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In a sample of credit card holders, the mean monthly value of credit card purchases was 358 with a sample variance of 49, which means their sample standard deviation would be 7.
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Assume that the population distribution is normal, answer the following rounding your answers to two decimal places where appropriate.
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So part a, suppose the sample results were obtained from a random sample of 15 credit card holders.
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Find a 95 % confidence interval.
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Well, 15 is a small sample, so we're going to use a t distribution for this.
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So t of 0 .05 divided by 2 with a degree of freedom of 14 means that our critical value is going to be 2 .14.
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So we're going to take 358 plus or minus 2 .145 times 7 divided by the square root of 15.
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And that's going to give us 358 plus or minus a margin of error of 3 .88 for an interval of 354 .1 -12 to 361 .88.
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Suppose the sample results were obtained from a random sample of 22 credit card holders...