STATISTICAL INFERENCE learning journal 4
.docx
School
University of the People**We aren't endorsed by this school
Course
CS 1281
Subject
Statistics
Date
Dec 18, 2024
Pages
5
Uploaded by EarlIbex1928
1 University of the PeopleStatistical InferenceHitesh VermaDecember 12, 2024
2PART 1 a.Null hypothesis (Ho): There is no significant difference between the average student scores on reading and writing examsAlternative hypothesis (Ha): The student grades on reading and writing exams are different from each other. b.Requisites to pass the test Randomization: From the population of seniors, 250 individuals should be randomly chosen to be participants. Independence: One student’s grade should not be impacted by another student’s grade.Normality: The distribution of the scores should follow the normal distribution. c.We can determine whether there is strong evidence that the average scores on the two exams have different distributions by doing a paired t-test. F t = (X-bar - μ) / (s / sqrt(n))
3The sample mean difference score found in the X-bar ( -0.545) μ is the predicted population mean difference, expected to be 0 in the null hypothesis. The difference standard deviation is s = 8.887. The sample size is 250 /9 n)t= (-0.545 - 0) / (8.887 / sqrt (250))Degrees of Freedom (n-1) (250-1) = 249 When comparing the p-value of 0.39 to the significance level ( x=0.05), we can conclude.If the p-value is less than the significance level, we reject the null hypothesis. d.The type of error that might have occurred is Type II or a false negative. This means that we fail to reject the null hypothesis when it is in reality false. In this case, we should conclude that there is no significant difference between student’s average scores on reading and writing exams. e.Based on the hypothesis test, the p-value is 0.39 is greater than 0.05, so we fail to reject the null hypothesis. Therefore, we cannot conclude that there is a difference between the average scores on both exams. PART 21.Null hypothesis (Ho): There is no difference between vehicles with manual and automated gearboxes in terms of average city mileage.
4Alternative hypothesis (Ha): There is a significant difference in the average fuel efficiency of the two types of vehicles. 2.t = (X1-bar - X2-bar) / sqrt((s1^2 / n1) + (s2^2 / n2))X1-bar = 16.12 The sample mean or X2-bar = 19.85 Automatic vehicle standard deviation = 3.58 Manual car standard deviation = 4.51 The sample size for automated vehicles = 26 The sample size for manual autos = 26t = (16.12 - 19.85) / sqrt((3.58^2 / 26) + (4.51^2 / 26))3.df = (n1 + n2 - 2)(26 + 26 - 2 ) = 50 degrees of freedom 4.Since the p-value (0.0029) is smaller than the significance level (0.05), the null hypothesis is rejected. We conclude that there is a difference between the normal city mileage of vehicles with manual and automatic transmissions.
5ReferencesTaylor, C. (2019, February 13). What Is Paired Data?ThoughtCo. https://www.thoughtco.com/what-is-paired-data-3126311