Question

The average travel time from Newark, DE to Baltimore is being studied, call it random variable T. The researcher times 50 cars making the trip and finds the sample mean time to be = 58 min. Suppose the standard deviation of the trip travel time is known to be σ_T = 15 min.

1. Compute a two-sided 99% confidence interval of the mean travel time.
2. How many additional car trips would have to be observed so that the average travel time could be estimated to within ±1 min. with 99% confidence?

Answer 1

Part a

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± Z*σ/sqrt(n)

From given data, we have

Xbar = 58

σ = 15

n = 50

Confidence level = 99%

Critical Z value = 2.5758

(by using z-table)

Confidence interval = Xbar ± Z*σ/sqrt(n)

Confidence interval = 58 ± 2.5758*15/sqrt(50)

Confidence interval = 58 ± 5.4642

Lower limit = 58 - 5.4642 = 52.54

Upper limit = 58 + 5.4642 = 63.46

Confidence interval = (52.54, 63.46)

Part b

The sample size formula is given as below:

n = (Z*σ/E)^2

We are given

σ = 15

Confidence level = 99%

Critical Z value = 2.5758

(by using z-table)

Margin of error = E = 1

The sample size is given as below:

n = (Z*σ/E)^2

n = (2.5758*15/1)^2

n = 1492.81777

Required sample size = 1493