Large-Sample Test for a Population Mean On June 14, 2016, Bargain Airlines claim

Question

Large-Sample Test for a Population Mean On June 14, 2016, Bargain Airlines claimed that their average fare from Philadelphia to any destination in Europe was significantly less than $1150. A random sample of 125 flights offered by Bargain Airlines to any destination in Europe was selected. The average fare from the sample of 125 flights was $950, with a standard deviation of $150. Test Bargain Airline's claim at a.05 significance level. The alternative hypothesis isa following terms: right, left, two) What is the value of the population mean that is used in the null and alternative hypotheses? $ tailed test. (Fill in the blank with one of the The value of the test statistic z round to the 100th place). The p-value is

Explanation / Answer

Solution:

Given information
n = 125, x? = 950 , s= 150
Test Hypothesis
H0: ?0 < 1150
Ha: ?0 ? 1150

Test statitsic
z = x? - ?0/(s/?n)
= (950-1150)/150/sqrt(125)
= -14.9071

P-value = 0.00005 (Using P-value calculator)
The P-value < 0.0001
we reject the null hypothesis because | t | ? | t(124),0.05 |
Also p-value is less than level of significance.

Conclusion:
There is enough evidence to to support the claim made by Bargain Airlines.
There is sufficient sample evidence to claim that the average fair from Philadelphia to any destination in Europe is significantly less than $1150.

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