Question

A used car dealer says that the mean price of a three-year-old sport utility vehicle in good condition is $18,000. A random sample of 20 such vehicles has a mean price of $18,450 and a standard deviation of $1930. At α=0.08, can the dealer’s claim be supported? No, since the test statistic of 1.04 is close to the critical value of 1.24, the null is not rejected. The claim is the null, so is supported Yes, since the test statistic of 1.04 is not in the rejection region defined by the critical value of 1.85, the null is not rejected. The claim is the null, so is supported Yes, since the test statistic of 1.04 is in the rejection region defined by the critical value of 1.46, the null is rejected. The claim is the null, so is supported No, since the test statistic of 1.04 is in the rejection region defined by the critical value of 1.85, the null is rejected. The claim is the null, so is not supported

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 18000(Claim)
Alternative hypothesis: u 18000

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.08. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 431.561

DF = n - 1

D.F = 19
t = (x - u) / SE

t = 1.043

t_critical = + 1.85

Rejection region is - 1.85 < t < 1.85.

Interpret results. Since the t-value (1.043) is not in the critical region, we cannot reject the null hypothesis.

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 19 degrees of freedom is less than -1.043 or greater than 1.043.

Thus, the P-value = 0.31.

Interpret results. Since the P-value (0.31) is greater than the significance level (0.08), we cannot reject the null hypothesis.

Do not reject null hypothesis.