# homework作业 | assignment – HUDM4122: LECTURE 17

### HUDM4122: LECTURE 17

homework作业 | assignment – , 该题目是值得借鉴的assignment代写的题目 #### (also the Large-sample C.I. for p )

``````(copyright  1999-2021 by James E. Corter)
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Last Time: _ Distribution of X (practice by doing problems)

Today:

1. Large sample confidence intervals for – known (large samples) ==> use z distribution
2. Large-sample confidence intervals for p, a population proportion
3. Give homework assignment on confidence intervals READING:

Last time, we were talking about the distribution ofX. We’re interested inX because it is our best point estimate of .

TWO USES OF THIS :

USE 1) to test hypotheses about

e.g. H 0 : = 100 (IQ: Is this classroom different from norm?) Ha: 100

H 0 : 0 (Weight gain in rats: affected by a nutritional supplement?) Ha: > 0

Remember:

Example. Take samples of size 16 from normal population with = & =10. What is X& X?

Answer. X=X=64 X= / nx =10 / 16= 2.

We now know enough to test hypotheses about means or to do confidence intervals (assuming that either we know or we can estimate it very exactly).

### E(X)= =X

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##### Var(X)= = /nX ( / n)X=

We are in the realm of problems called " Large-sample Theory ". But please be aware that the practical definition of what constitutes a large sample varies in different contexts, for different approximations in estimation.

For example, we have results stating that for "large" samples:

1. The distribution of X will be approximately normal, whether X is or not. ( Central Limit Theorem). (here, a common interpretation of "large": n >20)
2. You can always assume that is known, because you can estimate it very exactly. (Law of Large Numbers) (here, large can be taken as meaning > 60)

If you KNOW X ~ N & you know (or willing to assume you know) , then you can use the normal distribution to test hypotheses or construct confidence intervals.

USE 2) Confidence Intervals for the Mean (Large Samples)

Assume there is a RV defined on a population such that X ~ N(,)

However, in the usual research situation we don’t know . We come along, trying to estimate . We gather some data (a large sample) & calculateX Can we place limits on what has to be, given our data?

? ?

X=

We can find an interval aroundX in which must lie (with, say, 95% confidence) this

interval can be described as X-E to X+E. The question is: what is E = to? By reasoning about the sampling distribution of X, you can show that the interval we

want is defined by X-E to X+E, where E = 2 ().

We can do this, because is distributed normally (and assuming we know )

Note: X N

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=. Also note that the 95% C.I. corresponds to an error rate of 1-95% =
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``````The Formula for the 95% C.I. is given by:
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#### (X z+/2 ,X z ) where z z XX /2 /2==.05/2 z.

5%, that is, = 1 – .95 = .05. so /2 = .025 and / 2 = 1.

Graphical explanation: consider the sampling distribution of X around the true (but unknown) mean :

How wide is the interval that will contain 95% of the sample values of X? (= E) I.e., What X/2corresponds to Z/2? X/2 = + (Z/2)( X ) = + E

What X-/2corresponds to Z (^) – /2? X/2 = – (Z/2)( X ) = – E So, we would be surprised if X were more than E = (Z/2)( X ) = (Z/2)(/n) units from (or if were more that z X units from X). Therefore, conclude that there is a 95% coverage probability that the interval (X Z/ X ) contains (across many such samples and CIs). So we can be 95% confident that any single CI contains the true values of . EXAMPLE: Assume we give a test to n=85 students, and find that X=63.1, s = 12.9. Because this is a large sample (n>60), we can ASSUME that our value for s is sufficiently close to so that we can assume s = 12.9 = . Then we can calculate the standard error of the mean: X= / nx = 12. 9 85 = 1. Then, 95% CI for = X / 2 X = 63.1 (1.96)(1.40) = 63.1 2. GENERAL FORMULA:

#### (1- )% C.I. for = X / 2 X = X ( 2)()

So, using the data from the above example: For the 99% C.I., = .01 and /2 = .005, so / 2 = 2.

99% CI for = X / 2 X = 63.1 (2.58)(1.40) = 63.1 3.

EXAMPLE. We give a test of personality trait X to a sample of 75 students. We find

that = 45.7, s = 12.6. Give the 90% confidence interval for .

We have a large sample (N=75), so ASSUME s = . So s = = 12.6.

compute the standard error of the mean: =

###### = 12.6 / 8.66 = 1.

90% CI –> = .10 /2 = .05 /2 = 1.

90% CI for = (/2)( )

= 45.7 (1.65)(1.455) = 45.7 2.40 = (43.3, 48.1)

#### Interpretation: We can be 90% confident that the true value of falls between 43.3 and

###### 48.1.

[coverage probability interpretation: It is also true that 90% of confidence intervals

#### constructed by this method will cover the true value of .]

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Large-Sample Confidence Intervals for a Population Proportion, p

The Sampling Distribution of the Sample Proportion:

#### Remember our estimator =

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#### =

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#
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When N is large (>40), the distribution of , the sample proportion, is approx. normal. In fact, we have a new version of the Central Limit Theorem:

Central Limit Theorem:

As N infinity, the distribution of the statistic

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approaches the standard normal
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(i.e., the z distribution)

#### where the standard error of is given by: =

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Therefore we can use the Z distribution to construct confidence intervals for p, the unknown population proportion.

Key idea: the standard error of the sample proportion can be shown to be equal to

#### .. ==

^ Because we usually do not know the true p and q =(1- p ), our best estimate of the standard error is:

.. =

#### ^

Therefore, we construct the (1-)% C.I. for p using the formula:

( 1 )% C. I. =( 2 )(..) = (

`````` 2
)(

)^
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#### (Note that we ASSUME that = p and q = q , which will be reasonably good

approximations as long as we have a large sample)

Example: We take a sample of size N=120. Of these respondents, 83 say they consume at least one alcoholic drink per week. What is the 95% C.I. for the proportion of alcohol users in this population?

For the 95% C.I., =.05, so /2=.025. Therefore, Z/2 = 1.

#### = 83/120 = .692, q = 1- = 1-.692 =.

Then the estimated standard error is

###### =. 001776=.

And the 95% C.I. for p = 2