Statistics Question: Setting jup a right-tailed decision rule?

Question

I'm having trouble figuring out which formula to use for this. I think I'm supposed to use a Z test statistic. But I'm lost from there.

The mean arrival rate of flights at O'Hare Airport in marginal weather is 195 flights per hour w/a historical standard deviation of 13 flights. To increase arrivals, a new air traffic control pocedure is implemented. In the next 30 days of marginal weather the mean arrival rate is 200 flights/hour. (a) Set up a right-tailed decision rule at Type I error level of significance = .025 to decide whether there has been a significant increase in the mean # of arrivals per hour. Carry out the test & make the decision. Would the decision be different if you used level of significance = .01? What assumptions are you making, if any? Flights are: 210, 215, 200, 189, 200, 213, 202, 181, 197, 199, 193, 209, 215, 192, 179, 196, 225, 199, 196, 210, 199, 188, 174, 176, 202, 195, 195, 208, 222, 221

Answer 1

Hypothesis Test for mean:



Assuming you have a large enough sample such that the central limit theorem holds, or you have a sample of any size from a normal population with known population standard deviation, then to test the null hypothesis

H0: μ ≤ Δ or

H0: μ ≥ Δ or

H0: μ = Δ

Find the test statistic z = (xbar - Δ ) / (sx / √ (n))



where xbar is the sample average

sx is the sample standard deviation, if you know the population standard deviation, σ , then replace sx with σ in the equation for the test statistic.

n is the sample size



The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.



H1: μ > Δ; p-value is the area to the right of z

H1: μ < Δ; p-value is the area to the left of z

H1: μ ≠ Δ; p-value is the area in the tails greater than |z|



If the p-value is less than or equal to the significance level α, i.e., p-value ≤ α, then we reject the null hypothesis and conclude the alternate hypothesis is true.



If the p-value is greater than the significance level, i.e., p-value > α, then we fail to reject the null hypothesis and conclude that the null is plausible. Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible.



The hypothesis test in this question is:



H0: μ ≤ 195 vs. H1: μ > 195



The test statistic is:

z = ( 200 - 195 ) / ( 13 / √ ( 30 ))

z = 2.106625



The p-value = P( Z > z )

= P( Z > 2.106625 )

= 0.01757504



Since the p-value is less than the significance level we reject the null hypothesis and conclude the alternate hypothesis μ > 195 is true.



now, using the fixed level idea for hypothesis testing we would reject the null hypothesis if the test statistic is greater than 1.96



the value of xbar for rejection is xbar > 195 + 1.96 * 13/sqrt(30) = 199.652



since the mean of the sample is 200 we can reject the null.



in the case of a 1% level test we would fail to reject. We can find this be looking for the new critical region or note that the p-value is greater than 1% and thus we cannot reject the null hypothesis at the 1% level.

Answer 2

The critical region for testing H0: μ=195 vs. H1: μ > 195 is



xbar > 195 + (13/√30) z_(1-α)



so for α = 0.025, the critical region is



xbar > 195 + (13/√30)(1.96) = 199.652



With the observed mean of 200, we would reject the null hypothesis.



I leave the α=0.01 calculation and conclusion to you.



One assumption being made is that the new procedure has the same standard deviation as the old procedure, since we're using the old procedure's standard deviation in the calculation.

Answer 3

RE: Statistics - Decision Rule, Pooled Proportion? The null and alternate hypotheses are: H0: π1 ≤ π2 H1: π1> π2 A sample of 100 observations from the first population indicated that X1 is 70. A sample of 150 observations from the second population revealed X2 to be 90. Use the .05 significance level to test the hypothesis. (a) State the...