Respuesta :
We assume the lunch prices we observe are drawn from a normal distribution with true mean [tex] \mu [/tex] and standard deviation 0.68 in dollars.
We average [tex] n=45 [/tex] samples to get [tex] \bar{x} [/tex].
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
[tex]\sigma = 0.68 / \sqrt{45} = 0.101 [/tex]
Our goal is to come up with a confidence interval (a,b) that we can be 90% sure contains [tex] \mu [/tex].
Our interval takes the form of [tex]( \bar{x} - z \sigma, \bar{x} + z \sigma ) [/tex] as [tex] \bar{x} [/tex] 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
[tex]( \bar{x} - 1.65 (.101), \bar{x} + 1.65 (.101) ) [/tex]
in other words a margin of error of
[tex] \pm 1.65(.101) = \pm 0.167 [/tex] dollars
That's around plus or minus 17 cents.
Answer
0.1668 (using round up rule)
Step-by-step 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
