Question

6. Give a correct interpretation of a 99% confidence level of 0.328 <p < 0.486. (1 point) We are 99% confident that the interval frorn 0 328 to 0 486 actually does contain the true value of the population mean μ There is a 99% chance that the true value ofμ will fall between 0 328 and 0 486. 99% of sample means fall between 0328 and 0486. We are confident that 99% of values between 0.328 and 0 486 represent the true value of the population mean . 7. Randomly selected statistics students participated in an experiment to test their ability to point) determine when 1 min (or 60 seconds) had passed Forty students yielded a sample mean of 58.3 sec. Assume that the population standard deviation is 95 second, and that there is a 95% confidence interval estimate of the population mean of all students 55.4 sec < μ <61.2 sec Give a correct interpretation of the confidence interval. we are 95% confident that the interval from 55.4 seconds to 61.2 seconds actually does contain the true value ofA. @ There is a 95% chance that the true value ofμ will fall between 55.4 seconds and 61.2 seconds 95% of sample means fall between 55.4 seconds and 61.2 seconds. we are confident that 95% of values between 55.4 seconds and 61.2 seconds represent the true value of the population mean

Answer 1

Ans 6. The correct option is A. i.e., We are 99% confident that the interval from 0.328 to 0.486 actually does contain the true value of the population mean $\mu$.

Formula for CI:     $\bar{x}\pm z\left ( \frac{\sigma }{\sqrt{n}} \right )$

Ans 7. The correct option is A. i.e., We are 95% confident that the interval from 55.4 seconds to 61.2 seconds actually does contain the true value of $\mu$.

For the given question we have to construct a 95% confidence interval that will contain the true population mean $\mu$, i.e., mean for the ability to determine when 1 min(or 60 sec) had passed.

Formula for CI:

$\bar{x}\pm z\left ( \frac{\sigma }{\sqrt{n}} \right )$

we have given , $n=40,\:\:\bar{x}=58.3,\:\:\sigma =9.5$

$CI=\bar{x}\pm z\left ( \frac{\sigma }{\sqrt{n}} \right )$  

$CI=58.3\pm 1.96\left ( \frac{9.5}{\sqrt{40}} \right )$

$CI=58.3\pm 2.9$

$CI=(58.3-2.9,\:\:58.3&plus;2.9)$

$CI=(55.4,\:\:61.2)$

Ans 8. The correct option is A.   i.e., the Margin of error=4.0486

The formula for Confidence Interval is: $\bar{x}\pm z\left ( \frac{\sigma }{\sqrt{n}} \right )$

where the term    $z\left ( \frac{\sigma }{\sqrt{n}} \right )$     is known as the Margin of error.

Given: $\sigma =16,\:\:n=60$ , Z=1.96(for 95% CI)

$Margin \:\:of\:\:Error=z\left ( \frac{\sigma }{\sqrt{n}} \right )=1.96\left ( \frac{16}{\sqrt{60}} \right )$

$Margin \:\:of\:\:Error=1.96*2.06559$

$Margin \:\:of\:\:Error=4.048576\approx 4.0486$

$Margin \:\:of\:\:Error=4.0486$

Ans 9. The correct option is B.   i.e., 555< $\mu$<569,   Confidence interval(90%) for population mean.

Given: $\sigma =26.1,\:\:n=35,\:\:\bar{x}=562$ , Z=1.65(for 90% CI)

95% Confidence Interval for the population mean is:

$\bar{x}\pm z\left ( \frac{\sigma }{\sqrt{n}} \right )$

$562\pm 1.65\left ( \frac{26.1}{\sqrt{35}} \right )$

$(562\pm 7.2793)\approx 562\pm 7.2$

$(562-7.2 ,\:\:562&plus;7.2)$

$(555,\:\:569)$ is 95% CI for population mean $\mu$.

Ans 10. The correct option is (D).210

We actually have to find the sample size(n) in here. We have given population standard deviation $(\sigma )$, Margin of error$(ME)$.

$ME=0.1, \:\:\sigma =0.88, \:\:Z=1.65$ (for 90% Confidence Interval).

So, from the formula we used for Margin of Error above we can calculate the value of n.

$Margin \:\:of\:\:Error(ME)=z\left ( \frac{\sigma }{\sqrt{n}} \right )$

rewrite the formula for n,

$ME=z\left ( \frac{\sigma }{\sqrt{n}} \right )$

$\sqrt{n}=\frac{z*\sigma }{ME}$

$n=\left ( \frac{z*\sigma }{ME} \right )^{2}$

$n=\left ( \frac{1.65*0.88}{0.1} \right )^{2}$

$n=\frac{131769}{625}$

$n=210.8304\approx 210$

$n= 210$