UNIVERSITI UTARA MALAYSIA SESSION 2023/2024 (A231) SQQS3113 MULTIVARIATE ANALYSIS INDIVIDUAL ASSIGNMENT 2 (10%) PREPARED FOR: SIR BAHTIAR JAMILI BIN ZAINI PREPARED BY NAME MATRIC NO. KHAIRINA BINTI KHAIRUDDIN 288221 SUBMISSION DATE: 15 December 2023
Question 1: (10 MARKS) Perspiration from 20 healthy females was analyzed. Three components, X1 = sweat rate, X2 = sodium content, and X3 = potassium content, were measured, and the results, which we call the sweat data, are presented in Table below: Test the hypothesis H0: µ’ = [4, 50, 10] against H1: µ’ = [4, 50, 10] at level significance α = 0.10. Sweating is a vital natural mechanism that assists in regulating body temperature. It involves the release of a saline fluid from sweat glands, commonly known as perspiration. The evaluation of three key components plays a crucial role: 1. Sweat rate 2. Sodium 3. Potassium
Objective: The hypothesis of interest is that women meet healthy females with the condition. This null hypothesis would be rejected if women fail to meet the good condition on any one or more to be healthy females. In mathematical notation, the null hypothesis is the population mean vector, µ equals the hypothesised mean vector µ as shown below: 0 : 0 µ = µ0 From the question, the H0: µ’ = [4, 50, 10] and H1: µ’ ≠ [4, 50, 10] Descriptive statistics : Descriptive statistics are tools used to summarise and explain data in a simple way. They include things like averages, ranges, and graphs that help people quickly understand the main points of a set of information. The goal is to provide a clear and straightforward overview without making predictions about a larger group by using R programming. The summary From the summary, notice that the standard deviations of potassium are large relative to their respective means. This indicated a high variability of sodium content produced in sweat perspiration
The variance The sample variance was given by the diagonal elements. Therefore, the covariance for all there are ● Covariance between sweat rate 11 = 2. 879368 ● Covariance between sodium is 22 = 199. 7884 ● Covariance between potassium 33 = 3. 637658 We can see that the highest is sodium with 199.7884. Therefore it is the high variability of sodium content produced in sweat perspiration to healthy females. The correlations From the sample correlations, here we can see that the correlation between each of the variables and themselves are all equal to one, and the off-diagonal elements give the correlation between each of the pairs of variables. Generally, we look for the strongest correlation first. The result above suggests that sweat rate and sodium are positively associated while if it is related to the potassium it will be all negatively associated. Another method to measure association is the coefficient of determination by simply equal to the square of the correlation. For example, the coefficient of determination between sodium and sweat rate is 2 = (0. 4173499) 2 = 0. 174180939 This says that about 17.42% of the sodium is explained by the variation in the sweat rate.
STEP 1 => IDENTIFY THE HYPOTHESIS TESTING , 0 1 : 0 µ = µ0 : 1 µ ≠ µ0 Our null hypothesis is rom the question, the H0: µ’ = [4, 50, 10] and H1: µ’ ≠ [4, 50, 10] The recommended requirement to healthy and sample mean are given below VARIABLE RECOMMENDATION NEED MEAN Sweat rate . x1 4 4.64 Sodium . x2 50 45.50 Potassium . x3 10 9.97
STEP 2 => FIND THE TEST VALUE The formula for hotelling’s 2 The answer hotelling’s T-square (test value) is 2 = 9. 738773
STEP 3 => FIND THE CRITICAL VALUE OR find the critical value by using formula refer to the table F distribution at α = 0. 10 (20−1)3 20−3 3,20−3 = 57 12 3,17 = (3. 3529)(2. 4374) = 8. 17245 Therefore, we can see that for an 0.10 level test, the critical value is approximately 8.17245 CV = 8.17245
STEP 4 => COMPARE When the test value of 9. 738773 is greater than critical value 8.17245, we can say that to reject the null hypothesis that the average recommendation to get the healthy females STEP 5 => MAKE CONCLUSION For all healthy females, the average recommendation to get healthy females does not meet. Returning to the table of sample mean and recommendation need, it appears that females fail to meet the good condition because of the sodium and potassium , only exceed the sweat rate
Question 2: (10 MARKS) In the first phase of a study of the cost of transporting milk from farms to dairy plants, a survey was taken of firms engaged in milk transportation. Cost on x1 = fuel, x2 = repair, and x3 = capital, all measured on a per-mile basis, are presented in Table 1 for n1 = 36 gasoline and n2 = 23 diesel trucks. Test for the differences in the mean cost vectors. Use α = 0.01. From the equation, it is the problem of the cost of transporting milk from farm to dairy plant that wants to compare two population meanings which are gasoline and diesel trucks. Let ~ and represent the distribution of the two population ( µ1 , ∑ ) ~ ( µ2 , ∑ ) On the basic of independent samples on x and 1 , 2 , . . . . . . . . . , 1 on y, we wish to test the hypothesis 1 , 2 , . . . . . . . . . , 2 Objective: To determine is that the two trucks are the same or different in the cost of transporting milk from farm to dairy plant. This is essential for the company to calculate the cost of the fuel, repair and capital to minimise the total expense of transportation.
STEP 1 => IDENTIFY THE HYPOTHESIS TEST 0 1 Consider testing the null hypothesis that the two populations have identical population mean vectors. 0 : µ1 = µ2 1 : µ1 ≠ µ2 The null hypothesis is satisfied if and only if the population means are identical for all of the variables. The alternative is that at least one pair of these means is different. This is expressed below
STEP 2 => FIND THE TEST VALUE The formula for hotelling’s 2 The sample covariance matrices for the two trucks are Therefore, the pooled estimate of the covariance matrix Σ is given by The hotelling’s test statistic is defined by 2 The answer hotelling’s T-square (test value) is 2 = 50. 91279
STEP 3 => FIND THE CRITICAL VALUE Therefore the critical value is 12.93096 = 12. 93096 STEP 4 => COMPARE The test value of 50. 91279 is greater than critical value of 12. 93096, therefore we can reject the null hypothesis that the mean vector for the same cost. STEP 5 => MAKE CONCLUSION Since test value > critical value , therefore we reject the null hypothesis. It shows that there are different costs shown by two modes of transportation which are diesel and gasoline trucks.