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 7 cars has a mean gas mileage of 20 mpg and a standard deviation of 4 mpg. at α = 0.025, test the company’s claim. assume the population is normally distributed.

which sampling distribution should be used and why?

a. use a t - sampling distribution because the population is normal, and σ is known.
b. use a normal sampling distribution because the population is normal and σ is known.
c. use a t - sampling distribution because n < 30.
d. use a normal sampling distribution because n > 30.
e. use a normal sampling distribution because the population is normal, and σ is unknown.
f. use a t - sampling distribution because the population is normal, and σ is unknown.

state the appropriate hypotheses to test.

a. ( h_0: mu geq 23 )( h_a: mu < 23 )
b. ( h_0: mu leq 23 )( h_a: mu > 23 )
c. ( h_0: mu = 23 )( h_a: mu
eq 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 - 1.98 (round to two decimal places as needed.)

what is the critical value?

the critical value is □ (round to three decimal places as needed.)

Explanation:

Step1: Determine the sampling distribution

Since the population is normal, \(n = 7<30\), and \(\sigma\) is unknown, we use a \(t -\)sampling distribution.

Step2: State the hypotheses

The company claims \(\mu\geq23\). The null hypothesis \(H_0:\mu\geq23\) and the alternative hypothesis \(H_a:\mu < 23\) (because we believe the claim is incorrect in the direction of being less).

Step3: Calculate the test - statistic

The formula for the \(t -\)test statistic is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\). Here, \(\bar{x} = 20\), \(\mu = 23\), \(s = 4\), \(n = 7\).

$$t=\frac{20 - 23}{4/\sqrt{7}}=\frac{- 3}{4/\sqrt{7}}=\frac{-3\sqrt{7}}{4}\approx - 2.29$$

Step4: Find the critical value

The significance level \(\alpha=0.025\), and the degrees of freedom \(df=n - 1=7-1 = 6\). Using the \(t -\)distribution table for a one - tailed test, the critical value \(t_{0.025,6}=- 2.447\)

Answer:

The sampling distribution: Use a \(t -\)sampling distribution because the population is normal and \(\sigma\) is unknown.
The hypotheses: \(H_0:\mu\geq23\), \(H_a:\mu < 23\)
The standardized test statistic: \(-2.29\)
The critical value: \(-2.447\)