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. (Give answers to 2 decimal places) Store's Card Major Credit Card Sample size 64 49Sample mean $140 $125Population variance $100 $641. A point estimate for the difference between the mean purchases of the users of the two credit cards is:.2. At 95% confidence, the margin of error is:.3. A 95% confidence interval estimate for the difference between the average purchases of the customers using the two different credit cards is___ to ___.
4. The test statistic for an alpha of .05 is:.

Step-by-step explanation:

1) Point estimate is the difference between the sample means. Therefore,

A point estimate for the difference between the mean purchases of the users of the two credit cards is = 140 - 125 = $15

2) Margin of error = z√(σ²/n1 + σ2²/n2)

Where

z is the z score from the 95% confidence level. From the normal distribution table, z = 1.96

s1 and s2 are standard deviation for both customers respectively.

Standard deviation = √variance

σ1 = √100 = 10

σ2 = √641 = 25.32

Margin of error = 1.96√(10²/64 + 25.32²/49 = 7.5

At 95% confidence, the margin of error is 7.5

3) The confidence interval for the difference of two population means is expressed as point estimate ± margin of error

Confidence interval = 15 ± 7.5

The upper boundary for the confidence interval is

15 - 7.5 = 7.5

The lower boundary for the confidence interval is

15 + 7.5 = 22.5

A 95% confidence interval estimate for the difference between the average purchases of the customers using the two different credit cards is $7.5 to $22.5

4) Since the population standard deviations are known, we would use the formula to determine the test statistic(z score)

z = (x1 - x2)/(√σ1²/n2 + σ2²/n2)

z = (140 - 125)/√10²/64 + 25.32²/49

z = 1.02

The test statistic for an alpha of .05 is 1.02