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2 years ago

6

Liam wants to estimate the percentage of people who lease a car. He surveys 240 individuals and finds that 54 lease a car. Find

the margin of error for the confidence interval for the population proportion with a 95% confidence level. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576

1 answer:

scoundrel [369]2 years ago

6 0

Answer:

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

The confidence interval is 95% and the significance level is $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ and the critical value would be:

$z_{\alpha/2}= 1.96$

And the margin of error would be:

$ME = 1.96 \sqrt{\frac{0.225 (1-0.225)}{240}}= 0.0528$

Step-by-step explanation:

We have the following info given:

$n =240$ the sample size selected

$X=54$ the number of people who lease a car

$\hat p =\frac{54}{240}= 0.225$ the estimated proportion of people who lease a car

The margin of error is given by:

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

The confidence interval is 95% and the significance level is $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ and the critical value would be:

$z_{\alpha/2}= 1.96$

And the margin of error would be:

$ME = 1.96 \sqrt{\frac{0.225 (1-0.225)}{240}}= 0.0528$

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Step-by-step explanation:

Data given and notation

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$s=1.4$ represent the sample standard deviation

$n=12$ sample size

$\mu_o =10.1$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the technique performs differently than the traditional method, the system of hypothesis would be:

Null hypothesis: $\mu =10.1$

Alternative hypothesis: $\mu \neq 10.1$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{10.5-10.1}{\frac{1.4}{\sqrt{12}}}=0.990$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=12-1=11$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(11)}>0.990)=0.343$

Conclusion

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