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nadya68 [22]
3 years ago
15

Consider random samples of size 86 drawn from population A with proportion 0.43 and random samples of size 60 drawn from populat

ion B with proportion 0.15
(a) Find the standard error of the distribution of differences in sample proportions A - B Round your answer for the standard error to three decimal places. Standard error=
(b) Are the sample sizes large enough for the Central Limit Theorem toa
Yes or No?
Mathematics
1 answer:
Elan Coil [88]3 years ago
5 0

Answer:

a) The standard error is s = 0.071.

b) Yes, as both sample sizes are above 30.

Step-by-step explanation:

To solve this question, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Samples of size 86 drawn from population A with proportion 0.43

This means that n = 86, p = 0.43. So:

s_A = \sqrt{\frac{0.43*0.57}{86}} = 0.0534

Samples of size 60 drawn from population B with proportion 0.15:

This means that n = 60, p = 0.15. So:

s_B = \sqrt{\frac{0.15*0.85}{60}} = 0.0461

(a) Find the standard error of the distribution of differences in sample proportions A - B Round your answer for the standard error to three decimal places. Standard error=

This is:

s = \sqrt{s_A^2 + s_B^2}

s = \sqrt{(0.0534)^2 + (0.0461)^2}

s = 0.071

The standard error is s = 0.071.

(b) Are the sample sizes large enough for the Central Limit Theorem. Yes or No?

Yes, as both sample sizes are above 30.

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From the question we are told that

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=> P(|p-\^{p}| >  0.03)  =  1 -  P(-2.88  \le Z \le 2.88 )

=>   P(|p-\^{p}| >  0.03)  =   1 - [P(Z \le 2.88) - P(Z \le -2.88)]

From the z-table  

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So

     P(|p-\^{p}| >  0.03)  =   1 - [0.9980 - 0.0020]

=>   P(|p-\^{p}| >  0.03)  =   0.0040

     

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