Question

            Confidence Intervals

                                                    

1.         I want to estimate the average money spent on Xmas toys each week in December. I take a

            sample of 40 individuals and determine that the mean spent is $210 with a standard

            deviation of $32.00

            a.         For a 95% confidence interval, estimate the interval within which the true population

                        would fall.

b.         If I reduced the interval size to just 90%, what would that interval be?

c.         For the 90% confidence interval, calculate the maximum error of the estimate/mean?

2.   The MBTA wants to estimate the average amount of money an individual spends at MBTA parking lots each week. A sample of 64 commuters was taken and the average amount paid was $44 dollars, with a standard deviation of $3.00.

a.   What is the best estimated of the amount paid for parking?

b.         Construct a 90% confidence interval of the population mean.

Answer 1

Answer)

1)

Given mean = 210

Standard deviation = 32

Sample size (n) = 40

As here population standard deviation is unknown, we will use t distribution to estimate the interval

A)

First we need to estimate the degrees of freedom, degrees of freedom is equal to n-1, 40-1, = 39

And confidence level is 95%

For degrees of freedom 39 and confidence level of 95%

Critical value t is 2.023

Now we need to find the margin of error

MOE = t*(standard deviation/√n)

= 2.023*(32/√40)

= 10.2356603304

Confidence interval is given by,

Mean - MOE, Mean + MOE

(199.764339669, 220.235660330)

We are 95% confident that true population mean lies in between 199.76 and 220.24

B)

In case of 90%, critical value is 1.685

Therefore,

MOE = 1.685*(32/√40)

= 8.52550057181

Confidence interval

= (mean - moe, mean + moe)

= (201.47449942819, 218.52550057181)

We are 90% confident that true population mean lies in between 201.47 and 218.53

C)

We know that error = t*(standard deviation/√n)

t for 90% confidence level and degrees of freedom 39 is 1.685

Standard deviation = 32 and n =40

Error = 1.685*(32/√40)

= 8.52550057181

2)

Answer)

A) best point estimate is the sample mean:

Which is $44

B)

As population standard deviation is unknown, therefore we will use t distribution

Degrees of freedom is equal to n-1, 64-1 = 63

And confidence level is 90

From t distribution critical value for 63 degrees of freedom and 90% confidence level is

= 1.669

Margin of error (MOE) = t*(standard deviation/√n)

MOE = 1.669*(3/√64)

= 0.625875

Confidence interval is given by

(Mean - MOE, mean + MOE)

(44-0.625875, 44+0.625875)

( 43.374125, 44.625875)

We are 90% confident that true population mean lies in between 43.37 and 44.63.