Question

While researching the cost of school lunches per week across the state, you use a sample size of 45 weekly lunch prices. The standard deviation is known to be 68 cents. In order to be 90% confident, what is the margin of error in units of dollars?

Answer 1

We assume the lunch prices we observe are drawn from a normal distribution with true mean
`\mu` and standard deviation 0.68 in dollars.

We average
`n=45` samples to get
`\bar{x}`.

The standard deviation of the average (an experiment where we collect 45 samples and average them) is the square root of n times smaller than than the standard deviation of the individual samples. We'll write

`\sigma = 0.68 / √(45) = 0.101`

Our goal is to come up with a confidence interval (a,b) that we can be 90% sure contains
`\mu`.

Our interval takes the form of
`( \bar{x} - z \sigma, \bar{x} + z \sigma )` as
`\bar{x}` is our best guess at the middle of the interval. We have to find the z that gives us 90% of the area of the bell in the "middle".

Since we're given the standard deviation of the true distribution we don't need a t distribution or anything like that. n=45 is big enough (more than 30 or so) that we can substitute the normal distribution for the t distribution anyway.

Usually the questioner is nice enough to ask for a 95% confidence interval, which by the 68-95-99.7 rule is plus or minus two sigma. Here it's a bit less; we have to look it up.

With the right table or computer we find z that corresponds to a probability p=.90 the integral of the unit normal from -z to z. Unfortunately these tables come in various flavors and we have to convert the probability to suit. Sometimes that's a one sided probability from zero to z. That would be an area aka probability of 0.45 from 0 to z (the "body") or a probability of 0.05 from z to infinity (the "tail"). Often the table is the integral of the bell from -infinity to positive z, so we'd have to find p=0.95 in that table. We know that the answer would be z=2 if our original p had been 95% so we expect a number a bit less than 2, a smaller number of standard deviations to include a bit less of the probability.

We find z=1.65 in the typical table has p=.95 from -infinity to z. So our 90% confidence interval is

`( \bar{x} - 1.65 (.101), \bar{x} + 1.65 (.101) )`

in other words a margin of error of

`\pm 1.65(.101) = \pm 0.167` dollars

That's around plus or minus 17 cents.

Answer 2

0.1668 (using round up rule)

Explanation:

The given values are

sigma=.68

n=45 weekly lunches

a 90% confident

1) 90%=.9

2) 1-.9=.1

3) .1/2=.05

4) 1-.05=.95

5)critical value or z sub alpha over 2 =1.645 according to a z-scale

6) formula needed: critical value times sigma divided by the square root of n equals E

7) 1.645x.68/sqrt(45)=0.1667

with the round up rule it is 0.1668 without the rule it is just 0.1667