female StatCrunchU students and µ2 is the mean credit card debt for all male StatCrunchU students. (directions) Since the numbers are dollars, round to two decimal places when you enter your answer. Flag this Question Question 42 pts The lower limit on the confidence interval is

Answer 1

Answer:

The 95% confidence interval for the difference between means is (-2164.21, -299.13).

The lower limit on the confidence interval is -$2164.21.

The upper limit on the confidence interval is -$299.13.

Step-by-step explanation:

The sample data is:

Gender   Mean          Std. dev.     n

Female    2577.75      1916.29     67

Male        3809.42     2379.47     33

We have to calculate a 95% confidence interval for the difference between means, with a T-model.

The sample 1, of size n1=67 has a mean of 2577.75 and a standard deviation of 1916.29.

The sample 2, of size n2=33 has a mean of 3809.42 and a standard deviation of 2379.47.

The difference between sample means is Md=-1231.67.

$M_d=M_1-M_2=2577.75-3809.42=-1231.67$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1916.29^2}{67}+\dfrac{2379.47^2}{33}}\\\\\\s_{M_d}=\sqrt{54808.468+171572.045}=\sqrt{226380.513}=475.795$

The t-value for a 95% confidence interval is t=1.96.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=1.96 \cdot 475.795=932.54$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = -1231.67-932.54=-2164.21\\\\UL=M_d+t \cdot s_{M_d} = -1231.67+932.54=-299.13$

The 95% confidence interval for the difference between means is (-2164.21, -299.13).