3. In a old income survey study for a state, if a randomsample of 64 will be obs

Question

3. In a old income survey study for a state, if a randomsample of 64 will be observed, and that we do overall average household ncome, = $45,000 and s.d-S 1 6,000 (a) Now based on this information, what can you say about the sampling distribution ofthe sample means, if the samples were collected from 64 households? (b) Is the average householdincome of$52,000 from64 households an indicationof anunusually high average? [Hint: compute probability that X >52000 and compare with0 025] () Find a 95thpercentile of average household incomes from 64 households

Explanation / Answer

The sampling distribution of a statistic is the distribution of that statistic, considered as a random variable, when derived from a random sample of size n. It may be considered as the distribution of the statistic for all possible samples from the same population of a given size. If population is follow normal with mean mu and variance s.d^2. The sample mean from sample of size n follow the normal distribution with mean mu and variance s.d.^2/n.

a) Ans.

Therefore, the sampling distribution of the sample mean where the sample were collected from 64 households follow normal distribution with mean=$45,000 and variance=16000^2/64.

b) Ans:

Let, X_~N(mu, s.d.^2/n),

Then,       (X_-mu)/(s.d./sqrt(n))~N(0,1)

That is (X_-45000)*8/16000~N(0,1)

Again, let Y=( X_-45000)*8/(16000)= (52000-45000)*8/16000=3.5

The P(Y>0.4375) is calculated by using the normal table and give 0.0002. The calculated p-value is less than 0.025 level of significance. Hence, the average HHs income $52,000 from 64 HHs is an indication of an usually high average.

c) Ans:

P(( X_-45000)*8/(16000)<Y)=0.95

The Y value at the normal table which give the 0.95 probability is 1.645

Therefore, ( X_-45000)*8/(16000)=1.645

=> X_ - 45000=3290

=>X_=48290

Therefore, the 95th percentile of the average HHs income from 64 households is $48290

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