In: Statistics and Probability
Approximately how much people spend at chipotle for one order:
10.5 |
10 |
10 |
10 |
11 |
10 |
12.5 |
8 |
15 |
0 |
12.75 |
11 |
20 |
15 |
10 |
10 |
15 |
10 |
12 |
13.5 |
15 |
10 |
10 |
11.5 |
14.75 |
15 |
18 |
10.5 |
11 |
12 |
14 |
14.5 |
15 |
13 |
11 |
8.5 |
10 |
16 |
11 |
10 |
10 |
10 |
10 |
10.5 |
11 |
8 |
12 |
12 |
9 |
11.5 |
Create a null and alternative hypothesis using the data given, national average per order at chipotle is $11
Is the sample random?
Put data into a box and whisker plot
Based on the data create a confidence interval with 95% confidence
Based on the data do you accept or reject null hypothesis at 5%significance level
The data provided is :
Data | |
10.5 | |
10 | |
10 | |
10 | |
11 | |
10 | |
12.5 | |
8 | |
15 | |
0 | |
12.75 | |
11 | |
20 | |
15 | |
10 | |
10 | |
15 | |
10 | |
12 | |
13.5 | |
15 | |
10 | |
10 | |
11.5 | |
14.75 | |
15 | |
18 | |
10.5 | |
11 | |
12 | |
14 | |
14.5 | |
15 | |
13 | |
11 | |
8.5 | |
10 | |
16 | |
11 | |
10 | |
10 | |
10 | |
10 | |
10.5 | |
11 | |
8 | |
12 | |
12 | |
9 | |
11.5 | |
Count | 50 |
Mean | 11.62 |
StDev | 3.03434423 |
Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho:
Ha:
Random
Yes the sample seems random
Box and whisker plot
Confidence interval
The critical value for α=0.05 and df=n−1=49 degrees of freedom is . The corresponding confidence interval is computed as shown below:
Accept the null hypothesis since 11 is contained in the above interval.
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