Experts have predicted that approximately 1 in 12 tractor-trailer units will be involved in an accident this year. One reason for this is that 1 in 3 tractor-trailer units has an imminently hazardous mechanical condition, probably related to the braking systems on the vehicle.

1. If you wanted to calculate a 99% confidence interval that is no wider than 0.06, how many tractor-trailers would you need to sample?

2. Suppose a sample of 124 tractor-trailers is taken and that 45 of them are found to have a potentially serious braking system problem. Find a 90% confidence interval for the true proportion of all tractor-trailers that have this potentially serious problem.

3. Interpret your confidence interval from 2

1. 564 tractor trailers

2. (0.2919, 0.4339)

3. It should be noted that there is 90% confidence that the true population proportion lies between 0.2919 and 0.4339.

Step-by-step explanation:

1)

Proportion= P = 0.8333333333333 (1/12)

Margin error= 0.06/2 = 0.03

Confidence level= 99

Significance level = α= (100 - 99)%= 1%= 0.01

α/2 = 0.01/2 = 0.0005

Sample size= n

= p(1 - p) (Z*/E)

=0.8333333333333 x (2.576/0.03)^2

=563.15

Approximately

=564

2) Sample size n= 124

Sample number of event x =45

Sample proportion = p= x/n

=45/124

= 0.3629

Standard error =√p(1 - p) /n

√(0.3629× (1 - 0.3629)/ 124

= 0.0432

Confidence level= 90

Significance level α= (100-90)% =0.01

Critical value Z* = 1.645

Margin of error = Z×Standard error= 1.645 × 0.0432= 0.07103

Lower limit = p- margin error= 0.3629 - 0.07103= 0.2919

Upper limit = p+margin error= 0.3629 + 0.07103= 0.4339

The answer is (0.2919, 0.4339)

3) It should be noted that there is 90% confidence that the true population proportion lies between 0.2919 and 0.4339.

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step-by-step explanation: