Question

For the next 3 questions, please read the description below. Researchers want to estimate the mean monthly electricity bill in a large urban area using a simple random sample of 100 households. Their calculation shows that the sample standard deviation is $15.50.

*Assume that the population standard deviation is unknown.What is the error of estimate for a 99% confidence interval?

answer choices 3.67 4.07 3.99 4.73

*Assume that the population standard deviation is known to be $30. The average monthly electricity bill of the 100 households is $120. What is the upper bound for a 90% confidence interval?

answer choices 124 122 125 123

*Assume that the population standard deviation is unknown. If the researchers want to decrease the error of estimate, which of the following number of households is more likely to be their sample size?

answer choices 30 100 90 120

Answer 1

n = sample size = 100

Sample standard deviation= s = 15.50

1) Assume that the population standard deviation is unknown.

So we use t interval.

Confidence level = c = 0.99

Degrees of freedom = n - 1 = 100 - 1 = 99

t critical value for alpha = 0.01 and df = 99 from excel using function:

= TINV(0.01,99)

= 2.626 (Round to 3 decimal)

Margin of error (e) :

e = 2.626 * 1.55

e = 4.07 (Round to 2 decimal)

Margin of error = 4.07

2)

n = 100

Sample mean =

Population standard deviation =

Here Population standard deviation is known so we use z interval.

Confidence level = c = 0.90

Upper Bound

where zc is z critical value for (1+c)/2 = (1+0.90)/2 = 0.95

zc= 1.645 (From statistical table of z values, average of 1.64 and 1.65, (1.64+1.65)/2 = 1.645)

= 120 + 1.645 * 3

= 120 + 4.935

= 124.935

= 125 (Round to nearest integer)

Upper bound = 125

If the researchers want to decrease the error of estimate then he has to increase sample size.

For sample size n = 100, margin of error = 4.07 with 99% confidence.   

Here sample size greater than 100 is 120.

Number of households is more likely to be their sample size = 120