Question

The management of a department store is interested in estimating the difference between the mean credit purchases of customers using the store's credit card versus those customers using a national major credit card. You are given the following information. Assume the samples were selected randomly. Store's Card Major Credit Card Sample size 64 49 Sample mean $140 $125 Population standard deviation $10 $8 Refer to Exhibit 10-6. At 95% confidence, the margin of error is

Answer 1

The margin of error is of $3.3124.

Explanation:

Subtraction of normal variables:

When we subtract normal variables, the mean is the subtraction of the mean, while the standard deviation is the square root of the sum of the variances.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Store's Card:

Sample of 64, mean of 140, standard deviation of 10, sample of 64, standard error of
`s = (10)/(√(64)) = 1.25`

Major Credit Card:

Sample of 39, mean of 125, Standard deviation of 8, Sample of 49, standard error of
`s = (8)/(√(49)) = 1.14`

Subtraction:

Mean: 140 - 125 = 15

Standard deviation:

`\sigma = √(1.25^2+1.14^2) = 1.69`

Confidence interval:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1 - 0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.025 = 0.975`, so Z = 1.96.

Now, find the margin of error M as such

`M = zs`

Then

`M = 1.96*1.69 = 3.3124`

The margin of error is of $3.3124.