Confidence interval for telephone spending using MegaStat

[mexicnfmle]

Homework Statement

The Warren County Telephone Company claims in its annual report that “the typical customer spends $60 per month on local and long distance service.” A sample of 12 subscribers revealed the following amounts spent last month.

$64 $66 $64 $66 $59 $62 $67 $61 $64 $58 $54 $66

a. What is the point estimate of the population mean?


b. Develop a 90 percent confidence interval for the population mean.

c. Is the company’s claim that the “typical customer” spends $60 per month reasonable?
Justify your answer using the confidence interval.

Homework Equations

Sample population? 12
Mean=$63
Confidence Interval 90%

The Attempt at a Solution

I know I can use megastats to solve this problem, I am just drawing a blank on which formula to use. WHen I click on Descriptive statistics, it comes back saying "not enough data selected". I tried Confidence Interval and I tried both:
Sample size-mean
sample size- p

I must be entering something wrong somewhere.
Can someone just tell me which formula to use? I am drawing a blank..

Thanks,
Emily

 

[mfb]

You can approximate the distribution by a Gaussian distribution with the same mean and variance, assuming that's a reasonable approximation.
With just 12 entries using the data directly won't give a good estimate (what is half a subscriber?).

 

[FactChecker]

Although the number of samples is small for real accuracy, this is just an exercise to learn the concepts. If you are allowed to use other calculators in this class, try using this.

 

[pasmith]

I think the amount of time a customer spends on the phone each month probably follows an exponential distribution, and asusming they are charged at a fixed rate per minute the amount they spend will also follow an exponential distribution.

However as this is an introductory exercise I think the assumption is that the amount spent is normally distributed. If the population follows an $N(\mu,\sigma^2)$ distribution then the mean of a sample of size $n$ follows an $N(\mu, \sigma^2/n)$ distribution. Thus with probability 0.9 the sample mean $\bar{x}$ and population mean satisfy $$\bar{x} - \frac{\sigma}{\sqrt{n}}\Phi^{-1}(0.95) \leq \mu \leq \bar{x} + \frac{\sigma}{\sqrt{n}}\Phi^{-1}(0.95).$$ An unbiased estimate for the population variance $\sigma^2$ is $$\frac{1}{n - 1}\sum_i(x_i - \bar{x})^2$$ which is $n/(n-1)$ times the sample variance.

 

[BWV]

If the time spent on the phone is normal, how do you restrict t from taking a negative value?

 

[FactChecker]

If the time spent on the phone is normal, how do you restrict t from taking a negative value?

Looking at the data, I would say that the mean and standard deviation indicate that there is only a tiny bit of the normal distribution that would be negative. The difference between a normal model and an always-positive Gamma distribution model is negligible for this case.
That is assuming that the lower limit is at $x=0$. Suppose there is a fixed cost such as at 52, so $x \ge 52$. Then, for this sample mean and variance, the normal distribution would violate the constraint with a higher probability and would be significantly worse than a Gamma.

 

[mfb]

0 is over 10 standard deviations away (<<10^-23), so it's completely negligible.

 

[pasmith]

However as this is an introductory exercise I think the assumption is that the amount spent is normally distributed. If the population follows an $N(\mu,\sigma^2)$ distribution then the mean of a sample of size $n$ follows an $N(\mu, \sigma^2/n)$ distribution. Thus with probability 0.9 the sample mean $\bar{x}$ and population mean satisfy $$\bar{x} - \frac{\sigma}{\sqrt{n}}\Phi^{-1}(0.95) \leq \mu \leq \bar{x} + \frac{\sigma}{\sqrt{n}}\Phi^{-1}(0.95).$$ An unbiased estimate for the population variance $\sigma^2$ is $$\frac{1}{n - 1}\sum_i(x_i - \bar{x})^2$$ which is $n/(n-1)$ times the sample variance.

Correction: the quantity $(\bar{x} - \mu)/(S/\sqrt{n})$ where $S^2 = \frac{1}{n - 1}\sum_i(x_i - \bar{x})^2$ follows a $t$ distribution with $(n-1)$ degrees of freedom, so we need the inverse CDF of that rather than the inverse CDF of the normal distribution.

 

[WWGD]

Correction: the quantity $(\bar{x} - \mu)/(S/\sqrt{n})$ where $S^2 = \frac{1}{n - 1}\sum_i(x_i - \bar{x})^2$ follows a $t$ distribution with $(n-1)$ degrees of freedom, so we need the inverse CDF of that rather than the inverse CDF of the normal distribution.

Doesn't the $t$ approach a normal for n close to 15?