Question

Problem 2: For a recent year, the mean fare to fly from Charlotte, North Carolina to Chicago Illinois, on a discount ticket was $290. A random sample of 12 round trip discount fares on this route last month shows: 286 290 285 291 287 275 262 289 274 276 297 262 At the 0.01 significance level, can we conclude that the mean fare has decreased? What is the p value? 1. State the Null and Alternate hypothesis: 2. State the test statistic: 3. State the Decision Rule: 4. Show the calculation: 5. What is the interpretation of the sample data? 6. Show the P value

Answer 1

Answer:

Step-by-step explanation:

n = 12

The mean of the set of data given is

Mean, x = (286 + 290 + 285 + 291 + 287 + 275 + 262 + 289 + 274 + 276 + 297 + 262)/12 = 281.2

Standard deviation = √(summation(x - mean)²/n

Summation(x - mean)² = (286 - 281.2)^2 + (290 - 281.2)^2 + (285 - 281.2)^2 + (291 - 281.2)^2 + (287 - 281.2)^2 + (275 - 281.2)^2 + (262 - 281.2)^2 + (289 - 281.2)^2 + (274 - 281.2)^2 + (276 - 281.2)^2 + (297 - 281.2)^2 + (262 - 281.2)^2 = 1409.68

Standard deviation, s = √(1409.68/12) = 10.8

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

1) For the null hypothesis,

H0: µ = 290

For the alternative hypothesis,

H1: µ < 290

This is a left tailed test.

2) Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 12,

Degrees of freedom, df = n - 1 = 12 - 1 = 11

t = (x - µ)/(s/√n)

Where

x = sample mean = 281.2

µ = population mean = 290

s = samples standard deviation = 10.8

t = (281.2 - 290)/(10.8/√12) = - 2.82

Test statistic = - 2.82

3) The decision rule is to reject the null hypothesis if the significance level is greater than the p value.

4) We would determine the p value using the t test calculator. It becomes

p value = 0.0083

5) Since alpha, 0.01 > than the p value, 0.0083, then we would reject the null hypothesis. Therefore, At a 1% level of significance, the sample data showed significant evidence that the mean fare has decreased.