Question

a car company says that the mean gas mileage for its luxury sedan is at least 23 miles per gallon (mpg). you believe the claim is incorrect and find that a random sample of 6 cars has a mean gas mileage of 21 mpg and a standard deviation of 3 mpg. at \\( \alpha = 0.10 \\), test the companys claim. assume the population is normally distributed. click here to view the t - distribution table. click here to view page 1 of the normal table. click here to view page 2 of the normal table. which sampling distribution should be used and why? a. use a t - sampling distribution because \\( n<30 \\). b. use a normal sampling distribution because \\( n>30 \\). c. use a t - sampling distribution because the population is normal, and \\( \sigma \\) is unknown. d. use a normal sampling distribution because the population is normal, and \\( \sigma \\) is unknown. e. use a t - sampling distribution because the population is normal, and \\( \sigma \\) is known. f. use a normal sampling distribution because the population is normal, and \\( \sigma \\) is known. state the appropriate hypotheses to test. a. \\( h_{0}: \mu \geq 23 \\) \\( h_{a}: \mu<23 \\) b. \\( h_{0}: \mu = 23 \\) \\( h_{a}: \mu \
eq 23 \\) c. \\( h_{0}: \mu \leq 23 \\) \\( h_{a}: \mu>23 \\) d. \\( h_{0}: \mu \
eq 23 \\) \\( h_{a}: \mu = 23 \\) what is the value of the standardized test statistic? the standardized test statistic is (round to two decimal places as needed.)

Explanation:

Step1: Calculate the standardized test statistic

The formula for the \(t\)-test statistic is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\).
Here, \(\bar{x} = 21\), \(\mu = 23\), \(s = 3\), \(n=6\).
Substitute the values into the formula: \(t=\frac{21 - 23}{3/\sqrt{6}}\).
First, calculate the denominator \(3/\sqrt{6}\approx1.2247\).
Then, \(t=\frac{- 2}{1.2247}\approx - 1.63\).

Answer:

\(-1.63\)