Practical Application Scenario 2 In this practical application scenario, you hav

ID: 3022392 • Letter: P

Question

Practical Application Scenario 2

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.

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

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

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.

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

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

Question:

Practical Application Scenario 2

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.

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

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

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

Answer:

Description of Question:

We have data of 100 respondents on mall’s food court convenience interview. And we have given average spending of customer in mall’s food court per visit was $18.75 in 2011. Now there is a statement after seeing the descriptive results that- average amount spent by customer in mall’s food court per visit has increased. Now we just need to test that statement.

Null hypothesis-

A null hypothesis is typically the standard assumption and is defined as the prediction that there is no interaction between variables. The null hypothesis reflects that there will be no observed effect for our experiment. In a mathematical formulation of the null hypothesis there will typically be an equal sign. This hypothesis is denoted by H0. The null hypothesis is what we attempt to find evidence against in our hypothesis test. We hope to obtain a small enough p-value that it is lower than our level of significance alpha and we are justified in rejecting the null hypothesis. If our p-value is greater than alpha, then we fail to reject the null hypothesis. Here, in our case we are taking null hypothesis such that mean spending is $18.75 (as null hypothesis is giving the neutral statement), i.e.

H0: µ = $18.75

Alternative hypothesis-

Alternative hypothesis, is also known as the research hypothesis, defined as the prediction that there is a measurable interaction between variables. The alternative or experimental hypothesis reflects that there will be an observed effect for our experiment. In a mathematical formulation of the alternative hypothesis there will typically be an inequality, or not equal to symbol. This hypothesis is denoted by either Ha or by H1. The alternative hypothesis is what we are attempting to demonstrate in an indirect way by the use of our hypothesis test. If the null hypothesis is rejected, then we accept the alternative hypothesis. Here, in our case we are taking alternate hypothesis as mean spending has increased. i.e. H1: µ $18.75

Procedures in doing a t-test

1. Determine H0 and H1

2. Set the criterion § Look up tcrit, which depends on alpha and df

3. Collect sample data, calculate x and s , i.e. mean and standard deviation from data.

4. Calculate the test statistic.

5. Reject H0 if tobt is more extreme than tcrit

Now in our case, we hypothesized about response of interview on spending in food court of mall with a mean spending of $18.75. However, we do not know the population standard deviation. This tells us we must use a t-test instead of a z-test for testing one sample test.

Step 1: State the hypotheses

H0: µ = 68,

H1: µ $18.75

Step 2: Set the criterion

Here, in our case the following criteria has been set.

Step 3: Collect sample data, calculate x and s-

From our dataset, after calculation we get sample mean is 19.94, with a standard deviation (s) of 4.2159.

Below is the table for descriptive statistics of variable spending in mall’s food court.

Mean

19.94

Standard Error

0.421593897

Median

Mode

Standard Deviation

4.215938972

Sample Variance

17.77414141

Kurtosis

-0.813023969

Skewness

0.176441243

Range

Minimum

Maximum

Sum

1994

Count

100

Confidence Level(95.0%)

0.836533757

Step 4: Calculate the test statistic

So, here sample mean is = 19.94, observed mean = 18.75 and standard error of mean = 0.421593.

Now by putting the values in the formula of t-statistic, we get-

t = (19.94-18.75)/0.421593

t = 2.8226

Step 5: Reject H0 if tobt is more extreme than tcrit

Hence even after changing the level of significance there is no change in hypothesis decision.