The manufacturer of an airport baggage scanning machine claims it can handle an

Question

The manufacturer of an airport baggage scanning machine claims it can handle an average of 530 bags per hour. A sample of 25 randomly selected hours showed an average of 515 bags with a sample standard deviation of 40.

At 5% level of significance, please test whether the manufacturer’s claim is overstated. Please redo the test at 10% level of significance? Show all the necessary steps and explain what your conclusion means in each case.

Did you arrive at the same conclusion? What happens to the probability of type II error when you do the test at 10% instead of 5%? Please explain the difference between conducting the test at 5% and 10% levels of significance.   

Explanation / Answer

we know that the test stat is given as

t = (xbar-mu)/(sd/sqrt(n))

here xbar= 515 , sd = 40 , n = 25 and mu = 530

so we calculate the test stat as

t = (515-530)/(40/sqrt(25)) = -1.875

now we check the p values from the t table as

alpha = 0.05 , 1 tail
alpha = 0.10 and 1 tail

the df = n-1 = 25-1 = 24

The P-Value is .036506.

The result is significant at p < .05

The result is significant at p < .10.

so using the p value approach in both the cases the p value is less than the alpha , hence we reject the null hypothesis in favor of alternate hypothesis

Yes , we arrive at the same conclusion as the p value is less than the alpha value in both the cases . Only the alpha value changes in the case , however the result does not change

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