In this practical application scenario, you have just been hired away from the M

Question

In this practical application scenario, you have just been hired away from the Macy's linen department buyer position by the Mall of Elbonia (MoE), as the new food court manager. Why? Rumor had it you just earned your MBA. One week ago, the MoE conducted convenience interviews in the mall's food court. One hundred interviews were gathered. The results are summarized in the document titled MoE Interviews, provided in the Updates and Handouts section.

In examining the interview results, you find that each customer is listed in a row and each column contains data on:

·         The customer's gender. If you need this data, use the coding convention: 0 for female, 1 for male.

·         How much time the customer spent in the mall.

·         How much money he or she spent on food.

·         The customer's rating of the mall's friendliness and attractiveness.

After applying the descriptive data measures, the mean is particularly telling; you believe that you can happily tell the food court owners that the average amount mall customers spend on food during a visit has increased. A 2011 study found that customers spent an average of $18.75 per visit. You realize this is a one sample test of hypothesis situation. Use the data in the MoE Interviews document and the following outlined process to be certain that your assertions are correct and that the average amount mall customers spend on food during a visit to your mall has increased.

1.        Determine the null hypothesis, via both an explanation and a math equation.

2.        Determine the alternative hypothesis, via both an explanation and a math equation.

3.        Determine whether to accept or reject the null hypothesis. You may solve the equation manually or use the Hypothesis Tester – Single Sample document provided in the Resources or by using SAS Enterprise Guide.

4.        Determine if the p-value indicates acceptance or rejection of the null. Use alpha = .05.

5.        Report the rejection or acceptance of the null in terms of the scenario results. For this scenario, write a three-sentence paragraph that details why you can be statistically confident that the average amount a food court customer spends has increased, decreased, or remained the same, and what would happen if alpha was 0.01 or 0.1.

Mall of Elbonia Interview Results Customer Gender Clothing Food Time in Mall Friendliness Attractiveness 1 1 $65 $16 51 8 9 2 0 $139 $17 40 5 8 3 0 $186 $22 231 3 1 4 1 $125 $24 131 5 8 5 1 $193 $17 20 3 3 6 0 $186 $20 204 2 1 7 1 $60 $18 61 4 3 8 1 $231 $18 64 1 5 9 1 $83 $20 84 5 7 10 0 $157 $20 161 5 5 11 1 $185 $26 148 7 6 12 1 $125 $18 67 10 9 13 1 $174 $17 231 6 4 14 0 $125 $26 165 1 5 15 0 $110 $18 59 9 5 16 0 $164 $18 16 2 3 17 0 $192 $15 174 3 1 18 1 $165 $28 41 7 1 19 0 $123 $18 337 5 4 20 1 $189 $19 63 9 10 21 1 $115 $16 57 8 9 22 0 $116 $27 242 7 8 23 0 $188 $13 12 8 6 24 1 $146 $24 116 9 7 25 1 $184 $24 46 9 7 26 0 $121 $19 146 7 8 27 0 $169 $15 152 4 4 28 1 $95 $16 61 1 1 29 1 $135 $24 71 1 9 30 1 $157 $26 226 9 10 31 0 $163 $12 48 7 10 32 1 $134 $19 289 10 9 33 0 $179 $24 81 6 5 34 0 $144 $21 170 3 5 35 0 $180 $20 53 10 5 36 1 $92 $15 161 9 9 37 1 $154 $19 34 7 8 38 0 $180 $26 111 5 7 39 1 $168 $13 53 6 5 40 1 $117 $25 125 1 1 41 1 $193 $17 101 10 8 42 1 $96 $24 130 9 8 43 1 $118 $13 108 4 1 44 1 $214 $28 109 6 9 45 1 $169 $18 141 2 1 46 1 $113 $16 185 7 9 47 1 $199 $26 95 10 9 48 0 $64 $13 136 5 2 49 0 $177 $23 45 8 4 50 1 $149 $24 102 6 5 51 1 $130 $15 81 9 7 52 0 $186 $24 205 5 8 53 0 $180 $17 26 1 4 54 0 $93 $24 237 1 3 55 1 $114 $15 229 9 8 56 0 $195 $20 39 9 10 57 0 $185 $21 125 5 3 58 1 $81 $15 318 5 9 59 1 $223 $29 64 1 3 60 1 $164 $16 124 2 3 61 0 $119 $17 127 8 7 62 1 $165 $20 89 7 8 63 1 $146 $19 98 3 7 64 1 $87 $24 145 6 8 65 1 $158 $20 124 8 10 66 0 $187 $26 118 7 6 67 0 $113 $12 161 8 2 68 0 $159 $27 42 3 1 69 0 $174 $18 39 2 2 70 0 $169 $21 210 6 1 71 1 $140 $17 242 2 6 72 0 $126 $23 170 1 9 73 0 $177 $18 76 9 8 74 1 $124 $21 149 8 9 75 1 $181 $17 161 7 7 76 1 $124 $26 24 7 8 77 0 $145 $13 77 6 8 78 0 $186 $21 239 3 1 79 1 $118 $24 87 9 4 80 1 $141 $19 20 2 2 81 1 $154 $16 90 1 9 82 0 $161 $20 52 5 4 83 0 $92 $26 97 2 7 84 0 $183 $20 42 10 8 85 0 $215 $15 222 8 9 86 1 $104 $19 167 2 1 87 1 $178 $21 36 8 7 88 1 $155 $18 87 8 7 89 0 $104 $19 213 6 7 90 0 $171 $26 74 10 8 91 0 $172 $17 235 8 1 92 0 $88 $19 56 1 3 93 1 $208 $22 67 9 9 94 0 $98 $19 215 1 1 95 0 $174 $21 46 1 1 96 0 $149 $22 333 10 7 97 0 $140 $23 39 5 1 98 1 $170 $13 39 4 5 99 0 $177 $27 57 1 3 100 0 $79 $17 235 6 8

Explanation / Answer

1.        Determine the null hypothesis, via both an explanation and a math equation.

H0 : mu = $18.75   

H0 : average amount a food court customer spends is $18.75

2.        Determine the alternative hypothesis, via both an explanation and a math equation.

H1 : mu > $18.75

H1 : average amount a food court customer spends has increased.

Here variable under consideration is food.

sample size is 100 and sample standard deviation is unknown so we use t-test for testing.

Assume alpha = 0.05

The test statistic is,

t = (Xbar - mu) / (sd/sqrt(n) )

where, mu is population mean = 18.75

Xbar is sample mean of food.

sd is standard deviation of food.

n is the sample size.

Here we use one sample t-test.

This test we can done in MINITAB.

Steps :

Enter data in MINITAB sheet --> Stat --> Basic Statistics --> 1-Sample t --> Samples in columns : Food --> click on Perform hypothesis test --> Hypothesized mean : 18.75 --> Options --> Confidence level : 95.0 --> Alternative : select greator than --> ok --> ok

Output :

One-Sample T: Food

Test of mu = 18.75 vs > 18.75

95%
Lower
Variable N Mean StDev SE Mean Bound T P
Food 100 19.9400 4.2159 0.4216 19.2400 2.82 0.003

Test statistic = 2.82

P-value = 0.003

P-value < alpha (0.05)

Reject H0 at 5% level of significance.

Conclusion : Average amount a food court customer spends has increased.

If alpha = 0.01 and 0.1 the result and conclusion is same as alpha = 0.05

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