Sampling distribution, Type II errors, and one-tailed tests

kgkiwi11

For the past year, I have spent about 4.00$a day for lunch, give or take a quarter or so. (a) draw a rough sketch of this distribution of daily expenditures (b) if, without looking at the bill, I paid for my lunch with a 5$ bill and recieved 0.75$in change, should I worry that i was overcharged (c) explain logic in part (b) This has to do with sampling distribution but I have zero idea how to go about doing this. I was also asked what a "Type II error" would be in this question, and why I might want to adopt a one-tailed test, which tail I should choose, and what would happen if I chose the wrong tail. I know Z scores are probably used, but I can't figure out how to set this all up. trinker ggplot2orBust For the past year, I have spent about 4.00$ a day for lunch, give or take a quarter or so.
Hint: this is your mean and standard deviation

Go from there and if you need more help come back, report what you've tried and ask for further assistance. If you figure it out let us know that too.

kgkiwi11

New Member
The mean is $4; there are 2 standard deviations, both$.25, correct? Because it says give or take a quarter or so? Would you have advice on what the easiest way to convert $4.25 into a z score is? I'm trying to figure out how many SDs above or below the mean$4.25 is.

trinker

ggplot2orBust
The mean is $4; there are 2 standard deviations, both$.25, correct?
Yes but not the 2 standard deviations. .25 is just the standard deviation which is a measure of how much your payment varried from the mean. 2 standard deviations would actually be $$\pm .50$$ cents (which 95% of the population would fall into (so 95% of the time you paid between 3.50 and 4.50. You may want to check out some googling on the central limit theorem.

Would you have advice on what the easiest way to convert \$4.25 into a z score is?
Check out this wiki reading on standard (z scores) (LINK)

kgkiwi11

New Member
Alright, what I have right now is z=(4.25-4.00)/0.25 = 1
P( z > 1.5 ) = .159
If alpha is generally set at .05, then P>alpha
This means that I was most likely overcharged, correct?

trinker

ggplot2orBust
You getting more of it.

P( z > 1.5 ) = .159
Where did the 1.5 come from? I think you meant 1

If alpha is generally set at .05, then P>alpha. This means that I was most likely overcharged, correct?
We reject the null hypothesis when p is less than alpha. What does that mean for you?

kgkiwi11

New Member
I am a little confused, I was just told by somebody that the standard deviation for this equation is 0.083; this makes the z-score = 3.012. Do you have any idea how they got the number 0.083? Is this correct?

trinker

ggplot2orBust
I am a little confused, I was just told by somebody that the standard deviation for this equation is 0.083
No. who told you that? If it was someone in your class did you ask them why?

kgkiwi11

New Member
It was somebody who worked for Just Answers; I asked why and didn't get a good explanation, but oh well, I'll dismiss it. Still need to work on the original problem, I'll let you know what I come up with and see what advice you might have and go from there.