Final answer:
To find a 98% confidence interval for the difference in means between the amount of cash carried by young men and young women, we can use the formula CI = (XM - XF) ± Z * sqrt((sm^2 / n1) + (sf^2 / n2)). Substituting the given values into the formula and determining the critical value Z for a 98% confidence level, the confidence interval is (−0.89, 15.63).
Step-by-step explanation:
To find a confidence interval for the difference in means between the amount of cash carried by young men and young women, we can use the formula:
CI = (XM - XF) ± Z * sqrt((sm^2 / n1) + (sf^2 / n2))
where XM is the mean amount of cash carried by young men, XF is the mean amount of cash carried by young women, sm is the standard deviation of cash carried by young men, sf is the standard deviation of cash carried by young women, n1 is the sample size of young men, n2 is the sample size of young women, and Z is the critical value corresponding to the desired confidence level.
Substituting the given values into the formula, we have:
CI = (78.25 - 70.88) ± Z * sqrt((21.27^2 / 50) + (14.44^2 / 60))
Now we need to determine the critical value Z for a 98% confidence level. Looking up in the standard normal distribution table, the Z value corresponding to a 98% confidence level is approximately 2.33.
Substituting this value into the formula, we have:
CI = (78.25 - 70.88) ± 2.33 * sqrt((21.27^2 / 50) + (14.44^2 / 60))
Simplifying the equation, we find:
CI = 7.37 ± 2.33 * sqrt(9.0261 + 3.5738)
CI = 7.37 ± 2.33 * sqrt(12.5999)
CI = 7.37 ± 2.33 * 3.55
CI = 7.37 ± 8.26
Therefore, the 98% confidence interval for the difference in means between the amount of cash carried by young men and young women is (−0.89, 15.63).