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The gas mileage for a certain model of car is known to have a standard deviation of 4 mi/gallon. A simple random sample of 49 cars of this model is chosen and found to have a mean gas mileage of 27.5 mi/gallon. Construct a 97% confidence interval for the mean gas mileage for this car model.

Respuesta :

Answer:

27.5+/-1.24 mi/gallon.

= ( 26.26, 28.74) mi/gallon.

Therefore, the 95% confidence interval (a,b)= ( 26.26, 28.74) mi/gallon.

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 27.5 mi/gallon.

Standard deviation r = 4 mi/gallon

Number of samples n = 49

Confidence interval = 97%

z-value (at 95% confidence) = 2.17

Substituting the values we have;

27.5+/-2.17(4.0/√49)

27.5+/-2.17(0.571428571428)

27.5+/-1.24 mi/gallon.

= ( 26.26, 28.74) mi/gallon.

Therefore, the 95% confidence interval (a,b)= ( 26.26, 28.74) mi/gallon.

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