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The gas mileage for a certain model of car is known to have a standard deviation of 6 mi/gallon. A simple random sample of 49 cars of this model is chosen and found to have a mean gas mileage of 28.4 mi/gallon. Construct a 89% confidence interval for the mean gas mileage for this car model.

Respuesta :

Answer:

The answer is "(27.030,29.770)"

Step-by-step explanation:

[tex](\bar{x}) = 28.4[/tex]

[tex](n)= 49[/tex]

[tex](\sigma) = 6[/tex]

[tex](CI) = \bar{x}\pm z^*_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}[/tex]

[tex]z^*_{\frac{\alpha}{2}}\ for \ 89\%\ confidence = 1.598 \\\\\text{Using z table }\\\\ NORM.S.INV(1-\frac{(1-0.89)}{2})[/tex]

[tex]CI = 28.4 \pm 1.598 \times \frac{6}{\sqrt{49}}=(27.030,29.770)[/tex]

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