Questions and Answers #5 Data Distributions

Questions and Answers Sheet 5 Data Distributions Question #1 We are interested in significant differences (level 𝛼 = 0.05 ) between the expected values 𝜇1 and 𝜇2 of two populations. Which of the following statements on statistical tests are correct? a) We will formulate the null hypothesis as 𝜇1 = 𝜇2 . b) A t-test can always be applied in this situation. c) A p-value is the probability that the null hypothesis is correct, given the observed data. d) If we obtain a p-value of 0.04, we will reject (level 𝛼 = 0.05 ) the null hypothesis. Answer: Step 1 (a) p-value doesn't tells us about alternative hypothesis and the null hypothesis so it is not easy to say by looking into only p-value and decide about the null and alternative hypothesis statement. But here we are going to check the difference between the population mean so null hypothesis can be this type. So option A is correct. (b) P-value also fail to State that which type of test statistic is used.generally of the population is normal but the population standard deviation is not known to us then t-statistic is used. So statement B is incorrect. Step 2 (c) P- value is the evidence against the null hypothesis.lower the p-value greater the chance of rejecting the null hypothesis. In other words p-value is the probability that null hypothesis is correct. high the pvalue higher the chance of accepting the null hypothesis. So statement C is correct. (d) if P-value < level of significance We reject the null hypothesis. If P-value > level of significance. We accept the null hypothesis. Here P-value (0.04) < level of significance(0.05) Hence we reject the null hypothesis. So statement D is correct. Hence the correct choice are A, C and D Question #2 Given the following data: 6.61 6.65 8.27 7.69 4.70 6.89 7.61 5.93 7.50 8.74 5.96 7.25 6.99 7.92 7.44 5.75 7.26 6.86 7.48 5.52 6.36 6.94 5.38 6.41 5.10 7.47 5.20 7.70 8.88 5.86 Obtain: A) The frequency distribution table. B) The variance of a population Note: Class width = 4.2 Answer: 7.56 6.79 5.48 6.67 7.49 8.04 Step 1 A) From the given data values, The minimum value is 4.70 and the maximum value is 8.88. Class width is 4.2 Let us consider the number of classes as 3. Therefore, the frequency distribution is, class interval 0 − 4.2 4.2 − 8.4 8.4 − 12.6 frequency 0 34 2 36 Step 2 B) Variance of a population: class interval Midpoint , x 0 − 4.2 2.1 4.2 − 8.4 6.3 8.4 − 12.6 10.5 frequency 0 34 2 N =  f = 36  x f 0 214.2 21 f  x = 235.2  x2  f 0 1349.46 220.5 f  x 2 = 1569.96 Therefore, Population variance = [∑ 𝑓 ⋅ 𝑥 2 − (∑ 𝑓 ⋅ 𝑥)2/𝑁]/𝑁 = [1569.96 − 235.22/36]/36 = 0.9256 Question #3 The display provided from technology available below results from using data for a smartphone carrier's data speeds at airports to test the claim that they are from a population having a mean less than 6.00 Mbps. Conduct the hypothesis test using these results. Use a 0.05 significance level. 1. Identify the null and alternative hypotheses, 2. Identify the test statistic, 3. Identify the P-value, 4. State the final conclusion that addresses the original claim. 𝜇 < 6.00 𝑡 = −0.295669 𝑝 = 0.384526 𝑥 = 5.75 𝑆𝑥 = 5.347659 𝑛 = 40 Answer: Step 1 1. The null and alternative hypothesis is obtained below: The claim of test is the population having a mean less than 6.00 mbps. Here, less than represent the one tail test. Null hypothesis: 𝐻0 : 𝜇 = 6.00 Alternative hypothesis: 𝐻1 : 𝜇 < 6.00 Step 2 2. The test statistics: From the display provided from technology results, the value of test statistics (t) is −0.295669. Thus, the value of test statistics is −0.295669. Step 3 3. The P- value: From the display provided from technology results, the P-value (p) is 0.384526. Thus, the value of P-value is 0.384526. Step 4 4. Conclusion: Use the level of significance 𝛼 = 0.05 . The P-value (0.384526) is greater than level of significance (0.05). By the decision rule, we should fail to reject (𝐻0 ) . Therefore, the sample provided not sufficient evidence that the population having a mean less than 6.00 mbps. Question #4 A study tested the claim that heights of Americans and heights of Austrians have different variances with 𝑠 = 7.48092 cm for Americans and 7.14756 cm for Austrians. The sample sizes are 𝑛1 = 145 and 𝑛2 = 157 . When using the F test with these data, is it correct to reason that there is no need to check for normalcy because n1 is greater than 30, and n2 is greater than 30? Answer: Step 1 It is given that the sample variances are 7.48092 cm and 7.14756 cm with the corresponding to sample sizes are 145 and 157, respectively. Step 2 The assumptions of F test is that the distribution are randomly and normally distributed. Now, either the distribution should be symmetric or the sample size should be large enough to have the normal distribution. In this case, both of the samples are greater than 30 or large enough. Thus, it is correct reason that there is no need to check for normality because n1 is greater than 30, and n2 is greater than 30. Question #5 A local manufacturer of copying equipment includes a repair warranty with each copier. Virtually all of his equipment exhibit an increasing failure rate. Based on historical repair data, the failure rate for Model 25cc7 is accurately described by the function 𝑟(𝑡) = 2.7786𝑡 1.3 , where t is measured in months of continuous operation. What is the probability that the time between two successive failures of this equipment exceeds two months of operation? Answer: Step 1 𝑟(𝑡) = 2.7786𝑡 1.3 𝛼𝛽𝑡 𝛽−1 = 2.7786𝑡 1.3 𝛽 − 1 = 1.3 𝛽 = 2.3 2.7786 𝛼= 2.3 = 1.2081 Step 2 The probability that the time between two successive failures of this equipment exceeds two months of operation is calculated as follows: 2 𝑃 (𝑇 > ) = 𝑃(𝑇 > 0.1667) 12 𝛽 = 𝑒 −𝛼𝑡 = 𝑒 −1.2081𝑡 = 0.9806 2.3 Question #6 Based on an analysis of sample data, an article proposed the pdf 𝑓(𝑥) = 0.65𝑒 −0.65(𝑥−1) when 𝑥 ≥ 1 as a model for the distribution of 𝑋 = time (sec) spent at the median line. (a) What is the probability that waiting time is at most 4 sec? More than 4 sec? at most 4 sec 𝑃(𝑋 ≤ 4) =? more than 4 sec 𝑃(𝑋 > 4) =? (b) What is the probability that waiting time is between 3 and 7 sec? Answer: Step 1 The pdf of X is given below: 0.65e −0.65( x −1) f ( x) =  0 , x 1 , otherwise a) The probability that waiting time is at most 4 sec is given below: 4 P ( x  4) =  0.65e −0.65( x −1) dx = 1 −0.65(𝑥−1) 4 [−𝑒 ]1 −0.65(𝑥−1)4 1 = −𝑒 = −𝑒 −0.65(4−1) − (𝑒 −0.65(1−1) ) = 1 − 0.1423 = 0.8577 The probability that waiting time is more than 4 sec is given by: 𝑃(𝑥 > 4) = 1 − 𝑃(𝑥 ≤ 4) = 1 − 0.8577 = 0.1423 Step 2 b) The probability that waiting time is at most 4 sec is given below: 7 𝑃(𝑥 ≤ 4) = ∫3 0.65𝑒 −0.65(𝑥−1) 𝑑𝑥 7 = [−𝑒 −0.65(𝑥−1) ]3 = −𝑒 −0.65(7−1) − (−𝑒 −0.65(3−1) ) = 0.2725 − 0.0202 = 0.2523 Therefore, the indicated probability is 0.2523. Question #7 The recommended daily dietary allowance for zinc among males older than age 50 years is 15𝑚𝑔/𝑑𝑎𝑦. An article reports the following summary data on intake for a sample of males age 65 − 74 years: 𝑛 = 117, 𝑥 = 12.1, and 𝑠 = 6.98 . Does this data indicate that average daily zinc intake in the population of all males age 65-74 falls below the recommended allowance? (Use 𝛼 = 0.05 .) State the appropriate null and alternative hypotheses. 1) 𝐻0 : 𝜇 = 15 𝐻𝑎 : 𝜇 > 15 2) 𝐻0 : 𝜇 = 15 𝐻𝑎 : 𝜇 ≤ 15 3) 𝐻0 : 𝜇 = 15 𝐻𝑎 : 𝜇 < 15 4) 𝐻0 : 𝜇 = 15 𝐻𝑎 : 𝜇 ≠ 15 Answer: Step 1 Given, sample size (𝑛) = 117 sample mean (𝑥) = 12.1 standard deviation (𝑠) = 6.98 𝛼 = 0.05 Step 2 𝐻0 : 𝜇 = 15 𝐻𝑎 : 𝜇 < 15 Question #8 The following sample data represents the age of 30 lottery winners. 18 27 29 32 34 35 37 38 43 46 48 50 51 52 54 57 57 58 59 60 62 63 67 67 69 71 72 72 75 81 a) Estimate the sample standard deviation of this data using the formula 𝑆 = 𝑅4𝑆 = 𝑅4 where SS is the standard deviation and RR is the range. b) Use Excel to get the sample standard deviation. Answer: Step 1 a) The value of range is obtained below: From the given information, the minimum value of the data set is = 18 , the maximum value of the data set is = 81 . The range formula is given below: Range = Maximum value − Minimum value = 81 − 18 = 63 Thus, the value of the range is 63. Step 2 𝑅 By using the formula 𝑠 = to find the sample standard deviation is given below: 𝑆= = 63 𝑅 4 4 4 = 15.75 Thus, the sample standard deviation is 15.75. b) By using the Excel formula to get the sample standard deviation is given below. First need enter the given data into the excel spreadsheet. The excel formula for calculate the sample standard deviation is = STDEV (number 1, number 2…) The sample standard deviation is: = 𝑆𝑇𝐷𝐸𝑉(18,27,29 … ) Thus, the sample standard deviation (s) is 16.1425. Question #9 A recent claim has been made that people who have iPads spend more time on their iPads than people who have laptops spend on their laptops. After all calculations were performed, the study notes the z statistic to be -2.81 compared to significance level of 𝛼 = 0.01 . Is there truly a significant difference between the two data sets? Since, p-value ? the significance level, 𝛼 , we can conclude that there ? a truly significant difference between the two data sets. Answer: Step 1 Test is that whether there is truly significant difference between the two data sets. The test statistic is 𝑍 = −2.81 Level of significance = 0.01 Step 2 The p-value for 𝑍 = −2.81 and for two-tailed test is defined as P-value = 2 × 𝑃(𝑍 < −2.81) = 0.0050 (from standard normal table) Since p-value is less than the significance level, we can conclude that there is a truly significant difference between the two data sets Question #10 The table shows data for bus arrivals at one stop. By how much time can you expect a bus to be late? Minutes Late 0 1 2 3 4 5 Number of Days 8 7 3 4 2 1 time=? min Answer: Step 1: Given: A data table regarding bus timings. Minutes Late 0 1 2 3 4 5 Number of Days 8 7 3 4 2 1 To find : Expected minutes of late Step 2: Let x be the number of minutes late. x 0 1 2 3 4 5 TOTAL frequency ( f ) 8 7 3 4 2 1 25 f x 0 7 6 12 8 5 38 Step 3: The expected number of late minutes is obtained by, Expected number of late minutes = Expected number of late minutes = ∑ 𝑓×𝑥 ∑𝑓 38 25 Expected number of late minutes = 1.52 minutes Thus, the expected minutes of late is 1.52 minutes Question #11 For two data sets, X and Y, if the 𝐶𝑉𝑥 > 𝐶𝑉𝑦 then - The dataset X is more consistent than Y - The dataset X an Y are equally consistent - There is no cinsistency between X and Y - The dataset Y is more consistent than X Answer: Step 1 Coefficient of variation CV is used for comparing the variability of two datasets. Step 2 A dataset X is said to be more consistent than dataset Y if 𝐶𝑉𝑥 < 𝐶𝑉𝑦 . Since 𝐶𝑉𝑋 > 𝐶𝑉𝑌 then the dataset Y is more consistent than dataset X. So the correct choice is The dataset 6 is more consistent than X. Question #12 During the first 13 weeks of the television season, the Saturday evening 8:00 P.M. to 9:00 P.M. audience proportions were recorded as ABC 29%, CBS 28% , NBC 24% , and independents 19%. A sample of 300 homes two weeks after a Saturday night schedule revision yielded the following viewing audience data: ABC 95 homes, CBS 64 homes, NBC81 homes, and independents 60 homes. Test with 𝑎 = 0.05 to determine whether the viewing audience proportions changed. Round your answers to two decimal places. Test statistic = Answer: Given data, - Total number of house for sample are 300. - ABC 95 homes, CBS 64 homes, NBC 81 homes, and independents 60 homes. Test with 𝑎 = 0.05 . - The Saturday evening 8:00 P.M. to 9:00 P.M. audience proportions were recorded as ABC 29%, CBS 28%, NBC 24% , and independents 19% Step 1 Audience proportions were recorded as ABC 29% , CBS 28% , NBC 24% , and independents 19%. Taking, Null Hypothesis 𝐻0 : 𝑃𝐴𝐵𝐶 = 0.29, 𝑃𝐶𝐵𝑆 = 0.28, 𝑃𝑁𝐵𝐶 = 0.24 and 𝑃𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 = 0.19 Alternative Htpothesis: The population proportion are not, 𝐻1 : 𝑃𝐴𝐵𝐶 = 0.29, 𝑃𝐶𝐵𝑆 = 0.28, 𝑃𝑁𝐵𝐶 = 0.24 and 𝑃𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 = 0.19 Here, Sample size is 100. Step 2 Category ABC CBS NBC Independent Proportion Observated frequency(f) Expected frequency(e) 0.29 0.28 0.24 0.19 1 95 64 81 60 300 0.29  300 = 87 0.28  300 = 84 0.24  300 = 72 0.19  300 = 57 f − e ( f − e) 2 8 −20 9 3 64 400 81 9 ( f − e) 2 e 1.73 4.76 1.125 0.158 7.773 Test statistics: 𝑥2 = ∑ (𝑓−𝑒)2 𝑒 = 7.77 Question #13 There is a fire sensor system in a building. In long series of experiments, it was found that the accuracy of the fire sensor system can be described as follows: - In the case of a fire, the sensor system will be activated with probability 0.97. - In a night without fire, the sensor system may be activated with probability 0.01. The probability of a fire in this specific area is founded 2 397 from the past data. The fire sensor system is activated. Find the probability that there was actually a fire. Answer: Step 1 Given Information : In the case of a fire, the sensor system will be activated with probability 0.97 In a night without fire, the sensor system may be activated with probability 0.01. 2 The probability of a fire in this specific area is founded = = 0.005 397 First we will draw a probability contingency table with all the information provided: There is fire Sensor system is activate Sensor system is not activate Total 0.97  0.005 = 0.0049 0.005 − 0.0049 = 0.0001 2 = 0.005 397 1 − 0.005 = 0.995 1 There is no fire 0.01  (1 − 0.005) = 0.00995 Total 0.01485 0.995 − 0.00995 = 0.98505 0.98515 Step 2 Now, if The fire sensor system is activated. Find the probability that there was actually a fire We will use the principle of conditional probability: Sensor system is activated There is fire 0.97  0.005 = 0.0049 There is no fire 0.01  (1 − 0.005) = 0.00995 Total 0.01485 𝑃(There is fire|sensor system is activated) = 𝑃(There is fire∩sensor system is activated) P(sensor system is activated) 𝑃(There is fire|sensor system is activated) = 0.0049 0.01485 = 0.329 probability that there was actually a fire = 0.329 Question #14 A data set contains the observations 7,4,2,3,1. Find (∑ 𝑥)2 . Answer: Step 1 Data are provided, 7,4,2,3,1 Step 2 To find (∑ 𝑥)2 ∑ 𝑥 = Sum of all observations (∑ 𝑥) = (7 + 4 + 2 + 3 + 1) = 17 (∑ 𝑥)2 = (17)2 = 289 Therefore, (∑ 𝑥)2 = 289 Question #15 For healthy adults, the mean pH level is 𝜇 = 7.4 , a new drug for arthritis has been developed. However, it is throught that this drug may change blood pH. A random sample of 33 patients with arthritis took the drug for three months. Blood tests showed that the sample mean from these 33 patients was a pH of 8.0 and it is known that 𝜎 = 2.3 . (Note, there is no unit for this type of data). Use 𝛼 = 0.05 level of significance to test the claim that the drug has increased the mean pH level of the blood. If we use the classical approach, what would the criterion (step 3) be? 1) −1.645 or less 2) −2.33 or less 3) 1.645 or more 4) 2.33 or more Answer: Step 1 Null and Alternate Hypothesis: 𝐻0 : 𝜇 = 7.4 𝐻𝐴 : 𝜇 > 7.4 Level of significance = 0.05 Step 2 As we know the population standard deviation, we will use z-distribution Using z-distribution Critical region for two tailed test = [1.645, ∞) Therefore, Answer: 1.645 or more Question #16 a) Calculate 𝑆𝑆𝑥𝑥 , 𝑆𝑆𝑦𝑦 , and 𝑆𝑆𝑥𝑦 (Leave no cells blank - be certain to enter ''0'' wherever required. Round your answers to 2 decimal places.) College Student Weekly Hourse Worked (X) Weekly Pay (Y) ( xi − x ) 2 11 103 16 187 16 216 16 157 32 262 x y SS xx Earnings Dollars( n = 5) ( yi − y ) 2 ( x1 − x )( yi − y ) SS yy SS xy b) )Calculate the sample correlation coefficient. Check your work by using Excel's function =CORREL(array1,array2). (Round your answer to 4 decimal places.) Answer: Step 1 Given data of College student weekly earnings Hours Worked (X) Weekly Pay (Y) 11 103 16 187 16 216 16 157 32 262 The average hours worked and average weekly pay are calculated as shown below ∑𝑋 𝑋̅ = 𝑛 11+16+16+16+32 𝑋̅ = = 18.2 5 ∑𝑌 𝑌̅ = 𝑛 103+187+216+157+262 𝑌̅ = 5 Using the average values the missing values in table are calculated as shown below College Student Weekly Earnings Dollars(n = 5) Hourse Worked (X) Weekly Pay (Y) ( xi − x )2 ( yi − y ) 2 ( x1 − x )( yi − y ) 11 103 (11 − 18.2) 2 = 51.84 (103 − 185) 2 = 6724 (11 − 18.2)(103 − 185) = 590.4 16 187 (16 − 18.2) 2 = 4.84 (187 − 185) 2 = 4 (16 − 18.2)(187 − 185) = −4.4 2 2 16 216 (16 − 18.2) = 4.84 (216 − 185) = 961 (16 − 18.2)(216 − 185) = −68.2 16 157 (16 − 18.2) 2 = 4.84 (157 − 185) 2 = 784 (16 − 18.2)(157 − 185) = 61.6 2 2 32 262 (32 − 18.2) (262 − 185) = 5929 (32 − 18.2)(262 − 185) = 1062.6 x = 18.2 y = 185 S xx = 256.8 S yy = 14402 S xy = 1642 x y SS xx SS yy SS xy Step 2 b) The correlation coefficient is calculated using the below mentioned formula 𝑟= 𝑟= 𝑆𝑥𝑦 √𝑆𝑥𝑥 𝑆𝑦𝑦 1642 √256.8×14402 = 0.853816 using the excel function CORREL(X data,Y data) we get the sample correlation coefficient as 0.853815903 Rounding to 4 decimal places we get same value of r in both methods. 𝑟 = 0.8538