Hypothesis Tester - Single Sample Hypothesis Test for a Population Mean, Sigma Known If the population standard deviation is known, we can directly calculate the standard deviation of the sampling distribution (the standard error of the estimate), and use the standardized normal distribution to get a z-multiple, using the Excel function NORMSINV. We can then calculate p for each of the three possible test conditions, and compare it to each level of alpha to see whether the null hypothesis should be rejected. Inputs: Hypothesized population mean: 295 Population standard deviation (sigma): 12 Sample size (n): 50 Sample mean (x-bar): 297.6 <-- Input the appropriate number for your situation. <-- Input the appropriate number for your situation. <-- Input the appropriate number for your situation. <-- Input the appropriate number for your situation. Intermediate Calculations: Standard error of the estimate: 1.69705627 Test statistic (z): 1.53206469 Results: One-tailed, H0: Mu =>295, p= 0.9372 For the Alpha level given, H0 should be: 0.01 Not rejected One-tailed, H0: Mu <=295, p= 0.0628 Not rejected Two-tailed, H0: Mu = 295, p = 0.1255 Not rejected Alpha: End of worksheet ate number for your situation. ate number for your situation. ate number for your situation. ate number for your situation. he Alpha level given, H0 should be: 0.05 Not rejected 0.1 Not rejected Not rejected Rejected Not rejected Not rejected Hypothesis Tester - Single Sample Hypothesis Test for a Population Mean, Sigma Unknown If the population standard deviation is not known, we must use the sample standard deviation as an estimate and use it to calculate the standard deviation of the sampling distribution (the standard error of the estimate). We also use the t-distribution to get a multiple corresponding to the desired confidence level, using the Excel function TINV. We can then calculate p for each of the three possible test conditions, and compare it to each level of alpha to see whether the null hypothesis should be rejected. Inputs: Hypothesized population mean: 7 Sample standard deviation (s): 1.05 Sample size (n): 60 Sample mean (x-bar): 7.25 <-- Input the appropriate number for your situation. <-- Input the appropriate number for your situation. <-- Input the appropriate number for your situation. <-- Input the appropriate number for your situation. Intermediate Calculations: Standard error of the estimate: 0.1355544 Test statistic (t): 1.8442778 Degrees of freedom (d.f.): 59 Results: One-tailed, H0: Mu =>7, p= 0.9649 End of worksheet For the Alpha level given, H0 should be: Alpha: 0.01 0.05 Not rejected Not rejected One tailed, H0: Mu <=7, p= 0.0351 Not rejected Rejected Two-tailed, H0: Mu = 7, p = 0.0702 Not rejected Not rejected our situation. our situation. our situation. our situation. H0 should be: 0.1 Not rejected Rejected Rejected Hypothesis Tester - Single Sample Hypothesis Test for a Population Proportion From a sample proportion we can calculate the standard deviation of the sampling distribution (the standard error of the estimate) and use the standardized normal distribution to get a z-multiple, using the Excel function NORMSINV. We can then calculate p for each of the three possible test conditions, and compare it to each level of alpha to see whether the null hypothesis should be rejected. Inputs: Hypothesized population proportion: 0.35 Sample proportion (p-bar): 0.4 Sample size (n): 30 <-- Input the appropriate number for your situation. <-- Input the appropriate number for your situation. <-- Input the appropriate number for your situation. Intermediate Calculations: Standard error of the estimate: 0.0871 Test statistic (z): 0.5741692518 Results: One tailed, H0: P => 0.35, p = 0.7171 For the Alpha level given, H0 should b 0.01 Not rejected One tailed, H0: P <= 0.35, p = 0.2829 Not rejected Two-tailed, H0: P = 0.35, p = 0.5659 Not rejected Alpha: End of worksheet te number for your situation. te number for your situation. te number for your situation. he Alpha level given, H0 should be: 0.05 Not rejected 0.1 Not rejected Not rejected Not rejected Not rejected Not rejected Mall of Elbonia Interview Results Customer 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 Gender 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 0 0 1 0 1 1 0 0 1 1 0 0 1 1 1 0 1 0 0 0 1 1 0 1 1 1 1 1 Clothing $65 $139 $186 $125 $193 $186 $60 $231 $83 $157 $185 $125 $174 $125 $110 $164 $192 $165 $123 $189 $115 $116 $188 $146 $184 $121 $169 $95 $135 $157 $163 $134 $179 $144 $180 $92 $154 $180 $168 $117 $193 $96 $118 Food $16 $17 $22 $24 $17 $20 $18 $18 $20 $20 $26 $18 $17 $26 $18 $18 $15 $28 $18 $19 $16 $27 $13 $24 $24 $19 $15 $16 $24 $26 $12 $19 $24 $21 $20 $15 $19 $26 $13 $25 $17 $24 $13 Time in Mall 51 40 231 131 20 204 61 64 84 161 148 67 231 165 59 16 174 41 337 63 57 242 12 116 46 146 152 61 71 226 48 289 81 170 53 161 34 111 53 125 101 130 108 Friendliness 8 5 3 5 3 2 4 1 5 5 7 10 6 1 9 2 3 7 5 9 8 7 8 9 9 7 4 1 1 9 7 10 6 3 10 9 7 5 6 1 10 9 4 Attractiveness 9 8 1 8 3 1 3 5 7 5 6 9 4 5 5 3 1 1 4 10 9 8 6 7 7 8 4 1 9 10 10 9 5 5 5 9 8 7 5 1 8 8 1 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 1 1 1 1 0 0 1 1 0 0 0 1 0 0 1 1 1 0 1 1 1 1 0 0 0 0 0 1 0 0 1 1 1 0 0 1 1 1 0 0 0 0 1 1 1 $214 $169 $113 $199 $64 $177 $149 $130 $186 $180 $93 $114 $195 $185 $81 $223 $164 $119 $165 $146 $87 $158 $187 $113 $159 $174 $169 $140 $126 $177 $124 $181 $124 $145 $186 $118 $141 $154 $161 $92 $183 $215 $104 $178 $155 $28 $18 $16 $26 $13 $23 $24 $15 $24 $17 $24 $15 $20 $21 $15 $29 $16 $17 $20 $19 $24 $20 $26 $12 $27 $18 $21 $17 $23 $18 $21 $17 $26 $13 $21 $24 $19 $16 $20 $26 $20 $15 $19 $21 $18 109 141 185 95 136 45 102 81 205 26 237 229 39 125 318 64 124 127 89 98 145 124 118 161 42 39 210 242 170 76 149 161 24 77 239 87 20 90 52 97 42 222 167 36 87 6 2 7 10 5 8 6 9 5 1 1 9 9 5 5 1 2 8 7 3 6 8 7 8 3 2 6 2 1 9 8 7 7 6 3 9 2 1 5 2 10 8 2 8 8 9 1 9 9 2 4 5 7 8 4 3 8 10 3 9 3 3 7 8 7 8 10 6 2 1 2 1 6 9 8 9 7 8 8 1 4 2 9 4 7 8 9 1 7 7 89 0 90 0 91 0 92 0 93 1 94 0 95 0 96 0 97 0 98 1 99 0 100 0 End of worksheet $104 $171 $172 $88 $208 $98 $174 $149 $140 $170 $177 $79 $19 $26 $17 $19 $22 $19 $21 $22 $23 $13 $27 $17 213 74 235 56 67 215 46 333 39 39 57 235 6 10 8 1 9 1 1 10 5 4 1 6 7 8 1 3 9 1 1 7 1 5 3 8 Running head: HYPOTHESIS TESTING 2 Hypothesis Testing 2 Michelle Mc Donald- Upshaw MBA-FP6018 December 13, 2016 Professor: Jeffrey Edwards 1 HYPOTHESIS TESTING 2 2 Hypothesis Testing 2 Practical Application Scenario Food Court Owners, Recently convenience interviews were conducted in the food court of one hundred customers. The survey related to the money and time that customers spend at the mall. The information compiled produced the following results: Number of male customers 51 Number of female customers 49 Total average spent on clothes $148.71 Average males spent on clothes $145.29 Average females spent on clothes $152.27 Total average spent on food $19.93 Average males spent on food $19.91 Average females spent on food $19.96 Total average time spent in the mall 120.03 min Average time males spent in the mall 110.07 min Average time females spent in the mall 130.41min Total average friendliness of the mall 5.59 Average friendliness ranking by males 6.00 Average friendliness ranking by females 5.16 Total average attractiveness of the mall 5.65 Average attractiveness ranking males 6.41 HYPOTHESIS TESTING 2 3 Average attractiveness ranking females 4.86 The data shows that females spent more money on clothes, food, and time in the mall. While their male counterparts found the mall to be more friendly and attractive. In a previous study in 2011 it was concluded that food purchase was at an average of $18.75 per visit. However, in this survey it shows an average increase of $19.93 per visit. This survey was taken from a single hypothesis where the sigma is known. The current survey of one hundred customers and the known sigma produced the following results: Null Hypothesis: The number of customers who spent an average of $18.75 per visit (H0: = $18.75). Alternative Hypothesis: The number of customers who spent an average that was more than $18.75 per visit (Ha: = $18.75). The following test statistics was used in this hypothesis test: Z = n (Xbar-)/s (used mainly in large samples) Xbar = the sample mean, s = sample deviation & n= sample size. Xbar= 20, s=4 and n=100 Z=100 (20 - 18.75)/4 = Z = 3.125 At = 0.05, Z 0.05 = 1.65 The null hypothesis was rejected because Z > Z 0.05 Inputs: Hypothesized population mean: Population standard deviation (sigma): Sample size (n): Sample mean (x-bar): 18.75 4 100 20 HYPOTHESIS TESTING 2 4 Intermediate Calculations: Standard error of the estimate: 0.4 Test statistic (z): 3.125 Results: Alpha : For the Alpha leve One-tailed, H0: Mu =>18.75, p= 0.9988 0.01 Not rejected 0.05 Not rejected One-tailed, H0: Mu <=18.75, p= 0.0012 Rejected Rejected Two-tailed, H0: Mu = 18.75, p = 0.0023 Rejected Rejected The sample was analyzed to determine the null hypothesis which is the average price spent by customers in the food court is less than or equal to $18.75. The equation is demonstrated as H0: P <= $18.75. The alternate hypothesis was the average price spent by customers in the food court is greater than or equal to $18.75. The equation is H1: P=> to $18.75. The data concludes that the p- value =0.0012. Therefore, the null hypothesis can be rejected because the pvalue is less than the significance level. Conclusion A null hypothesis testing enables a researcher to determine the probability of change within a population. In this test the null hypothesis is rejected as the p value is less than .05 providing evidence against the null hypothesis. Statistically the information proves that the average amount of food court customers remained constant as the p value is less than the alpha value. If the Alpha was .01 the null hypothesis would still reject as the p value of 0.0012 is less HYPOTHESIS TESTING 2 5 than .01. Even if the Alpha was .10 the null hypothesis will still reject as the data is 90% accurate. With the information obtained from the survey it statically proves that there has been an increase in the amount of money spent on food purchase at the mall from the last survey in 2011. The recommendation to the food court owners is to continue to introduce variety and capitalize on the areas that female mall patrons find less appealing. These areas include the attractiveness and friendliness of the mall. This recommendation is based on the knowledge that female patrons spend the most money on food and once these areas improve there will be a continuous increase in food purchase per visit. Respectfully Submitted, Food Court Manager