A random sample of 36 shoppers spent an average of $24.35 per visit at Barnes & Noble with a standard deviation of $2.65. Find the 99% confidence interval of the true mean cost per visit.

The 99% confidence interval of the true mean of the cost per visit is `23.15<mu<25.55`

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We are given that a random sample of n=36 shoppers had a mean purchase of `bar(x)=$24.35` with a sample standard deviation of `s=$2.65` . (We are not told the population standard deviation.) We are asked to find the 99% confidence interval (`alpha=.01` ) of the true mean cost `mu` per...

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We are given that a random sample of n=36 shoppers had a mean purchase of `bar(x)=$24.35` with a sample standard deviation of `s=$2.65` . (We are not told the population standard deviation.) We are asked to find the 99% confidence interval (`alpha=.01` ) of the true mean cost `mu` per visit.

We are not told whether the underlying population is normally distributed, but the sample size is greater than 30, so we can compute the interval. Since we do not have the population standard deviation, we use the `t-"distribution"`

The interval begins with a point estimate. The mean of the sample is the best point estimate for our purposes. An error value is added to and subtracted from that point estimate to create the interval. The error consists of a factor computed from the confidence level called `t_(alpha/2)` and the standard error of the mean, which is `s/sqrt(n)` .

Thus, the interval is `bar(x) pm t_(alpha/2)(s/sqrt(n))` .

The degrees of freedom are 35 (d.f.=n-1; if you were told the mean of the 36 numbers and given 35 of the numbers, you would be able to get the 36th number). From a t-table with degrees of freedom 35 and alpha=.01, we get `t_(alpha/2)~~2.724` .

So, the interval is `24.35 pm 2.724(2.65/sqrt(36))=24.35 pm 1.2031`,

or `23.15<mu<25.55`.

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