Question

The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The Bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes. The population distribution is assumed to be normal. Construct a 90% confidence interval for the population mean time to complete the forms. State the confidence interval, sketch the graph, and calculate the error bound.

Answer 1

Answer:

$8.2-2.58\frac{2.2}{\sqrt{18}}=6.86$    

$8.2+2.58\frac{2.2}{\sqrt{18}}=9.54$    

So on this case the 90% confidence interval would be given by (6.86;9.54)    And the error is given by:

$ME= 2.58\frac{2.2}{\sqrt{18}} =1.338$

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".

$\bar X=8.2$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma=2.2$ represent the population standard deviation

n represent the sample size  

Solution to the problem

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

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.005$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

Now we have everything in order to replace into formula (1):

$8.2-2.58\frac{2.2}{\sqrt{18}}=6.86$    

$8.2+2.58\frac{2.2}{\sqrt{18}}=9.54$    

So on this case the 90% confidence interval would be given by (6.86;9.54)    And the error is given by:

$ME= 2.58\frac{2.2}{\sqrt{18}} =1.338$

Answer 2

Answer:

90% confidence interval for the population mean time = [7.944 , 8.456]

Step-by-step explanation:

We are given that the  Bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes.

Now, the pivotal quantity for 90% confidence interval for the population mean time to complete the forms is;

       P.Q. = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} }}$ ~ N(0,1)

where, Xbar = sample mean = 8.2 minutes

             $\sigma$ = population standard deviation = 2.2 minutes

             n = sample size = 200

So, 90% confidence interval for the population mean time, $\mu$ is ;

P(-1.6449 < N(0,1) < 1.6449) = 0.90

P(-1.6449 < $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} }}$ < 1.6449) = 0.90

P(-1.6449 * ${\frac{\sigma}{\sqrt{n} }}$ < ${Xbar-\mu}$ < 1.6449 * ${\frac{\sigma}{\sqrt{n} }}$ ) = 0.90

P(Xbar - 1.6449 * ${\frac{\sigma}{\sqrt{n} }}$ < $\mu$ < Xbar + 1.6449 * ${\frac{\sigma}{\sqrt{n} }}$ ) = 0.90

90% confidence interval for $\mu$ = [Xbar - 1.6449 * ${\frac{\sigma}{\sqrt{n} }}$ , Xbar + 1.6449 * ${\frac{\sigma}{\sqrt{n} }}$ ]

                                                 = [8.2 - 1.6449 * ${\frac{2.2}{\sqrt{200} }}$ , 8.2 + 1.6449 * ${\frac{2.2}{\sqrt{200} }}$ ]

                                                 = [7.944 , 8.456]

Therefore, 90% confidence interval for the population mean time to complete the forms is [7.944 , 8.456] .