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Semenov [28]
3 years ago
14

A telephone exchange operator assumes that 8% of the phone calls are wrong numbers. If the operator is right, what is the probab

ility that the proportion of wrong numbers in a sample of 421 phone calls would differ from the population proportion by greater than 3%? Round your answer to four decimal places.
Mathematics
1 answer:
Vera_Pavlovna [14]3 years ago
7 0

Answer:

The probability that the sample proportion differ from the population proportion by greater than 3% is 0.0241.

Step-by-step explanation:

Let <em>X</em> = number of phone calls that are wrong numbers.

The proportion of phone calls that are wrong numbers is, <em>p</em> = 0.08.

A sample of<em> </em><em>n</em> = 421 phone calls is selected to determine the proportion of wrong numbers in this sample.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of a Binomial distribution is:

P(X=x)={421\choose x}0.08^{x}(1-0.08)^{421-x}

Now, for the sample proportion to differ from the population proportion by 3% the value of the sample proportion should be:

\hat p-p=0.03\\\hat p-0.08=0.03\\\hat p=0.11                            \hat p-p=-0.03\\\hat p-0.08=-0.03\\\hat p=0.05

So when the sample proportion is less than 5% or greater than 11% the difference between the sample proportion and population proportion will be greater than 3%.

  • If sample proportion is 5% then the value of <em>X</em> is,

        X=np=421\times 0.05=21.05\approx21

        Compute the value of P (X ≤ 21) as follows:

       P(X\leq 21)=\sum\limits^{21}_{x=0}{{421\choose x}0.08^{x}(1-0.08)^{421-x}}=0.0106

  • If the sample proportion is 11% then the value of <em>X</em> is,

        X=np=421\times 0.11=46.31\approx47

        Compute the value of P (X ≥ 47) as follows:

       P(X\geq 47)=\sum\limits^{471}_{x=47}{{421\choose x}0.08^{x}(1-0.08)^{421-x}}=0.0135

Then the probability that the sample proportion differ from the population proportion by greater than 3% is:

P(\hat p-p>0.03)=P(X\leq 21)+P(X\geq 47)=0.0106+0.0135=0.0241

Thus, the probability that the sample proportion differ from the population proportion by greater than 3% is 0.0241.

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Answer:

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

And the confidence interval is given by:

(0.123, 0.177)

And for this case the interval contains the value 0.16, so then we can conclude at 5% of significance that the true proportion is not different from 0.16

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

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

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

And the confidence interval is given by:

(0.123, 0.177)

And for this case the interval contains the value 0.16, so then we can conclude at 5% of significance that the true proportion is not different from 0.16

3 0
3 years ago
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