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Mathematics, 22.06.2019 21:00 lily3934

In 2008 the better business bureau settled 72% of complaints it received (usa today, march 2, 2009). suppose you have been hired by the better business bureau to investigate the complaints it received this year involving new car dealers. you plan to select a sample of new car dealer complaints to estimate the proportion of complaints the better business bureau is able to settle. assume the population proportion of complaints settled for new car dealers is 0.72, the same as the overall proportion of complaints settled in 2008. round your answers to four decimal places. a. suppose you select a sample of 450 complaints involving new car dealers. show the sampling distribution of p.

Answers

ansver
Answer from: Derick21

Mean=p=0.75

sd=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{450}}=0.0204

Step-by-step explanation:

1) Data given

p=75\%=0.75 represent the population proportion of complaints settled for new car dealers

n=450 represent the sample of complaints involving new car dealers.

2) Find the distribution of \hat p

First we can begin with the expected value

E(\hat p)=p and that represent the mean

Now we can find the variance for \hat p

When we use a proportion p, when we draw n items each from a Bernoulli distribution. The variance of each Xi distribution is p(1−p) and hence the standard error is p(1−p)/n. for this reason the variance for \hat p is given by:

Var(\hat p)= \frac{p(1-p)}{n}

So then the deviation would be given by:

Sd(\hat p)=\sqrt{\frac{p(1-p)}{n}}

The sample distribution of the sample proportion \hat p is normal, so then we have this:

\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

3) Calculating the mean and standard deviation

We can replace the values given in order to find the mean and deviation:

Mean=p=0.75

sd=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{450}}=0.0204

ansver
Answer from: reghanhoward3

a) the sampling distribution of the sample proportion is approximately normal with mean 0.75 and standard deviation is 0.0204

b) the probability that the sample proportion will be within 0.04 of the population proportion is 0.95

c) sampling distribution of the sample proportion is approximately normal with mean 0.75 and standard deviation is 0.03061

d) the probability that the sample proportion will be within 0.04 of the population proportion is 0.8088

e) gain in precision is 0.1402.

Step-by-step explanation:

a) Let p represent the

Given that

population proportion of complaints settled for new car dealers p = 0.75.

and n = 450

mean of the sampling distribution of the sample proportion is the population proportion p

i.e  up° = p

mean of the sampling distribution of the sample proportion p° = 0.75

so standard error of the proportion is;

αp° = √(p( 1-p ) / n)

we substitute

αp° = √(0.75 ( 1-0.75 ) / 450)

=√(0.1875 / 450

= √0.0004166

= 0.0204

therefore the sampling distribution of the sample proportion is approximately normal with mean 0.75 and standard deviation is 0.0204

b)

(p° - p) is within 0.04

so lets consider

p ( -0.04 ≤ p° - p ≤ 0.04) = p ( ( -0.04/√(0.75 ( 1-0.75 ) / 450)) ≤ z ≤ ( 0.04/√(0.75 ( 1-0.75 ) / 450))

= p( -0.04/0.0204 ≤ z ≤ 0.04/0.0204)

= p ( -1/96 ≤ z ≤ 1.96 )

= p( z < 1.96 ) - p( z < -1.96 )

now from the S-normal table,

area of the right of z = 1.96 = 0.9750

area of the left of z = - 1.96 = 0.0250

p( -0.04 ≤ p°- p ≤ 0.04)  =  p( z < 1.96 ) - p( z < -1.96 ) = 0.9750 - 0.0250

= 0.95

therefore the probability that the sample proportion will be within 0.04 of the population proportion is 0.95

c)

population proportion of complaints settled for new car dealers p = 0.75.

n = 200

mean of the sampling distribution of the sample proportion p°.

i.e up° = p

mean of the sampling distribution of the sample proportion p° = 0.75

Sampling distribution of the sample proportion p is determined as follows

αp° = √(p( 1-p ) / n)

we substitute

αp° = √(0.75 ( 1-0.75 ) / 200)

=√(0.1875 / 200

= √0.0009375

= 0.03061

therefore sampling distribution of the sample proportion is approximately normal with mean 0.75 and standard deviation is 0.03061

d)

(p° - p) is within 0.04

so lets consider

p ( -0.04 ≤ p° - p ≤ 0.04) = p ( ( -0.04/√(0.75 ( 1-0.75 ) / 200)) ≤ z ≤ ( 0.04/√(0.75 ( 1-0.75 ) / 200))

= p( -0.04/0.03061≤ z ≤ 0.04/0.03061)

= p ( -1.31 ≤ z ≤ 1.31 )

= p( z < 1.31 ) - p( z < -1.31 )

now from the S-normal table,

area of the right of z = 1.31 = 0.9049

area of the left of z = - 1.31 = 0.0951

p( -0.04 ≤ p°- p ≤ 0.04)  =  p( z < 1.31 ) - p( z < -1.31 ) = 0.9049 - 0.0951

= 0.8098

therefore the probability that the sample proportion will be within 0.04 of the population proportion is 0.8088

e)  

From b), the sample proportion is within 0.04 of the population proportion; with the sample of 450 complaints involving new car dealers is 0.95.

sample proportion is within 0.04 of the population proportion; with the sample of 200 complaints involving new car dealers is 0.8098.

measured by the increase in probability, gain in precision occurs by taking the larger sample in part (b)

i.e

Gain in precision will be;

0.9500 − 0.8098

= 0.1402

therefore  gain in precision is 0.1402.

ansver
Answer from: KillerSteamcar

Explained below.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 \mu_{\hat p}= p

The standard deviation of this sampling distribution of sample proportion is:

 \sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}

(a)

The sample selected is of size n = 450 > 30.

Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{450}}=0.0204

So, the sampling distribution of sample proportion is \hat p\sim N(0.75,0.0204^{2}).

(b)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

P(p-0.04

                                          =P(-1.96

Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.95.

(c)

The sample selected is of size n = 200 > 30.

Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{200}}=0.0306

So, the sampling distribution of sample proportion is \hat p\sim N(0.75,0.0306^{2}).

(d)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

P(p-0.04

                                          =P(-1.31

Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.81.

(e)

The probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 450 is 0.95.

And the probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 200 is 0.81.

So, there is a gain in precision on increasing the sample size.

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