Question

For a recent year, the mean fare to fly from Charlotte, North Carolina, to Chicago, Illinois, on a discount ticket was $267. A random sample of 13 round-trip discount fares on this route last month shows: $321 $286 $290 $330 $310 $250 $ 270 $280 $299 $265 $291 $275 $281 At the 0.01 significance level, can we conclude that the mean fare has increased? What is the p-value? Click here for the Excel Data File a. What is the decision rule? (Round your answer to 3 decimal places.) Reject HO:us 267 and fail to reject H1:> 267 when the test statistic is X This is a numeric cell, so please enter numbers only. b. Compute the value of the test statistic. (Round your answer to 3 decimal places.) Value of the test statistic c. What is your decision regarding Ho? Reject Do not reject d. Estimate the p-value? The p- value is

Answer 1

Solution:-

Mean Sum S.D Count S.E Sample 288.30769230769: 3748 22.458794177280 13 6.2289487685596

a) Reject H0, if t > 2.681

p=0.01 4 -3 -2 -1 0 2 3 t%3D2.681

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

Null hypothesis: u < 267
Alternative hypothesis: u > 267

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. 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 = 6.2289
DF = n - 1

D.F = 12
b)

t = (x - u) / SE

t = 3.42

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.

Interpret results. Since the t-value (3.42) lies in the rejection region, hence we have to reject the null hypothesis.

c) Reject null hypothesis.

d)

P-value = P(t > 3.42)

P-value = 0.003