In this technical assignment, you will have the opportunity to apply the statistical knowledge that you’ve learned so far. The assignment will require you to use R Studio and import data. We recognize that doing both of these things may be new to you, so please make sure to ask your Lab TA, lecture TA, or Dr. Woodward any questions that may come up. Though the assignment is not due until Friday, it is strongly recommended that you start the assignment before then.
I worked by myself.
16 University of Minnesota students wanted to see if their daily average tik tok consumption was greater than that of the average UMN student. They happen to know that the population mean for college student daily tik tok consumption is 3.8 hours per day. The students also reported their daily TikTok consumption averages in the data below:
student<- c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16)
tiktokConsumption<- c(5.3,3.4,4.7,4.2,3.9,7.1,4.5,3.2,1.3,2.3,4.5,2.6,3.4,3.2,4.6,4.8)
tiktokDat<- cbind.data.frame(student, tiktokConsumption)
Does the sample of students have a greater average for tiktok consumption than an average UMN student?
The average consumption is less than or equal to 3.8.
The average consumption is greater than 3.8
Average tiktok consumption measured by hours
We will calculate a one sample t-test, to compare the sample mean to the population mean.
0.05, convention
qt(0.05, 2.8,lower.tail = TRUE )
## [1] -2.4235
t.test(tiktokDat$tiktokConsumption, mu = 3.8, alternative = "greater")
##
## One Sample t-test
##
## data: tiktokDat$tiktokConsumption
## t = 0.40681, df = 15, p-value = 0.3449
## alternative hypothesis: true mean is greater than 3.8
## 95 percent confidence interval:
## 3.34498 Inf
## sample estimates:
## mean of x
## 3.9375
10. What conclusions can we make? Interpret your findings in the context of the data provided.
We fail to reject the null hypothesis. This is becasue the p-value, which is 0.3449, is greater than the alpha value, which is 0.05. We cannot find reliable evidence to support the idea that the average sample of students tiktok consumption is greater than the average UMN student. t(15) = 0.04681, p = 0.3449
## Part 2:
Students and we want to know whether this sample of 8 students happen to study more than the average PSY 3801 student on Fridays. Based on recent data collected we know that the average PSY 3801 student’s studying time on Friday is 2.3 hours.
```r
participant<- c(1:8)
studyTime<- c(1,3.5,.6,2.1,1.4,5,.8, 2.3)
studyDat<- cbind.data.frame(participant, studyTime)
Does this sample of students spend more time studying on Fridays than the average PSY 3801 student?
The sample of students time studying on Friday is less than or equal to 2.3.
The sample of students time studying on Fridays is greater than 2.3.
Time spent studying.
Comparing the sample mean to the population mean using a one sample t-test.
0.05, convention
qt(0.05, 2.3, lower.tail = TRUE)
## [1] -2.676126
t.test(studyDat$studyTime, mu = 2.3, alternative = "greater" )
##
## One Sample t-test
##
## data: studyDat$studyTime
## t = -0.39695, df = 7, p-value = 0.6484
## alternative hypothesis: true mean is greater than 2.3
## 95 percent confidence interval:
## 1.073264 Inf
## sample estimates:
## mean of x
## 2.0875
We fail to reject the null hypothesis. This is because the p-value, which is 0.6484, is greater than the alpha value, which is 0.05. We cannot find reliable evidence to support the idea that the average sample of students study time on Fridays is greater than the average PSY2801 student. t(7) = -0.39695, p = 0.6484