Gender HeartRate
male 70
male 71
male 74
male 80
male 73
male 75
male 82
male 64
male 69
male 70
male 68
male 72
male 78
male 70
male 75
male 74
male 69
male 73
male 77
male 58
male 73
male 65
male 74
male 76
male 72
male 78
male 71
male 74
male 67
male 64
male 78
male 73
male 67
male 66
male 64
male 71
male 72
male 86
male 72
male 68
male 70
male 82
male 84
male 68
male 71
male 77
male 78
male 83
male 66
male 70
male 82
male 73
male 78
male 78
male 81
male 78
male 80
male 75
male 79
male 81
male 71
male 83
male 63
male 70
male 75
female 69
female 62
female 75
female 66
female 68
female 57
female 61
female 84
female 61
female 77
female 62
female 71
female 68
female 69
female 79
female 76
female 87
female 78
female 73
female 89
female 81
female 73
female 64
female 65
female 73
female 69
female 57
female 79
female 78
female 80
female 79
female 81
female 73
female 74
female 84
female 83
female 82
female 85
female 86
female 77
female 72
female 79
female 59
female 64
female 65
female 82
female 64
female 70
female 83
female 89
female 69
female 73
female 84
female 76
female 79
female 81
female 80
female 74
female 77
female 66
female 68
female 77
female 79
female 78
female 77
Let denote the mean heart rate of men and women respectively. To test:
Vs
The appropriate statistical test the above hypothesis, where we compare the means of continuous observations in two independent groups, would be an independent sample t-test. But before running the test, we must ensure that, along with the data being normal, the assumption of homogeneity of variance must be satisfied. We may test this assumption using F test for equality of variance:
Vs
Using excel:
We get the output:
We find that the test is significant (p-value = 0.006 < 0.05) at 5% level.We may reject the null and hence conclude that the homogeneity of variance assumption is violated.
Hence, instead of going for an independent sample t-test, we may go for Welch's t-test, which is robust to the violation.
Test statistic = 0.63
P-value = 0.53
Now, running an ANOVA on the same data, comparing the means of two groups:
F statistic = 0.40
P-value = 0.53
In Welch independent t-test, no assumption is made on the homogeneity of variance (or standard deviation) i.e. the test is robust to the assumption of homogeneity of variance. However, ANOVA requires this assumption to be satisfied.Hence, the correct option would be:
ANOVA requires the group standard deviations to be similar.
Gender HeartRate male 70 male 71 male 74 male 80 male 73 male 75 male 82...
Problem #1: Consider the below matrix A, which you can copy and paste directly into Matlab. The matrix contains 3 columns. The first column consists of Test #1 marks, the second column is Test # 2 marks, and the third column is final exam marks for a large linear algebra course. Each row represents a particular student.A = [36 45 75 81 59 73 77 73 73 65 72 78 65 55 83 73 57 78 84 31 60 83...
Use the accompanying data set on the pulse rates (in beats per minute) of males to complete parts (a) and (b) below. LOADING... Click the icon to view the pulse rates of males. a. Find the mean and standard deviation, and verify that the pulse rates have a distribution that is roughly normal. The mean of the pulse rates is 71.871.8 beats per minute. (Round to one decimal place as needed.) The standard deviation of the pulse rates is 12.212.2...
The given data is the grades for people in this class. The goal here is to determine the factors that effect student's Grade in the class. 4) Find the mean and median for the men's and the women's Quizzes. Gender Men Women 5) Test the claim that the majority of students at this class are women. F M F F M F F F F M M F F F M F F F F M M F F M...
1. Forecast demand for Year 4. a. Explain what technique you utilized to forecast your demand. b. Explain why you chose this technique over others. Year 3 Year 1 Year 2 Actual Actual Actual Forecast Forecast Forecast Demand Demand Demand Week 1 52 57 63 55 66 77 Week 2 49 58 68 69 75 65 Week 3 47 50 58 65 80 74 Week 4 60 53 58 55 78 67 57 Week 5 49 57 64 76 77...
Paste Font Alignment Number fc AVERAGE(A2:A6) A8 1 MALE FEMALE 123.4 132.8 68.8 121.8 73.8 33.1 53.1 78.1 55.5 81.2 4 104.12 MALE 83 120 98 116 76 70 98 FEMALE 85 65 69 61 62 81 81 60 70 87 82 72 19 90 76 79 71 60 78 73 64 73 78 60 68 Male Avg- 93.56 Female Avg- 69.35 1) Assume the cut-score (passing point) = 80 2) Calculate the percentage of each group passing 3) Calculate...
An experiment is conducted to determine if classes offered in an online format are as effective as classes offered in a traditional classroom setting. Students were randomly assigned to one of the two teaching methods. Data below. a. Test the claim that the standard deviations for the two groups are equal. What is the p-value of the test? b. Construct a 95% confidence interval on the difference in expected final exam scores between the two groups. Does the data support...
Use the Grouped Distribution method for the following exercise (see Self-Test 2-4 for detailed instructions), rounding each answer to the nearest whole number. Using the frequency distribution below (scores on a statistics exam taken by 80 students), determine:ion 1 of the preliminary test (scores on a statistics exam taken by 80 students), determine: 68 84 75 82 68 90 62 88 76 93 73 79 88 73 60 93 71 59 85 75 61 65 75 87 74 62 95...
Please compute your z -don't use a package State your conclusion in plain English - not just rejection 1 Students in the online class are suspicious that their schools 60 point loss may have caused their teacher to take out his frustration on the students by giving them a harder than usual exam. They manage to hack into his computer and get the following data: Scores for test Nov 2011 71 74 64 77 58 72 73 79 50 78...
Below is a sample of 25 exam scores. 80 79 69 71 74 73 77 75 65 52 81 84 84 79 70 78 62 77 68 77 88 70 75 85 84 1. Is the exam score significantly greater than 70 at the 5 percent level of significance? Follow and show the 7 steps for hypothesis testing. 2. Determine the p-value and interpret its meaning. 3. What assumption must you make about the population distribution in order to conduct...
A random sample of final examination grades for a college course follows.55 85 72 99 48 71 88 70 59 98 80 74 93 85 7482 90 71 83 60 95 77 84 73 63 72 95 79 51 8576 81 78 65 75 87 86 70 80 64Use a = 0.05 and test to determine whether a normal distribution should berejected as being representative of the population’s distribution of grades.