This statistics problem is insanely hard.

Toggle sidebar Toggle sidebar
M

ManBearPig

Diamond Member
Sep 5, 2000
9,173
6
81
A student suspected that the average cost of a Saturday night date was no longer $30.00. To test her hypothesis, she randomly selected 16 men from the dormitory and asked them how much they spent on a date last Saturday. She found that the average cost was $31.17. The standard deviation of the sample was $5.51.

a. At a = 0.05, is there enough evidence to support her claim?
b. At a = 0.10 is there enough evidence to support her claim?
c. Use a 95% confidence interval, to test the student?s hypothesis?
d. Use a 90% confidence interval, to test the hypothesis?
e. How do your conclusions in a and c relate? What is the relationship between b and d?

I found the critical value, the z score and z test value for 0.05 (for 0.1 itd be the same format), but ive got no effing idea how to find out if there is enough evidence or not...wtf does that mean? i assume i have to compare two of the numbers, but which?

THANKS

p.s. here are the values in case someone thinks im lying:

H0 = µ = $30
H1 = µ < $30

Z = (30-31.17)/(5.51) = -0.2123

Z test = (30-31.17) / ((5.51)/v16)) = -0.8494
 
Hacp

Hacp

Lifer
Jun 8, 2005
13,923
2
81
That was years ago lol. I hated business stat, especially those Z tables.
 
Leros

Leros

Lifer
Jul 11, 2004
21,867
7
81
The test is invalid because it does not guarantee that all the men she selected went on dates. Therefor the sampled data does not have any relevance on the question being asked.

And seriously, don't you need more than 40 samples for the formulas to work? Isn't there something about needing more than 40 samples?
 
M

ManBearPig

Diamond Member
Sep 5, 2000
9,173
6
81
Originally posted by: Leros
The test is invalid because it does not guarantee that all the men she selected went on dates. Therefor the sampled data does not have any relevance on the question being asked.
Click to expand...

haha, id love to be able to get away with that but im sure shed freak out.
 
Jadow

Jadow

Diamond Member
Feb 12, 2003
5,962
2
0
Jadow Date:

"Hey babe let's split a 6 piece bucket and get it on!"

results greasy sex FTW
 
S

Soccer55

Golden Member
Jul 9, 2000
1,660
4
81
Originally posted by: Kazaam
A student suspected that the average cost of a Saturday night date was no longer $30.00. To test her hypothesis, she randomly selected 16 men from the dormitory and asked them how much they spent on a date last Saturday. She found that the average cost was $31.17. The standard deviation of the sample was $5.51.

a. At a = 0.05, is there enough evidence to support her claim?
b. At a = 0.10 is there enough evidence to support her claim?
c. Use a 95% confidence interval, to test the student?s hypothesis?
d. Use a 90% confidence interval, to test the hypothesis?
e. How do your conclusions in a and c relate? What is the relationship between b and d?

I found the critical value, the z score and z test value for 0.05 (for 0.1 itd be the same format), but ive got no effing idea how to find out if there is enough evidence or not...wtf does that mean? i assume i have to compare two of the numbers, but which?

THANKS

p.s. here are the values in case someone thinks im lying:

H0 = µ = $30
H1 = µ < $30

Z = (30-31.17)/(5.51) = -0.2123

Z test = (30-31.17) / ((5.51)/v16)) = -0.8494
Click to expand...

First, you have your Z score formula wrong. You have (u - x)/sd when it should actually be (x-u)/sd [where u is the population mean and sd is the standard deviation]. Next, determine if you have a 2-tailed, right tailed, or left tailed test and determine the critical values for the 0.05 and 0.10 significance levels. To finally arrive at an answer for a and b, you need to take your test statistic and compare it to the critical values for the 0.05 and 0.10 significance levels. The test statistic being less than or greater than your critical value(s) will tell you whether or not you have enough evidence to reject the null hypothesis. For parts c and d, there are formulas to calculate the confidence interval limits, so you can just look those up in your textbook and it should be fairly straight forward from there.

PS - You will also want to determine whether this involves a standard normal distribution (Z-scores) or if you should use the T-distribution (T-scores).
 
R

rgwalt

Diamond Member
Apr 22, 2000
7,393
0
0
Sorry, I took real math classes, like calculus and differential equations.

:D

J/K, stats is a tough subject. I never wanted any part of it.
 
B

blinky8225

Senior member
Nov 23, 2004
564
0
0
First, you should be doing a two sided test.
H0: u == $30
H1: u =/= $30

Let your test statistic Z = (16)^(1/2) (X - 3000)/551, where X is average of the cost of the dates.

a) if you are going to test at level of significance 0.05, you will reject H0 if Z > 1.96 or Z < -1.96. This is not the case. Accept H0.
b) if you are going to test at level of significance 0.10, you will reject H0 if Z > 1.645 or Z < -1.645. Again this is not the case. Accept H0.
c) if you want to use a confidence interval of 95%, you find that the interval for the mean u is from 28.4701 to 33.8699. 30 falls into this interval. Again, accept H0.
d) if you want to use a confidence interval of 90%, you find that the interval for the mean u is from 28.9040 to 33.4360. 30 falls into this interval. Again, accept H0.
e) it turns out that A & C are the same test. B & D are also the same test.

So no, there is not enough evidence to support her claim using the above tests.
 
dbk

dbk

Lifer
Apr 23, 2004
17,685
10
81
Originally posted by: blinky8225
First, you should be doing a two sided test.
H0: u == $30
H1: u =/= $30

Let your test statistic Z = (16)^(1/2) (X - 3000)/551, where X is average of the cost of the dates.

a) if you are going to test at level of significance 0.05, you will reject H0 if Z > 1.96 or Z < -1.96. This is not the case. Accept H0.
b) if you are going to test at level of significance 0.10, you will reject H0 if Z > 1.645 or Z < -1.645. Again this is not the case. Accept H0.
c) if you want to use a confidence interval of 95%, you find that the interval for the mean u is from 28.4701 to 33.8699. 30 falls into this interval. Again, accept H0.
d) if you want to use a confidence interval of 90%, you find that the interval for the mean u is from 28.9040 to 33.4360. 30 falls into this interval. Again, accept H0.
e) it turns out that A & C are the same test. B & D are also the same test.
Click to expand...

wow..nicely done brah
 
B

blinky8225

Senior member
Nov 23, 2004
564
0
0
Originally posted by: dbk
Originally posted by: blinky8225
First, you should be doing a two sided test.
H0: u == $30
H1: u =/= $30

Let your test statistic Z = (16)^(1/2) (X - 3000)/551, where X is average of the cost of the dates.

a) if you are going to test at level of significance 0.05, you will reject H0 if Z > 1.96 or Z < -1.96. This is not the case. Accept H0.
b) if you are going to test at level of significance 0.10, you will reject H0 if Z > 1.645 or Z < -1.645. Again this is not the case. Accept H0.
c) if you want to use a confidence interval of 95%, you find that the interval for the mean u is from 28.4701 to 33.8699. 30 falls into this interval. Again, accept H0.
d) if you want to use a confidence interval of 90%, you find that the interval for the mean u is from 28.9040 to 33.4360. 30 falls into this interval. Again, accept H0.
e) it turns out that A & C are the same test. B & D are also the same test.
Click to expand...

wow..nicely done brah
Click to expand...
Thanks. If only I had questions like this on my statistics test last week, I might have actually passed.

<--Math major at Duke
 
M

ManBearPig

Diamond Member
Sep 5, 2000
9,173
6
81
are you freaking kidding? i thought she meant no more than instead of no longer, thats why i wasnt getting anything. CRAP.

thanks a lot guys!
 
M

ManBearPig

Diamond Member
Sep 5, 2000
9,173
6
81
can anyone point me in the right direction for this one please?

A film editor feels that the standard deviation for the number of minutes in a video is 3.4 minutes. A sample of 24 videos has a standard deviation of 4.2 minutes.
a. At a=0.05, is the sample standard deviation different from what the editor hypothesized?
b. At a=0.10, is the sample standard deviation different from what the editor hypothesized?
 
M

ManBearPig

Diamond Member
Sep 5, 2000
9,173
6
81
alright, ive got no effing idea how to do the chi squared one. i looked at the chi-squared distribution but i dunno how to apply this problem to that! damnit. at least its not worth tons of points.
 
B

blinky8225

Senior member
Nov 23, 2004
564
0
0
Brah, you're lucky that I'm bored tonight. The statistic X = n*v'/v has chi-squared distribution with n-1 degrees of freedom, where v' is the sample variance, and the v is the actual variance.

n = 24
v' = 4.2^2 = 17.64

H0: v == 3.4^2 = 11.56
H1: v =/= 11.56

You simply need to pick two values c1 and c2, where you reject if X > c2 or X < c1. You pick them so that P[X > c2 or X < c1] = a. You just have to put in random numbers for c1 and c2. I normal pick them so that they are centered around the mean of X, which is 23 in this case. There's other ways to do this, and you should follow the convention your book or professor uses. With my convention, you will reject H0 for both levels of significance 0.05 and 0.10. It's possible to pick an interval such that you would accept H0 at a = 0.05, though. You're pretty much always going to reject H0 at a = 0.10, though.

So, the director is probably wrong.
 
M

ManBearPig

Diamond Member
Sep 5, 2000
9,173
6
81
Originally posted by: blinky8225
Brah, you're lucky that I'm bored tonight. The statistic X = n*v'/v has chi-squared distribution with n-1 degrees of freedom, where v' is the sample variance, and the v is the actual variance.

n = 24
v' = 4.2^2 = 17.64

H0: v == 3.4^2 = 11.56
H1: v =/= 11.56

You simply need to pick two values c1 and c2, where you reject if X > c2 or X < c1. You pick them so that P[X > c2 or X < c1] = a. You just have to put in random numbers for c1 and c2. I normal pick them so that they are centered around the mean of X, which is 23 in this case. There's other ways to do this, and you should follow the convention your book or professor uses. With my convention, you will reject H0 for both levels of significance 0.05 and 0.10. It's possible to pick an interval such that you would accept H0 at a = 0.05, though. You're pretty much always going to reject H0 at a = 0.10, though.

So, the director is probably wrong.
Click to expand...

sweet jebus, i have to read that over a few times. i dont think its the same way as our professor taught us, but then, well...she didnt really teach us. thanks man.
 
B

blinky8225

Senior member
Nov 23, 2004
564
0
0
Originally posted by: Kazaam
sweet jebus, i have to read that over a few times. i dont think its the same way as our professor taught us, but then, well...she didnt really teach us. thanks man.
Click to expand...
Yeah, it's possible the question meant for you only to do a one-sided test, which would make it a lot easier. The question isn't very clear in this case. In the one-sided case, you would use:

H0: v =< 3.4^2 = 11.56
H1: v > 11.56

You would still reject H0 at both a = 0.05 and a = 0.1 in this case.
 
K

KayGee

Senior member
Sep 16, 2004
268
0
76
You can't use the normal distribution if the population standard deviation (sigma) is unknown. The general rule is to use the t-distribution if the sample size is less than 30 or sigma is unknown. Also, this is a two-tailed test.
 
You must log in or register to reply here.

TRENDING THREADS

COMPANY
RESOURCES
FOLLOW
Top Bottom