Question

(Please show if possible on excel .I believe i'm close but want to double check)

The Wall Street Journal reported that automobile crashes cost the United States $162 billion annually. The average cost per person for crashes in the Tampa, Florida, area was reported to be $1900. Suppose this average cost was based on a sample of 50 persons who had been involved in car crashes and that the population standard deviation is known to be $500. What is the 90% confidence interval of the average cost per person for crashes in Tampa? What would you recommend if the study required a margin of error of $150 or less?

Answer 1

a) Since the sample size is 50 which is larger than 30 hence we will use Z statistic for Confidence interval calculator.

Assumption:

Randomly selected sample

Approximately normal distribution, Now

The formula for estimation is:

μ = M ± Z(s_M)

where:

M = sample mean
Z = Z statistic determined by confidence level
s_M = standard error = √(s²/n)

M = 1900
Z= 1.645 calculated at 90 % confidence interval by Z table shown below.
s_M = √(500²/50) = 70.71

μ = M ± Z(s_M)
μ = 1900 ± 1.645*70.71
μ = 1900 ± 116.318

CI=[1738.682, 2016.318]

b) If a margin of Error required is $150 then by formula

$E=Z*\frac{s}{\sqrt{n}}$

at the same sample size by calculation

Now, Z =2.1231 which imply that 98 % confidence interval is required at $150 margin of error.