Confidence Intervals & Hypothesis Testing Assignment Guide

RSCH 665 4.2 Assignment 1 4.2 Assignment: Confidence Intervals and Introduction to Hypothesis Testing Po-Yun, Lai Embry Riddle Aeronautical University RSCH 665 Jan 30 th , 2024 Dr. Samuel Benavides

2 1. Calculate a 95% and 99% confidence interval for the following data set of college students’ numbers of hours per week working a part-time job. Note: this is a sample. Show your steps including the mean (5 pts), z-score (5 pts), standard deviation (5 pts), and standard error (5 pts) that you use. Explain what the Confidence Intervals mean with respect to this data set (15 pts each). (50 pts total) Step 1. Get mean, standard deviation and standard error (Stat > Summary Stats > Column). Summary statistics: Column n Mean Variance Std. dev. Std. err. Hours 20 14.7 43.063158 6.5622525 1.4673643 Mean=14.7 Standard Deviation=6.56 Standard Error=1.46 Step 2. Confidence Interval for 95% (Stat > Z-Stats > One Sample > With Data) One sample Z summary confidence interval: μ : Mean of population Standard deviation = 6.56 95% confidence interval results: Mean n Sample Mean Std. Err. L. Limit U. Limit μ 20 14.7 1.4668606 11.825006 17.574994 z = +/-1.96 for the 95% confidence interval Step 3. Confidence Interval for 99% (Stat > Z-Stats > One Sample > With Data) One sample Z confidence interval: μ : Mean of variable

3 Standard deviation = 6.56 99% confidence interval results: Variable n Sample Mean Std. Err. L. Limit U. Limit Hours 20 14.7 1.4668606 10.921618 18.478382 z = +/-2.576 for the 99% confidence interval Explain what the Confidence Intervals mean with respect to this data set. If you construct a 95% confidence interval, it means that if you were to take many samples and construct intervals in the same way, you would expect about 95% of those intervals to contain the true population parameter. In this scenario, a 95% confidence interval is between 11.82 to 17.57 weekly working hours, which means 95% of students’ weekly part-time working job is located within 11.82 to 17.57. For 99% confidence interval, however, is more strict. A 99% confidence interval would be wider, as it needs to capture a larger range of potential values. It provides a higher level of confidence but comes at the cost of a wider interval. In this example, 99% of students work between 10.92 to 18.47 hours weekly. 2. In an experimental treatment, a random individual scores 84 on a measurement. A sample shows people in general are normally distributed with a mean of 68 and a standard deviation of 9. The researcher predicts that the treatment will create an effect on the subject, but does not specify if the measurement will increase or decrease. Using a 95% confidence interval (p < .05), what conclusions should this researcher make about the treatment? Utilize the five-step hypothesis testing process and show all of your work. (25 pts)

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4 1) Restate the research question as a research hypothesis and a null hypothesis about the populations. (5 pts) Research hypothesis H1: People who do the treatment will increase or decrease their scores. Null Hypothesis H0: People who do the treatment will have no difference in increasing or decreasing their score. 2) Determine the characteristics of the comparison distribution. (5 pts) Comparison distribution: People who don’t do the treatment. Mean( μ) : 68 Standard Deviation( σ) : 9 Shape of distribution: normal curve 3) Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected. (5 pts) Significance level: 5% (p < .05) Z-score cutoff: -1.96 and +1.96 (every Z-score that is outside the red area will be Statistically significant) 4) Determine your sample’s score on the comparison distribution. (5 pts). Sample score=84

5 Sample Z-score= (84 – 68)/9=1.78 (Statistically significant for a cutoff Z-score of 1.96, we reject if score < -1.96 or score > 1.96) 5) Decide whether to reject the null hypothesis. (5 pts) A z-score analysis was conducted to examine if treatment would increase or decrease the test scores. The analysis was conducted using a confidence interval of 95% with a z-critical value of 1.96. The results of the z-score analysis produced a z-score value of 1.78. The sample z-score is within +/- 1.96, which stands for non-statistically significant in the treatment. We accept the null hypothesis: the treatment doesn’t increase or decrease the score. 3. For this activity, utilize the five-step process and the following data set. (25 pts) The population and sample data is:

6 • Population: Mean=78, SD=7 • Sample: (sample of one): Raw Score=95 Set up a one-tail Z test evaluating the improvement of a score, presenting the following: 1) Restate the research question as a research hypothesis and a null hypothesis about the populations. The population is the mean and SD for the FAA written test for Fundamental of Instruction (FOI). A flight school hires a professor with an education Ph.D. degree to instruct a new Certified Flight Instructor (CFI) class, and the sample is their score of the written test result. Flight school wants to examine if this is worth to invest in the future. Research hypothesis H1: The new education-specialized course will improve CFI students’ FOI test results. Null hypothesis H0: The new education-specialized course will have no difference in improving CFI students’ FOI test results. 2) Determine the characteristics of the comparison distribution. Comparison distribution: CFI students who don’t take the course. Mean(μ): 78 Standard Deviation(σ): 7 Shape of distribution: normal curve

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7 3) Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected. Researchers decided to use a standard 95% significance level. Z-score cutoff: +1.64 (every Z-score that is outside the red area will be statistically significant) 4) Determine your sample’s score on the comparison distribution. The sample raw score is 95. The sample z-score is (95-78)/7=2.42 5) Decide whether to reject the null hypothesis. A z-score analysis was conducted to examine if the new education-specialized course will improve CFI students’ FOI test results. 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 2.42. The sample z-score is greater than 1.64, which stands for statistically significant

8 in the education-specialized course. We reject the null hypothesis: The new education- specialized course will have no difference in improving CFI students’ FOI test results.

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