# Stat Problems Help (Homework)

#### Stats student 2008

##### New Member
1. A study in the journal of the American Medical Association examined people to see if they showed any signs of IRS (insulin resistance syndrome) involving major risk factors for Type 2 diabetes and heart disease. Among 102 subjects with high weekly dairy consumption, 24 were identified with IRS. In comparison, IRS was identified in 85 of 190 individuals with low dairy consumption. Is there strong evidence that IRS risk is different in people with high dairy consumption than those with low dairy consumption?

2. A jury list contains the names of all individuals who may be called for jury duty. The proportion of the available jurors on the list who are woman is .54. If 40 people are selected to serve as candidates for being picked on the jury, show all steps of significance test of the hypothesis that the selections are random with respect to gender. 8 of the 40 selected were woman. Use a significance level of 0.02.

HW word problems I am stuck on. I have an exam this week -- please help and do this so I get it! Thanks!

#### Dragan

##### Super Moderator
...the names of all individuals who may be called for jury duty. The proportion of the available jurors on the list who are woman is .54. If 40 people are selected to serve as candidates for being picked on the jury, show all steps of significance test of the hypothesis that the selections are random with respect to gender. 8 of the 40 selected were woman. Use a significance level of 0.02.

On #2 - a Chi-square test. (I'll let someone else do #1)

You have p=.54 under the null i.e. 40*.54 = 21.6. These are your expected frequencies.

Now, your observations are .8*40 = 32 are women and 8 are men.

Compute you Chi-Square statistic as: Chi^2 = Sum {[ O - E ]^2 / E} and compare it to the critical value on df=1 for alpha (0.02) = 5.41189

Reject the the null if your computed Chi-square statistic exceeds the critical value.

#### vinux

##### Dark Knight
Hi dragon,
I agree with your point. But this test will be an approximate level.
And in this situation , the one sample Proportion test will do

here H0 : Po= 0.54 Vs Po not =0.54
and we wanted to test that the proportion we got P^ = 8/40 = 0.20 is due to random? ( is it 0.8 or 8/40 ? I am taking it as 8/40)

and the test statistic is
z = (P^ -Po)/ {sqrt( Po* Qo / n)}
where Qo = 1- Po

calculate z value and it will be 4.31 ( check it)
and ta for significance level of 0.02 will be 2.32 [ ie P( -2.32 <Z <2.32 ) = .98]

So Reject the hypothesis. So the selection was not random..

and the Answer of the first question is Two sample proportional test.

Regards

#### Dragan

##### Super Moderator
Hi dragon,
I agree with your point. But this test will be an approximate level.
And in this situation , the one sample Proportion test will do

Vinux: Is my approximation too far off?

Specifically, I get a chi-square statistic of (about) 13.5704 with an associated p value of p=0.00023 which is obviously less than the critical alpha level of 0.02. i.e. reject the null as you indicated above.

#### Stats student 2008

##### New Member

3. The Lakeside Sand Co. has a contract with the San Diego River authority. The sand co. pays the river authority \$25 per truck load for sand. The contract states, "the truckloads may average no more than 25,000 lbs". To determine whether the sand co. is in compliance w/ the contract the river authority weighs 37 loads of sand. The ave. is 25,175 lbs with a standard deviation of 860 lbs. test if the co. is in compliance at the .05 level.

4. Last month a random sample of 1,000 subjects was interviewed and asked whether they think the president is doing a good job. This month the same subjects were asked this again. The results are: 450 said yes each time, 450 said no each time, 60 said yes on the first survey and yes on the second survey.

Find the test statistic and P-value for applying McNemar's test that the population proportion was the same each month, and interpret.

#### vinux

##### Dark Knight
Hi dragon,
Vinux: Is my approximation too far off?
Both are Actually the same( If it is two sided test)

Sum {[ O - E ]^2 / E} this can be written as
[ n*P^ - n*Po ]^2 / n*Po) + [ n*Q^ - n*Qo ]^2 / n*Qo)
=[ n*P^ - n*Po ]^2 /{ n *[ 1/Po + 1/Qo] } [ Since ( Q^ -Qo)**2 = ( P^ -Po)**2 ]
=[ n*P^ - n*Po ]^2 /{ n *Po Qo }

= [ P^ - Po ]^2 /{ Po Qo/n }
=z**2 where
z = (P^ -Po)/ {sqrt( Po* Qo / n)}
And Square of Standard normal is nothing but chisquare... So both are the same.
But I feel Proportional test is the more simpler term than chisquare. ( at least in the curriculam first will be proportional test and chisqure test will be later).
Anyway it was fun .
-------------------------------------
And Mr.Stats student 2008

Answer of third is onesided T test
and fourth .. are you sure about the statement
The results are: 450 said yes each time, 450 said no each time, 60 said yes on the first survey and yes on the second survey.

is it belonging to first 450?..

Regards

#### Dragan

##### Super Moderator
Both are Actually the same...

Vinux: Yes, I know, but remember, we're both computing of an approximate p-value - not exact. #### Stats student 2008

##### New Member
Are these answers working it out from beginning to end and then arriving at the answer? Cuz for the exam we're allowed 2 pages of notes and I'll be using these examples...

#### vinux

##### Dark Knight
Hi Stats student 200,
I think you are expecting a ready made answer so that you can copy and paste the solution. But I don't think you will get that in this forum. We have given the hints (for the second question I have given the solution too) and solutions will not be more than a page. I think you didn't read our replies. In that case you should have answered few of the queries

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#### Stats student 2008

##### New Member

Two television commercials are being developed for marketing a new product. Researchers constructed a 95% confidence interval for p1 - p2, where p1 represents the proportion Commercial A and would buy the product and p2 represents the proportion who saw Commercial B and would buy the product. The confidence interval was (-2.34, -1.07).

a: Based on the confidence interval, would you say that one commercial was more effective than the other? If so, which one? If not, why not?

Answer: Commercial B was more effective than Commercial A, because it came out negative and is higher than Commercial A. At 95% confidence we can say Comm. B will be 1.07 - 2.34% more effective than Comm. A.

b: Would a 99% confidence interval based on the same data be wider or narrower than this one?

Answer: Confidence interval will get wider. You have less error when you have a larger sample size. At 99% confidence interval will get wider because N is unknown.

#### Stats student 2008

##### New Member
A study in the journal of the American Medical Association examined people to see if they showed any signs of IRS (insulin resistance syndrome) involving major risk factors for Type 2 diabetes and heart disease. Among 102 subjects with high weekly dairy consumption, 24 were identified with IRS. In comparison, IRS was identified in 85 of 190 individuals with low dairy consumption. Is there strong evidence that IRS risk is different in people with high dairy consumption than those with low dairy consumption?

Answer:This is a five-step system solve.

Assumptions-- Categorical, random sample.

np > 15
n(1 - p) >15

Ho P1 = P2
Ha P1 = (does not equal) P2

x1 = 24
n1 = 102
x2 = 185
n2 = 190

Level of Significance = .05

=Since P-Value is less than .05 we reject null hypothesis.

There is strong evidence that there's a slight difference.

Z = -3.57
P = 3.54 x E -4
P (hat) 1 = .24
P (hat) 2 = .45
P (hat) = .37

#### Stats student 2008

##### New Member
A jury list contains the names of all individuals who may be called for jury duty. The proportion of the available jurors on the list who are woman is .54. If 40 people are selected to serve as candidates for being picked on the jury, show all steps of significance test of the hypothesis that the selections are random with respect to gender. 8 of the 40 selected were woman. Use a significance level of 0.02.

Null prop. = .54
p = (does not equal) .54

n = 40

1-prop-Z-test

P = 1.6
P (hat) = .2
Z = -4.32 X E -5

We reject null hypothesis because P-Value is less than significance level of .02.