Question

Suppose that an accounting firm does a study to determine the time needed to complete one person's tax forms. It randomly surveys 175 people. The sample mean is 22.8 hours. There is a known population standard deviation of 6.4 hours. The population distribution is assumed to be normal. NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
Construct a 90% confidence interval for the population mean time to complete the tax forms.
i. State the confidence interval. (Round your answers to two decimal places.)
ii. Calculate the error bound. (Round your answer to two decimal places.

Answer 1

Answer:

Step-by-step explanation:

Considering the central limit theorem, the distribution is normal since the number of samples is large. Also, the population standard

deviation is known. We would determine the z score.

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.90 = 0.1

α/2 = 0.1/2 = 0.05

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.05 = 0.95

The z score corresponding to the area on the z table is 2.05. Thus, confidence level of 90% is 1.645

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Confidence interval = mean ± z × σ/√n

Where

σ = population standard Deviation

Confidence interval = x ± z × σ/√n

x = 22.8 hours

σ = 6.4 hours

n = 175

i) Confidence interval = 22.8 ± 1.645 × 6.4/√175

= 22.8 ± 0.80

The lower end of the confidence interval is

22.8 - 0.80 = 22

The upper end of the confidence interval is

22.8 + 0.80 = 23.6

ii) error bound is the same as the margin of error

Error bound = 0.8