Statistics – Lab #6
Name:__________
Statistical Concepts: * Data Simulation * Discrete Probability Distribution * Confidence Intervals
Calculations for a set of variables
Answer:
Calculating Descriptive Statistics
Answer:
Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
Mean 20 0 3.560 0.106 0.476 2.600 3.225 3.550 3.775 4.500
Median 20 0 3.600 0.169 0.754 2.000 3.000 3.500 4.000 5.000
Calculating Confidence Intervals for one Variable
Answer:
Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
Mean 20 0 3.560 0.106 0.476 2.600 3.225 3.550 3.775 4.500
Median 20 0 3.600
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Is this estimate centered about the parameter of interest (the parameter of interest is the answer for the mean in question 2)?
The mean for the median column of the worksheet is 3.6Yes, the estimate is centered about the parameter of interest. |
5. Give the standard deviation for the mean and median column. Compare these and be sure to identify which has the least variability?
Standard Deviation for the mean column is 0.476Standard Deviation for the median column is 0.754Standard deviation for the mean column has least variability |
6. Based on questions 3, 4, and 5 is the mean or median a better estimate for the parameter of interest? Explain your reasoning.
Mean is the better estimate for the parameter of interest. The mean is more centered with least variability. |
7. Give and interpret the 95% confidence interval for the hours of sleep a student gets.
One-Sample T: Sleep Variable N Mean StDev SE Mean 95% CISleep 20 6.950 1.572 0.352 (6.214, 7.686)95% of students get between 6.2 hour to 7.69 hour of sleep. |
8. Give and interpret the 99% confidence interval for the hours of sleep a student gets.
One-Sample T: Sleep Variable N Mean StDev SE Mean 99% CISleep 20 6.950 1.572 0.352 (5.944, 7.956) 99% of students get between 5.9 hour to 7.95 hour of sleep. |
9. Compare the 95% and 99% confidence intervals for the hours of sleep a student gets. Explain
(2) Give that a sample of 25 had x = 75, and (x-x)² = 48 the mean and standard
1. By hand, compute the mean, median, and mode for the following set of 40 reading scores:
2. Calculate descriptive statistics for the variable where students flipped a coin 10 times. Pull up Stat > Basic Statistics > Display Descriptive Statistics and set Variables: to the coin. The output will show up in your Session Window. Type the mean and the standard deviation here.
7. The data set for this problem can be found through the Pearson Materials in the Student Textbook Resource Access link,
Sleep hours were then investigated to illuminate the impact it has on GPA. ANOVA (A-test #1) results indicated that subjects that got 0-4 hours of sleep had a GPA of (3.441 ∓ .168, n=6), 4-7 hours of sleep (3.364 ∓ .365, n=76), and finally 7-10 hours of sleep (3.374 ∓ .316, n=63)(p=.0919)(Figs._____). There was no statistical significance differences between group means by one-way ANOVA, F(2,142)=2.42,p=.0919 {F= F statistic, df (between)=2 , df (within)=142, F ratio=2.42}(.
A. SE for % = Sqrt ((True %)(100%-True %)/(Size of Sample)) Margin of error = 2% so Standard Error = 1% 1% = Sqrt ((5%)(100%-%%)/(Size of Sample) Size of the Sample = (5%*95%)/(.01^2) Size of the Sample = 475 Samples At 10 minutes each, 475 Samples = 4750 Minutes = 3.3 Days
In the article entitled "Sleep Duration and Self-Reported Type A Behavior: A Replication," Stuart McKelvie (1992) surveyed college students who receive eight hours or less of sleep per night. These surveys observed type A behavior students, which include students who are more devoted to studying, showing up for class, and being involved in other various activities. In a repeated study of the Hicks study from 1979, three hundred and eleven undergraduate students were divided into groups of thirty and completed a sleep survey. Regular sleepers indicated that they had received the same amount of sleep for one year and were satisfied with their sleeping patterns. From this group of students the survey study stated that seventy-two irregular sleepers received less than eight hours per night and seventy regular sleepers received more than eight hours per night. Between the two groups of students, test scores did not significantly differ. Both were satisfied with the amount of sleep that they receive each night. This is said to be the reason that their daily lives and grades are not significantly effected by the amount of sleep they receive. Research by Glass suggests the idea that students who lead a hectic college lifestyle on
b. Use a calculator or computer to compute the mean and standard deviation of the year
***The following four cases are based on a normal distribution of scores with Mean = 75 and the SD = 6.38.
Possible sources of information(at least 3) : Survey of at least 20 students and their sleeping habits (including hours per night) and GPA; Newspapers, magazines, Dictionary, encyclopedia, Internet
CHAPTER THREE DESCRIPTIVE STATISTICS: NUMERICAL MEASURES MULTIPLE CHOICE QUESTIONS In the following multiple choice questions, circle the correct answer. 1. Which of the following provides a measure of central location for the data? a. standard deviation b. mean c. variance d. range Answer: b 2.
In this case, the business goes with the mean data analysis; the customers coming in the hotel will be within the range of 20kms. If they go with median data analysis, range will only be with in 20 and 25kms. And if the go with the mode data analysis, the range could be anything from 5 to 40kms
From this number above, we may estimate the Mean, median and standard deviation as next: Mean: 16.66;
The purpose of this study was to observe the effects of sleep timing, sleep quality, and sleep duration on academic achievements in young adults. Those who participated in the study were thirty-six Italian seniors in high school. In order to determine their ideal sleep timing, they each filled out the Morningness-Eveningness Questionnaire for Children and Adolescents (MEQ-CA). For two non-consecutive weeks, students underwent actigraphy, which is a non-invasive method of monitoring human rest and activity cycles. This was done in a one month period, and was able to assess factors such as: habitual sleep timing though the midpoint (MS); habitual sleep quality through the parameter of sleep efficiency (SE); and habitual sleep duration through the parameter of total sleep time (TST) (Tonetti, Fabbri, Filardi, Martoni, Natale, 2015). At the end of each actigraphic-recording week, the students completed the Mini Sleep Questionnaire, and their school performance were assessed by the grades achieved after taking final exams.
Table 1 is showing that the mean score of young employees is 64.9, Standard Deviation is 12.97 and the mean score of middle-aged employees is 75.56, Standard Deviation is 17.2 on degree of freedom 58, t-value is 9.71 which are significant at 0.01.a