# URGENT HELP Required re. Statistical Inference - PLEASE!! ;-)

• May 7th 2007, 06:32 AM
Dybbuk
URGENT HELP Required re. Statistical Inference - PLEASE!! ;-)
Hello,

I've lost my revision notes for an exam tomorrow (I know, I know!) and I was hoping someone could help me with the following:

1. deduce the population mean (u)
2. demonstrate how to carry out a hypothesis test at the 0.95 probability level as to whether population mean is greater than 80.

---

This is for grouped data, and I have the following info:

n = 207

Arithmetic mean = 86.64
Median = 79.02
Standard deviation = 24.38
Co-efficient of variation = 28.14
Ql = 69.69
Qu = 102.92
Interquartile range = 86.31
Skewness = 0.94

sum of fx' = 17935
sum of fx'² = 1676975

Can anyone help??
• May 7th 2007, 12:48 PM
CaptainBlack
Quote:

Originally Posted by Dybbuk
Hello,

I've lost my revision notes for an exam tomorrow (I know, I know!) and I was hoping someone could help me with the following:

1. deduce the population mean (u)
2. demonstrate how to carry out a hypothesis test at the 0.95 probability level as to whether population mean is greater than 80.

---

This is for grouped data, and I have the following info:

n = 207

Arithmetic mean = 86.64
Median = 79.02
Standard deviation = 24.38
Co-efficient of variation = 28.14
Ql = 69.69
Qu = 102.92
Interquartile range = 86.31
Skewness = 0.94

sum of fx' = 17935
sum of fx'² = 1676975

Can anyone help??

If all of these are derived from the sample, then you cannot deduce the
population mean, at best you could estimate it.

Also the given coefficient of variation is impossible with mean and standard
deviation of the order of 86 and 24 respectivly.

RonL
• May 7th 2007, 12:53 PM
Dybbuk
Quote:

Originally Posted by CaptainBlack
If all of these are derived from the sample, then you cannot deduce the
population mean, at best you could estimate it.

Quote:

Originally Posted by CaptainBlack
Also the given coefficient of variation is impossible with mean and standard
deviation of the order of 86 and 24 respectivly.

Not sure I understand... please correct me if I'm wrong.

Coefficient of Variation (V%) = s/Xbar * 100 = 24.38/86.64 * 100 = 28.14
• May 7th 2007, 01:19 PM
CaptainBlack
Quote:

Originally Posted by Dybbuk
Not sure I understand... please correct me if I'm wrong.

Coefficient of Variation (V%) = s/Xbar * 100 = 24.38/86.64 * 100 = 28.14

Coefficient of variation=s/Xbar=0.2814 or 28.14%,

You did not mention percentages in you question, the default withou
qualification is a decimal fraction not a percentage. If you wish to give the
coefficient of variation as a percentage you have to say that is what it is.

RonL
• May 7th 2007, 01:24 PM
Dybbuk
Thank you, my mistake.

Regarding the population mean, I'm not very far into this but have the following:

H0 = u > 80
H1 = u (equal or less) 80
1.66 (1 sided 95% test)

Test statistic = 86.64 - 80 / SE = 3.9184972

where SE = 1.6945272 (24.38 / sqroot of 207)

Not quite sure where to go now with the test statistic...
• May 7th 2007, 01:35 PM
CaptainBlack
Quote:

Originally Posted by Dybbuk
Hello,

I've lost my revision notes for an exam tomorrow (I know, I know!) and I was hoping someone could help me with the following:

1. deduce the population mean (u)

Usual estimator of the population mean is the sample mean (though
caution is called for here as we have skewed data)

Quote:

2. demonstrate how to carry out a hypothesis test at the 0.95 probability level as to whether population mean is greater than 80.
Here we have a large sample so we assume that the distribution of the
sample means are normal with mean mu and SD sigma/sqrt(N), where mu is
the population mean, sigma the standard deviation of the population, and N
is the sample size. We also assume that we can use the standard deviation
s in place of the unknown population SD.

RonL
• May 7th 2007, 01:40 PM
Dybbuk
So, would I be correct in the following:

Confidence Interval = 83.32, 89.96
• May 7th 2007, 01:52 PM
CaptainBlack
Quote:

Originally Posted by Dybbuk
So, would I be correct in the following:

Confidence Interval = 83.32, 89.96

You will note that I have changed my post, as we dont want a 95%
confidence interval, but a one sided 95% test.

This is essentialy that the population mean should not be less than

86.64 - 1.645*s/sqrt(207) ~= 83.85,

where the 1.645 is the onesided 95% critical value for the standard normal
distribution.

RonL
• May 7th 2007, 01:57 PM
Dybbuk
Thanks.

Shouldn't the 1.645 value be 1.66 for a one tailed test at 95%?

BUT in relation to the question, would this imply that it is "95% probable" that the population mean is greater than 80? Would that be a reasonable way to phrase the answer?
• May 7th 2007, 08:51 PM
CaptainBlack
Quote:

Originally Posted by Dybbuk
Thanks.

Shouldn't the 1.645 value be 1.66 for a one tailed test at 95%?

My table gives 1.645, but it has been in error before, checking a second
source though confirms the 1.645.

Quote:

BUT in relation to the question, would this imply that it is "95% probable" that the population mean is greater than 80? Would that be a reasonable way to phrase the answer?
The wording of these things is always tricky. I think that this is close enough
to be acceptable. A more carefull wording might be "There is a probability of
95% that the interval [83.85, infty) contains the population mean.

RonL
• May 8th 2007, 06:07 AM
Dybbuk
Thanks for your help with this.

I got through the exam ok and I think generally I answered the questions well. Not 100% confident about "deducing population mean u" as I didn't have the notes for that, but in terms of hypothesis testing I probably didn't drop many points.

Thanks again.

BTW 1.645 was indeed correct (my notes suggest otherwise unfortunately!)