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A bank wonders whether omitting the annual credit card fee for customers who charge at least $3,000 in a year would increase the amount charged on their credit card. The bank makes an offer to an SRS of 500 existing credit card customers. It then compares how much these customers charge this year with the amount they charges last year. The mean increase is $565, and the standard deviation is $267.
Give a 95% confidence interval for the mean amount of the increase. Interpret your answer in simple English.

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Answer:

The 95% confidence interval for the mean amount of the increase is ($541.6, $588.4). This means that we are 95% sure that the mean amount of increase of all customers who charge at least $3,000 in a year is between these two values.

Step-by-step explanation:

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

[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 1.96\frac{267}{\sqrt{500}} = 23.4[/tex]

The lower end of the interval is the sample mean subtracted by M. So it is 565 - 23.4 = $541.6

The upper end of the interval is the sample mean added to M. So it is 565 + 23.4 = $588.4

The 95% confidence interval for the mean amount of the increase is ($541.6, $588.4). This means that we are 95% sure that the mean amount of increase of all customers who charge at least $3,000 in a year is between these two values.

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