rsch 665 5.2

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Feb 20, 2024

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Assignment 5.2 t-Test 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. Show your work A. Present the experimental data in a table. Present the table in accordance with the APA format. Table 1 Table Displaying Pilot's Reaction Time to Air Traffic Controller Instructions PilotTrial 1
TimeTrial 2 Time 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 RSCH 665 Assignment 4.23 B. Present a histogram of the experimental data. Present the histogram in accordance with the APA format. Figure 1 Histogram and Normal Distribution of Trial 1
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Time Figure 2 Histogram and Normal Distribution of Trial 2 Time RSCH 665 Assignment 4.24 C. What could be a research question for this experiment? In your own words, present a research question for this work. This research question will then be converted into the hypotheses. Research Question: Would pilots
who took a particular drug have decreased reaction time to air traffic controller instructions? Q4.Present an alternate hypothesis for this analysis. The alternate hypothesis should derive from the research question. Use proper notation for the hypothesis. Alternate Hypotheses: Pilots who took a particular drug
have decreased reaction time to air traffic controller instructions. Q5. Present a null hypothesis for this analysis. The null hypothesis should derive from the research question. Use proper notation for the hypothesis. Null Hypotheses: There is no difference in retention time to air traffic
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controller instructions between pilots who took a particular drug and those who did not. RSCH 665 Assignment 4.25 Q6. Determine the characteristics of the comparison distribution. Is the distribution normal? Why or why not. Include the skewness value in your discussion. This distribution is normal.
First, as mentioned in the problem, this population is normally distributed, and the sample size is large enough to be 32 (greater than 30), so this distribution can be considered normal. Also, we made a histogram of the treatment scores given in problem 2. And the shape of this histogram shows the
shape of the normal distribution. Finally, we know that the data are normal distribution through the skewness value. As can be seen in Table 2, the skewness of this data set is0.012, which is less than 1/2 (0.5). This also indicates that the data are normal. Table 2 Summary Statistics of Treatment Scores Q6.
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Identify the critical score on the comparison distribution. This is the critical score on the comparison distribution at which the null hypothesis should be rejected. If we calculate the critical score as given in the problem by using a 95% confidence interval, the resulting value is: Critical Score: 1.64 Figure 2 Critical
Score Using 95% Confidence IntervalColumnnMeanVa rianceStd. dev.Skewness Treatment Scores3251.218754.305 4435 2.07495630.01184 8156 RSCH 665 Assignment 4.26 Q7. Determine the result of the sample's test score. Show your work or provide a screenshot of the test
score from a software. The hypothesis test results are as follows: Z- Stat:3.98 P- value:<0.0001 Table 3 Hypothesis Test Results VariablenSample MeanStd. Err.Z-StatP- value Treatment Scores3251.218751.060 66023.9774756<0.0001 RSCH 665 Assignment 4.27 Figure 3 Hypothesis Test Results Q8. Decide
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whether to reject the null hypothesis. Explain why you decided to reject or not reject the null hypothesis. According to the results of the Z test, which can be seen in Table 3 and Figure 3, the null hypothesis was rejected. As a result of the test, the Z-score was 3.98, which is higher than the Z- critical of 1.64. If the
Z-score is greater than the Z-critical, it means that our null hypothesis is rejected, and the alternative hypothesis is supported. Additionally, the P-values could support these findings. Statistically, if the P- value is less than 0.05, we consider the test result to be RSCH 665 Assignment 4.28 statistically
significant. The p-value derived from the hypothesis test is less than 0.0001, so less than the significant p- value of 0.05. Therefore, the null hypothesis was rejected and the result is statistically significant. Q9. Write a sound conclusion of your analysis. The conclusion should include the specific type
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of analysis undertaken, the number of participants or data in the sample, the confidence interval, the critical value, the test sample value, a decision about rejecting the null hypothesis based on a comparison of the critical value to the test sample result, a statement about which hypothesis was
supported and a statement about the statistical significance of the analysis. A z-score analysis was conducted to examine if the experimental treatment on eye tracking would increase the memory retention performance. Random 32 people were chosen to participate in the treatment and the scores were recorded.
And the analysis was conducted using a confidence interval of 95% with a z-critical value of 1.64. The results of the z-score analysis produced a z- score value of 3.98. The value of 3.98 is greater than 1.64, and so the null hypothesis was rejected. These findings support the hypothesis that people who
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received experimental treatment on eye tracking will have increased memory retention performance. As this analysis used a confidence interval of 95%, or conversely p<0.05, the results of this analysis are statistically significant. Page 8 of 8