rsch 665

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Feb 20, 2024

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RSCH 665 - Assignment 5.2 t-Test Problems 1.A researcher wishes to determine if a particular drug affects pilot reaction time to air traffic controller instructions. The researcher has 10 pilots. The pilots are observed in normal performance and their reaction times are recorded. Then the pilots are administered the drug, observed again, and their
reaction times are recorded. The expectation is that the drug will reduce reaction time. PilotTrial 1 Time (sec)Trail 2 Time (sec) A.83.69 B.74.71 C.82.79 D.86.87 E.66.65 F.63.68 G.81.67 H.77.72 I.73.71 J.69.65 Conduct attest using the five-step hypothesis testing process and StatCrunch.Show your work. Step 1: Restate the question as a research hypothesis and null
hypothesis about the populations. H0: μ1 = μ2. Trail 1 time = trail 2 time. Ha: μ1 > μ2. Trail 1 time > trail 2 time. Step 2: Determine the comparison distribution. The comparison distribution is a t-distribution with N-1 degrees of freedom (10-1) = 9 df. Step 3: Determine the cutoff score where the null hypothesis will be rejected. This is a one-tail t-test, with 9 degrees of
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freedom, and we'll test at the .05 significance level. By using T-table, we can find cutoff score is 1.833. Step 4: Determine sample's score on the comparison distribution. Paired T hypothesis test: μ D = μ 1 - μ 2 : Mean of the difference between Trial 1 Time (sec) and Trail 2 Time (sec) H 0 : μ D = 0 H A : μ D > 0 All rights are reserved. The material contained herein is the copyright property of Embry-Riddle Aeronautical University, Daytona Beach, Florida, 32114. No part of
this material may be reproduced, stored in a retrieval system or transmitted in any form, electronic, mechanical, photocopying, recording or otherwise without the prior written consent of the University. College of Aeronautics| worldwide.erau.edu Sample statistics: SamplenMeanStd. dev. Trial 1 Time (sec)100.7540.077344827 Trail 2 Time (sec)100.7140.068345527 Hypothesis test results: DifferenceMeanStd. Err.DFT-StatCritical tP-value Trial 1 Time (sec) -Trail 2
Time (sec) 0.040.01891501592.11472 211.83311290.0318 T= mean difference / Std. err. = 0.04 / 0.189 = 2.11 Step 5: Decide whether to reject the null hypothesis. We reject the null hypothesis since the T- score is greater than the cutoff score ( 2.11>1.8.33). We conclude that the drug has an effect on reducing pilot's reaction time at
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significance level of 0.05. 2.A researcher wishes to determine if the college students with a "B" average have attain more flying hours by the end of their first semester of college than those that do not have a "B" average at the end of the first semester. The researcher surveys 20 college students that are also student pilots and records their flight hours as well as
their GPA (noted as "below B" and "B or better"), and, fortunately, there is an equal number "below B" and "B or better." The expectation is that those students with a B average or better will have more flying hours by the end of their first semester than those students that are below a "B" average. Here are the survey results: Below B flying hoursB or Better flying hours 4352
4132 3144 3658 3951 3730 3644 4155 4562 4152 Conduct attest using the five-step hypothesis testing process and StatCrunch. Show your work. All rights are reserved. The material contained herein is the copyright property of Embry-Riddle Aeronautical University, Daytona Beach, Florida, 32114. No part of this material may be reproduced, stored in a retrieval system or transmitted in any form, electronic, mechanical, photocopying, recording or otherwise without the prior written consent of the University. College of Aeronautics| worldwide.erau.edu
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Step 1: Restate the question as a research hypothesis and null hypothesis about the populations. H0: μ1 = μ2. Below B flying hours = B or better flying hours. Ha: μ1 < μ2. Below B flying hours < B or better flying hours. Step 2: Determine the characteristics of the comparison distribution. The comparison distribution is a t- distribution with N-1
degrees of freedom (20-2) = 18 df. Step 3: Determine the cutoff score where the null hypothesis will be rejected. This is a one-tail t-test, with 18 degrees of freedom, and we'll test at the .05 significance level. By using T-table, we can find cutoff score is -1.734. Step 4: Determine sample's score on the comparison distribution. Two sample T hypothesis test: μ 1 :
Mean of Below B flying hours μ 2 : Mean of B or Better flying hours μ 1 - μ 2 : Difference between two means H 0 : μ 1 - μ 2 = 0 H A : μ 1 - μ 2 < 0 (with pooled variances) Hypothesis test results: DifferenceSample Diff.Std. Err.DFT-StatCritical tP-value μ 1 - μ 2 - 93.571492118-2.5199552- 1.73406360.0107 T-score is -2.52. Step 5: Decide whether to reject the null hypothesis. We reject the null hypothesis since the
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T-score is less than the cutoff score (-2.52<- 1.73). We conclude that students with a B average or better have more flying hours than students that are below a "B" average at significance level of 0.05. All rights are reserved. The material contained herein is the copyright property of Embry-Riddle Aeronautical University, Daytona Beach, Florida, 32114. No part of this material may be reproduced, stored in a retrieval system or transmitted in a