Exercise r Easy-Fly Airline took a random sample of 25 fishts to determine if th

Question

Exercise r Easy-Fly Airline took a random sample of 25 fishts to determine if the mean time it takes Exercise 93: 1 for luggage to reach the travelers departing from a flight is less than 15 minutes. The sample mean was found to be 13.8 minutes with a sample standard deviation s of 4 minutes. Usin g a significance level of le? Hypothesis Test Step 1: State Ho, Ha HA Step 2: Specify level of significant Step 3: Compute Test statistic In Use t-Test for Mean when is unknown Step 4: Find the critical value and rejection rule The critical value = Rejection rule: Step 5.1: Determine whether to reject Ho Step 5.2: Write your conclusion in complete sentences

Explanation / Answer

n =25
Sample mean = 13.8min
Sample stdev = 4min

alpha = .05

Step1)
Hypothesis test:
Ho: Mu>=15 min
Ha: Mu<15 min

Step2)
5% is the level of significance

Step3) t = (xvar-Mu0)/(s/sqrt(n)) = (15-13.8)/(4/sqrt(25)) = 1.5
t = 1.5

Step4) Critical value of t is = 1.72
The P-Value is .073328.
The result is not significant at p < .05, since p-value is not in the critical region of <=.05

Rejection Rule:
5.1 We do not reject the null hypothesis
5.2 We conclude that the it does not take less than 15 minutes

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