Question

Information about the proportion of a sample that agrees with a certain statement is given below. Use StatKey or other technology to estimate the standard error from a bootstrap distribution generated from the sample. Then use the standard error to give a 95% confidence interval for the proportion of the population to agree with the statement. StatKey tip: Use ‘‘CI for Single Proportion" and then ‘‘Edit Data" to enter the sample information.
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In a random sample of 100 people, 35 agree.

Estimate the standard error.

Round your answer to three decimal places.

standard error = _____________

Find the 95% confidence interval.

Round your answers to three decimal places.

The confidence interval is ____________Entry field with incorrect answer now contains modified data to ______________Entry field with incorrect answer now contains modified data.

Answer 1

Answer:

$SE=\sqrt{\frac{\hat p (1-\hat p)}{n}}=\sqrt{\frac{0.35 (1-0.35)}{100}}=0.048$

If we replace the values obtained we got:

$0.35 - 1.96\sqrt{\frac{0.35(1-0.35)}{100}}=0.257$

$0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.443$

The 95% confidence interval would be given by (0.257;0.443)

Step-by-step explanation:

Notation and definitions

$X=35$ number of people that agree.

$n=100$ random sample taken

$\hat p=\frac{35}{100}=0.35$ estimated proportion of people that agree

$p$ true population proportion of peopl that agree

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:  

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

The standard error is given by:

$SE=\sqrt{\frac{\hat p (1-\hat p)}{n}}=\sqrt{\frac{0.35 (1-0.35)}{100}}=0.048$

If we replace the values obtained we got:

$0.35 - 1.96\sqrt{\frac{0.35(1-0.35)}{100}}=0.257$

$0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.443$

The 95% confidence interval would be given by (0.257;0.443)

Answer 2

Answer:

Step-by-step explanation:

From the given informtion

$n=100\\x=35$

The sample proportion is,

$P=\frac{x}{n}\\\\=\frac{35}{100}\\\\=0.350$

The standard error of the sample is,

$E=\sqrt{\frac{p(1-p)}{n}}\\\\=\sqrt{\frac{0.350(1-0.350)}{100}}\\\\=0.002275\\=0.002$

The 95% confidence interval for the proportion of population is,

p ± $Z_{\alpha/2}$ x E

=0.350 ± $Z_{0.05/2}$ x 0.002275

=0.350 ± 1.96 x 0.002275     $Z_{0.05/2}=1.96$( from std.normal table)

=0.350 ± 0.00459

=(0.354459, 0.345541)

=(0.354, 0.346)