Question

Data were collected on the amount that a sample of six moviegoers paid for two tickets with online service charges, large popcorn, and two medium soft drinks at a sample of six local cinemas:

$43.00      $33.75        $40.25        $35.05        $31.00        $36.15

Construct a 95% confidence interval estimate of the population mean price for two tickets with online service charges, large popcorn, and two medium soft drinks and answer the following questions:

a.   i.   What is the point estimate of the population mean? _____________

      ii. What is the standard deviation? _____________

b.   With 95% confidence, what is the margin of error for the estimation of the population mean?________

c.   State the t value appropriate in this context. ____________

d.   What is the 95% confidence interval for the population mean? _________________

e.   Suppose a 99% confidence interval is required.

      i.   What is the t value in this context? ______________

      ii. What is the margin of error for the estimation of the population mean? ________

Answer 1

Let X represent the data values.

We assume that the sample has been taken from a Normal Distribution (with unknown parameters)

The following table shows the calculations -

	X (in $)	X^2
	43.00	1849
	33.75	1139.0625
	40.25	1620.0625
	35.05	1228.5025
	31.00	961
	36.15	1306.8225
Total	219.2	8104.45

Total number of observations, n = 6

Answer a:

Mean of X, = 219.2/6 = 36.5333

i. = 36.5333 is the point estimate of the population mean

ii. Standard Deviation, s’ = {((x^2) - n(^2)) / (n - 1)}^0.5 = 4.3899

Answer b:
The Standard error of mean = s’ / (n^0.5) = 4.3899 / (6^0.5) = 1.7922

For 95% Confidence Interval, the value of = (100 - 95)% = 0.05

/ 2 = 0.025

The Critical / t - Value is t(0.025, 5) = 2.571

The Margin of Error for the estimation of the population mean = Critical value x Standard Error

                                                                                               = 2.571 x 1.7922 = 4.6077

Answer c:
The Critical Value for 95% Interval is t(0.025, 5) = 2.571

Answer d:

To calculate 100(1 - )% confidence limits to the population mean --

The 95% confidence interval is –

(( - Margin of Error), ( + Margin of Error))

= ((36.5333 - 4.6077), (36.5333 + 4.6077)) = (31.9256, 41.141)

Answer e:
The Standard Error of Mean = 1.7922

For 99% Confidence Interval, the value of = (100 - 99)% = 0.01

/ 2 = 0.005

The Critical / t - Value is t(0.005, 5) = 4.032

The Margin of Error for the estimation of the population mean = Critical value x Standard Error

                                                                                               = 4.032 x 1.7922 = 7.2262

(NOTE THAT: All the answers are rounded up to 4 decimal places)