Data Analysis: Foodland Supermarkets - Speech or Presentation Example

BU1007 BUSINESS DATA ANALYSIS Assignment This assignment consists of 3 questions given below (Answer all questions) Instructions Answers can be hand written or typed. 2. Use formulae, calculator and or / Excel as you see fit to do the actual calculations. 3. Keep Three decimal places for all your answers 4. You must show your complete work to receive full/partial credit for all the problems 5. Do not use red pen. The examiner will use this color for marking Answer all of the following three questions. Question 1. (10 Marks) (i)A study being undertaken to determine if the new cash registers installed at Foodland Supermarkets have sped up the checkout process. In the past the mean time spent in a checkout line was =11 minutes with standard deviation =3 minutes. A random sample of 100 customers is observed. If is in fact still 11, find the following results if the sample: a. exceeds 13 Z = = The probability that exceeds 13 will be about 0. b is less than 9 Z = = The probability that is less than 9 will be about 0. (ii) A government agency wants to estimate the mean amount of money spent per week for gasoline by truck drivers. A sample of 400 drivers is taken. If the population standard deviation is, what is the probability that the mean will be within $6 of the true mean ? – = ±$6, = $40, n = 400 Z = = Z = = Pr{} = 0.99865 – 0.00135 = 0.997 The probability that the mean will be within $6 of the true mean is 0.997. (iii) During the past twenty years, the average number of days of rainfall annually in a city was and . Find a 95% confidence interval of the population mean . , , n = 20 The population standard deviation is not known, therefore using t-distribution. Degrees of freedom, ν = 20 – 1 = 19 With degrees of freedom 19, the 95% confidence level critical vale of t is 2.093. 95% confidence interval of the population mean is = = (142.320, 151.680) A 95% confidence interval of the mean number of days of rainfall annually is 142.32mm to 151.68mm. (iv) In a random sample of 100 students at a particular university, 60 indicated that they favoured having the option of receiving pass-fail grade for elective courses. Obtain a 90% confidence interval for the proportion of the population of students who favour pass-fail grades for elective courses. Does this confidence interval contain the value? Explain why this particular value might be of interest. x = 60, n = 100 The sample proportion is p = x/n = 60/100 = 0.6 , for 90 percent confidence The 90 percent confidence interval for the proportion is = = (0.519, 0.681) A 90% confidence interval for the proportion of the population of students who favour pass-fail grades for elective courses is 0.519 to 0.681. This confidence interval does not contain the value. This suggests that more than 50% of the students favour pass-fail grades for elective courses. Question 2. (10 Marks) (i) Suppose 16 tax returns are randomly sampled by the Australian Taxation Office from the population of 1987 tax returns with adjusted gross income between $25000 and $30000. The interest deductions (in $) claimed on the returns are as follows: 2984 3050 3101 3415 2416 2910 3333 3002 1897 3872 3102 3222 2134 2806 2851 2999 In 1980 the average interest deductions for individuals in this tax bracket was $3011. Do the sample data provide sufficient evidence to conclude that in 1987 the average interest deduction claimed by taxpayers in this income bracket was different from 1980? Use. State all the assumptions you need to perform the test. Assuming that the sample comes from a normal population. Since, the population standard deviation is unknown, therefore using t-distribution to perform the test. The hypothesis test will one sample mean vs. hypothesized value two tail test. The null and alternate hypotheses are The significance level, α = 0.05. The degrees of freedom, ν = n – 1 = 16 – 1 = 15 At significance level of 0.05 with degrees of freedom of 15, the critical values of t are ±2.131. Therefore, the decision rule is Reject , if t > 2.131 or t < -2.131. Otherwise do not reject . Using Excel, the mean and sample standard deviation is 2,943.375 and 478.916, respectively. The test statistic is p-value (two-tailed) = 0.581 The test statistic -0.565 fall in-between non-rejection region. Therefore, do not reject . The sample data do not provide sufficient evidence to conclude that in 1987 the average interest deduction claimed by taxpayers in this income bracket was different from 1980. (ii) An economist claims that the unemployment rate for non-English speaking individuals in New York City is at least 30%. In a random sample of 400 non-English speaking residents of New York City, 90 are unemployed. Test against using 5% level of significance. x = 90, n = 400, p0 = 0.30 p = 90/400 = 0.225 The null and alternate hypotheses are The significance level, α = 0.05. The test will be hypothesis test for proportion vs. hypothesized value left tail test. At significance level of 0.05, the left tail critical value of z is -1.645. Therefore, the decision rule is Reject, if z < -1.645. Otherwise do not reject . The test statistic is = p-value (one-tailed, lower) = 0.0005 The test statistic -3.273 is less than critical value -1.645. Therefore, reject . The sample data provide sufficient evidence to conclude that the economist claims that the unemployment rate for non-English speaking individuals in New York City is at least 30 is true. (iii) Twenty babies are randomly selected for an experiment to determine if different brands of baby food affect a child’s weight. The babies are separated into 5 groups, and each group is fed a different diet. The weight gained in pounds by each baby after being on the diet four months is recorded in the following table: Baby Type of Diet 1 2 3 4 5 1 10 9 12 6 11 2 11 6 11 8 9 3 7 8 8 9 10 4 8 7 10 8 7 Use a 1% level of significance level to test if the type of diet affects weight gain. Clearly state the null and alternative hypotheses and state all the necessary assumptions to perform the test.(Use Excel to analyse the data) The null and alternate hypotheses are At least one mean is different from other. The significance level, α = 0.01. The test will be an Analysis of Variance (ANOVA) test. The assumptions for an Analysis of Variance (ANOVA) are: a) observations for the sample must be independent and normally distributed b) homogeneity of variance (equal variances across all the groups). There are c = 5 groups and n = 20 observations, so degrees of freedom for the F test are Numerator: d.f.1 = c − 1 = 5 − 1 = 4 (between treatments, factor) Denominator: d.f.2 = n − c = 20 − 5 = 15 (within treatments, error) At significance level of 0.01, the critical value of F is F4,15 = 4.893. Therefore, the decision rule is Reject, if F > 4.893. Otherwise do not reject . Using Excel for the calculations, results are given in below table. ANOVA: Single Factor SUMMARY Groups Count Sum Average Variance Group 1 4 36 9 3.333 Group 2 4 30 7.5 1.667 Group 3 4 41 10.25 2.917 Group 4 4 31 7.75 1.583 Group 5 4 37 9.25 2.917 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 20.5 4 5.125 2.064 0.137 4.893 Within Groups 37.25 15 2.483 Total 57.75 19         The test statistic is F = 2.064. The test statistic 2.064 is less than critical value 4.893. Therefore, do not reject . The sample data do not provide sufficient evidence to conclude that the type of diet affects weight gain. Question 3. (10 Marks) (i) An economist is interested in examining the relationship between an individual’s annual income in thousands of dollars (X) and the individual’s annual income tax payments in thousands of dollars (Y). The data are as follows: Y: 3.1 6.0 3.3 4.2 7.7 11.0 8.1 6.0 1.0 5.0 X: 20 40 27 36 54 90 74 60 18 45 Estimate the linear regression model. The income tax system is supposedly a progressive system: As income increases, people pay a higher proportion of their income in taxes. Do these data support the hypothesis that the tax system is progressive? Perform a suitable statistical test at 5% level of significance. The null and alternate hypotheses are The significance level, α = 0.05. As the income tax is supposedly a progressive system, therefore a right-tailed t-test for zero slope will be performed. The degrees of freedom, ν = n – 2 = 10 – 2 = 8. At the significance level of 0.05, the critical value of t with degrees of freedom of 8 for right tailed test is 1.860. Therefore, the decision rule is Reject, if t > 1.860. Otherwise do not reject . Using Excel, the test statistic is SUMMARY OUTPUT Regression Statistics Multiple R 0.952 R Square 0.907 Adjusted R Square 0.895 Standard Error 0.935 Observations 10 ANOVA   df SS MS F Significance F Regression 1 68.133 68.133 77.963 0.000 Residual 8 6.991 0.874 Total 9 75.124   Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 0.089 0.684 0.130 0.900 -1.490 1.667 X 0.117 0.013 8.830 0.000 0.087 0.148 The test statistic 8.830 is greater than critical value 1.860. Therefore, reject . The higher F statistic (p < .001) also shows that the overall regression model is significant. The sample data provide sufficient evidence to conclude that the tax system is progressive. The regression model is given by: Annual Income Tax Payment = 0.089 + 0.117 (Annual Income) (ii) A random sample of 62 job candidates were ranked on an interview and again on a written test. The sample rank correlation between the two rankings is . Using , test the null hypothesis against the alternative, . The null and alternate hypotheses are The significance level, α = 0.05. For a right-tailed test (Spearman rank correlation test) using ν = n − 1 = 62 − 1 = 61 degrees of freedom, the critical value of t is 1.670. Therefore, the decision rule is Reject, if t > 1.670. Otherwise do not reject . The test statistic is =1.581 The test statistic 1.581 is less than critical value 1.670. Therefore, do not reject. In conclusion, there is no relationship between the two rankings (interview and written test). Read More