Acredit reporting agency claims that the mean credit card debt in a town is greater than $3500. a random sample of the credit card debt of 20 residents in that town has a mean credit card debt of $3619 and a standard deviation of $391. at α=0.10, can the credit agency’s claim be supported, assuming this is a normally distributed data set?

the correct answer is the second option

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;