Admission in Indian

8/11/2019 Admission in Indian

1/31

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2/31

6.6: Small-sample inference for a proportion

7.1: Large sample comparisons for twoindependent sample means.

7.2: Difference between two large sampleproportions.

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3/31

So far, we have been making estimates andinferences about a single sample statistic

Now, we will begin making estimates andinferences for two sample statistics at once.

many real-life problems involve such comparisons two-group problems often serve as a starting point for

more involved statistics, as we shall see in this class.



4/31

Two independent random samples:

Two subsamples, each with a mean score for some othervariable

example: Comparisons of work hours by race or sex example : Comparison of earnings by marital status

Two dependent random samples : Two observations are being compared for each unit in

the sample example: before-and-after measurements of the same

person at two time points example: earnings before and after marriage husband-wife differen ces

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5/31

Hypothesis testing as we have done it so far:

Test statistic: z = (Y bar - o) / ( s / SQRT( n))

What can we do when we make inferences about adifference between population means ( 2 - 1)? Treat one sample mean as if it were o ?

(NO: too much type I error) Calculate a confidence interval for each sample mean

and see if they overlap?

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6/31

Is Y 2 Y1an appropriate way to evaluate 2 - 1?

Answer: Yes. We can appropriately define ( 2 - 1) as a

parameter of interest and estimate it in an unbiased waywith (Y 2 Y1) just as we would estimate with Y.

This line of argument may seem trivial, but it becomes

important when we work with variance and standarddeviations.



7/31

Comparing standard errors: A&F 213: formula without derivation

Is s 2Ybar2 - s 2Ybar1 an appropriate way to estimate2(Ybar2-Ybar1) ?

No!

2(Ybar2-Ybar1) = 2(Ybar2) - 2 (Ybar2,Ybar1) + 2(Ybar1)

Where 2 (Ybar2,Ybar1) reflects how much the observationsfor the two groups are dependent.

For independent groups, 2 (Ybar2,Ybar1) = 0,so 2(Ybar2-Ybar1) = 2(Ybar2) + 2(Ybar1)

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8/31

The parameter of interest is 2 - 1

Assumptions: the sample is drawn from a random sample of some sort,

the parameter of interest is a variable with an intervalscale,

the sample size is large enough that the samplingdistribution of Y bar2 Ybar1 is approximately normal.

The two samples are drawn independently



9/31

The null hypothesis will be that there is nodifference between the population means. Thismeans that any difference we observe is due torandom chance.

Ho: 2 - 1 = 0 (We can specify an alpha level now if we want)

Q: Would it matter if we used

Ho: 1 - 2 = 0 ?

Ho: 1 = 2 ?



10/31

The test statistic has a standard form: z = (estimate of parameter Ho value of parameter)

standard error of parameter

Q: If the null hypothesis is that the means are thesame, why do we estimate two different standard

deviations? 10

2

22

1

21

12 0)(

n

s

n

s

Y Y z

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11/31

P-value of calculated z: Table A

Stata: display 2 * (1 normal(z) ) Stata: testi (no data, just parameters) Stata: ttest (if data file in memory)



12/31

Step 5: Conclusion.

Compare the p-value from step 4 to the alpha levelin step 1.

If p < , reject H 0 If p , do not reject H 0

State a conclusion about the statistical significanceof the test.

Briefly discuss the substantive importance of yourfindings.



13/31

Do women spend more time on housework thanmen?

Data from the 1988 National Survey of Familiesand Households: sex sample size mean hours s.d men 4252 18.1 12.9 women 6764 32.6 18.2




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15/31

housework example with 99% interval:c.i.= (32.6 18.1) +/- 2.58*( ((12.9) 2/4252 + (18.2) 2/6764))

= 14.5 +/- 2.58*.30= 14.5 +/- .8, or (13.7,15.3)

By this analysis, the 99% confidence interval for thedifference in housew ork is 13.7 to 15.3 hours.

15

22

2

1

2

112.. n s

n

s z Y Y ic

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16/31

Immediate (no data, just parameters) ttesti 4252 18.1 12.9 6764 32.6 18.2, unequal

Q: why ttesti with large samples?

For the immediate command, you need the following: sample size for group 1 (n = 4252) mean for group 1 standard deviation for group 1 sample size for group 2 mean for group 2 standard deviation for group 2

instructions to not assume equal variance (, unequal) 16admission.edhole.com


17/31

. ttesti 4252 18.1 12.9 6764 32.6 18.2, unequal

Two-sample t test with unequal variances

------------------------------------------------------------------------------| Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]

---------+--------------------------------------------------------------------x | 4252 18.1 .1978304 12.9 17.71215 18.48785y | 6764 32.6 .221294 18.2 32.16619 33.03381

---------+--------------------------------------------------------------------combined | 11016 27.00323 .1697512 17.8166 26.67049 27.33597---------+--------------------------------------------------------------------

diff | -14.5 .2968297 -15.08184 -13.91816------------------------------------------------------------------------------Satterthwaite's degrees of freedom: 10858.6

Ho: mean(x) - mean(y) = diff = 0

Ha: diff < 0 Ha: diff != 0 Ha: diff > 0t = -48.8496 t = -48.8496 t = -48.8496

P < t = 0.0000 P > |t| = 0.0000 P > t = 1.0000



18/31

. ttest YEARSJOB, by(nonstandard) unequal

Two-sample t test with unequal variances------------------------------------------------------------------------------

Group | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]---------+--------------------------------------------------------------------

0 | 980 9.430612 .2788544 8.729523 8.883391 9.9778331 | 379 7.907652 .3880947 7.555398 7.144557 8.670747

---------+--------------------------------------------------------------------combined | 1359 9.005887 .2290413 8.443521 8.556573 9.4552---------+--------------------------------------------------------------------

diff | 1.522961 .4778884 .5848756 2.461045------------------------------------------------------------------------------

diff = mean(0) - mean(1) t = 3.1869Ho: diff = 0 Satterthwaite's degrees of freedom = 787.963

Ha: diff < 0 Ha: diff != 0 Ha: diff > 0

Pr(T < t) = 0.9993 Pr(|T| > |t|) = 0.0015 Pr(T > t) = 0.0007



19/31

. ttest conrinc if wrkstat==1, by(wrkslf) unequal

Two-sample t test with unequal variances------------------------------------------------------------------------------

Group | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]---------+--------------------------------------------------------------------self-emp | 190 48514.62 2406.263 33168.05 43768.03 53261.2

someone | 1263 34417.11 636.9954 22638 33167.43 35666.8

---------+--------------------------------------------------------------------combined | 1453 36260.56 648.5844 24722.9 34988.3 37532.82---------+--------------------------------------------------------------------

diff | 14097.5 2489.15 9191.402 19003.6------------------------------------------------------------------------------

diff = mean(self-emp) - mean(someone) t = 5.6636Ho: diff = 0 Satterthwaite's degrees of freedom = 216.259

Ha: diff < 0 Ha: diff != 0 Ha: diff > 0Pr(T < t) = 1.0000 Pr(|T| > |t|) = 0.0000 Pr(T > t) = 0.0000



20/31

In 1982 and 1994, respondents in the General Social Surveywere asked: Do you agree or disagree with this statement?Women should take care of running their homes and leaverunning the country up to men. Year Agree Disagree Total 1982 122 223 345 1994 268 1632 1900 Total 390 1855 2245

Do a formal test to decide whether opinions differed in thetwo years.



21/31


Assumptions:

the sample is drawn from a random sample of some sort, the parameter of interest is a variable with an interval

scale,

the sample size is large enough that the samplingdistribution of Pi hat2 Pi hat1 is approximately normal. The two samples are drawn independently



22/31

The null hypothesis will be that there is nodifference between the population proportions. Thismeans that any difference we observe is due torandom chance.

Ho: 2 - 1 = 0

(State an alpha here if you want to.)



23/31

The test statistic has a standard form:z = (estimate of parameter Ho value of parameter)

standard error of parameter

Where pi hat is the overall weighted average

This means we are assuming equal variance in the twopopulations. Q: why do we use an assumption of equal variance to

estimate the standard error for the t-test?

23

21

12

111

)

(

nn

z

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24/31

P-value of calculated z: Table A, or Stata: display 2 * (1 normal(z) ), or Stata: testi (no data, just parameters) Stata: ttest (if data file in memory)

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25/31

Conclusion:

Compare the p-value from step 4 to the alpha levelin step 1.

If p < , reject H 0 If p , do not reject H 0 State a conclusion about the statistical significanceof the test.

Briefly discuss the substantive importance of yourfindings.

25



26/31

1. Assumptions: random sample, interval-scale variable,sample size large enough that the sampling distribution of

2 - 1is approximately normal, independent groups

2. Hypothesis: H o: 2 - 1= 03. Test statistic:

z = (122/345 268/1900) /SQRT[(390/2245)*(1 - 390/2245)*(1/345 + 1/1900)]

= 9.59

4. p-value: p


27/31

confidence interval:

Notice that there is no overall weighted averagePi hat , as there is in a significance test forproportions.

Instead, we estimate two separate variances from theseparate proportions. Why?

27

222

1

11

12

)1()1(

.. n P P

n

P P z P P ic

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28/31

. prtesti 345 .3536 1900 .1411

STATA needs the following information: sample size for group 1 (n = 345) proportion for group 1 (p = 122/345) sample size for group 2 (n = 1900)

proportion for group 2 (p = 268/1900)

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29/31

. prtesti 345 .3536 1900 .1411

Two-sample test of proportion x: Number of obs = 345y: Number of obs = 1900

------------------------------------------------------------------------------Variable | Mean Std. Err. z P>|z| [95% Conf. Interval]

-------------+----------------------------------------------------------------x | .3536 .0257393 .3031518 .4040482y | .1411 .0079865 .1254467 .1567533

-------------+----------------------------------------------------------------

diff | .2125 .0269499 .1596791 .2653209| under Ho: .0221741 9.58 0.000------------------------------------------------------------------------------

Ho: proportion(x) - proportion(y) = diff = 0

Ha: diff < 0 Ha: diff != 0 Ha: diff > 0z = 9.583 z = 9.583 z = 9.583

P < z = 1.0000 P > |z| = 0.0000 P > z = 0.0000

Note the use of one standard error (unequal variance) for theconfidence interval, and another (equal variance) for thesignificance test.

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30/31

. prtest nonstandard if (RACECEN1==1 | RACECEN1==2), by(RACECEN1)

Two-sample test of proportion 1: Number of obs = 13892: Number of obs = 260


-------------+----------------------------------------------------------------1 | .2800576 .0120482 .2564436 .30367162 | .3538462 .0296544 .2957247 .4119676

-------------+----------------------------------------------------------------diff | -.0737886 .0320084 -.1365239 -.0110532

| under Ho: .0307147 -2.40 0.016------------------------------------------------------------------------------

diff = prop(1) - prop(2) z = -2.4024Ho: diff = 0

Ha: diff < 0 Ha: diff != 0 Ha: diff > 0Pr(Z < z) = 0.0081 Pr(|Z| < |z|) = 0.0163 Pr(Z > z) = 0.9919

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31/31

. gen byte wrkslf0=wrkslf-1(152 missing values generated)

. prtest wrkslf0 if wrkstat==1, by(sex)

Two-sample test of proportion male: Number of obs = 874female: Number of obs = 743


-------------+----------------------------------------------------------------

male | .8272311 .0127876 .8021678 .8522944female | .9044415 .0107853 .8833027 .9255802-------------+----------------------------------------------------------------

diff | -.0772103 .0167286 -.1099978 -.0444229| under Ho: .0171735 -4.50 0.000

------------------------------------------------------------------------------diff = prop(male) - prop(female) z = -4.4959

Ho: diff = 0

Ha: diff < 0 Ha: diff != 0 Ha: diff > 0Pr(Z < z) = 0.0000 Pr(|Z| < |z|) = 0.0000 Pr(Z > z) = 1.0000


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