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

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

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. =
`(Xbar-\mu)/((\sigma)/(√(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 <
`(Xbar-\mu)/((\sigma)/(√(n) ))` < 1.6449) = 0.90

P(-1.6449 *
`{(\sigma)/(√(n) )}` <
`{Xbar-\mu}` < 1.6449 *
`{(\sigma)/(√(n) )}` ) = 0.90

P(Xbar - 1.6449 *
`{(\sigma)/(√(n) )}` <
`\mu` < Xbar + 1.6449 *
`{(\sigma)/(√(n) )}` ) = 0.90

90% confidence interval for
`\mu` = [Xbar - 1.6449 *
`{(\sigma)/(√(n) )}` , Xbar + 1.6449 *
`{(\sigma)/(√(n) )}` ]

= [8.2 - 1.6449 *
`{(2.2)/(√(200) )}` , 8.2 + 1.6449 *
`{(2.2)/(√(200) )}` ]

= [7.944 , 8.456]

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

Answer 2

`8.2-2.58(2.2)/(√(18))=6.86`

`8.2+2.58(2.2)/(√(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(2.2)/(√(18)) =1.338`

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)(\sigma)/(√(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(2.2)/(√(18))=6.86`

`8.2+2.58(2.2)/(√(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(2.2)/(√(18)) =1.338`