Question

In: Statistics and Probability

Below are percentages for annual sales growth and net sales attributed to loyalty card usage at...

Below are percentages for annual sales growth and net sales attributed to loyalty card usage at 74 Noodles & Company restaurants.

Annual Sales Growth (px;) and Loyalty Card Usage (px; of Net Sales)
(n = 74 restaurants)
Store Growth% Loyalty% Store Growth% Loyalty%
1 -8.0 0.5 38 7.3 2.4
2 -7.5 2.5 39 7.5 1.6
3 -6.4 2.4 40 7.8 1.9
4 -5.2 1.8 41 8.0 2.1
5 -5.0 2.1 42 8.1 1.6
6 -1.7 1.6 43 8.1 1.3
7 -0.7 2.1 44 8.5 2.5
8 -0.5 2.2 45 8.5 2.3
9 -0.5 1.2 46 8.6 1.4
10 -0.5 2.6 47 8.7 0.8
11 0.3 2.6 48 8.8 2.7
12 0.9 0.8 49 9.0 2.1
13 0.9 1.9 50 9.1 1.4
14 1.2 1.3 51 9.2 2.1
15 1.7 2.2 52 10.5 2.0
16 1.8 2.4 53 10.8 1.7
17 1.9 2.5 54 10.8 1.4
18 2.0 1.9 55 11.0 0.9
19 4.0 0.8 56 11.3 1.8
20 4.3 2.1 57 11.4 1.2
21 4.5 1.4 58 11.6 0.7
22 4.7 2.2 59 11.8 1.5
23 4.9 1.5 60 11.8 2.1
24 5.2 1.8 61 13.5 1.2
25 5.2 2.4 62 14.1 1.5
26 5.3 1.6 63 14.3 1.9
27 5.4 1.2 64 15.1 0.7
28 5.5 2.0 65 15.5 2.1
29 5.6 2.6 66 15.9 1.6
30 5.7 2.0 67 16.0 0.9
31 5.9 1.5 68 16.2 2.6
32 6.0 1.9 69 16.2 1.4
33 6.4 2.3 70 17.8 2.2
34 6.4 0.6 71 18.8 2.1
35 6.6 1.9 72 18.9 1.3
36 6.6 2.3 73 19.8 0.6
37 6.7 1.2 74 24.0 1.7



(b) Find the correlation coefficient. (Round your answer to 3 decimal places. A negative value should be indicated by a minus sign.)
  
r            _________

(c-1) To test the correlation coefficient for significance at α = 0.01, fill in the following. (Use the rounded value of the correlation coefficient from part b in all calculations. For final answers, round tcalc to 3 decimal places and the p-value to 4 decimal places. Negative values should be indicated by a minus sign.)

tcalc
p-value

Solutions

Expert Solution

Answer(b):

The coefficient of correlation between two variables can be obtained by

Answer(c-1)

We have to test the below null hypothesis at α=0.01

H0: ρ=0

HA: ρ≠0

The test statistic to test this hypothesis is

tcalc = -1.723

The p-value for above test is 0.0892

p-value=0.0892

since the p-value is greater than 0.05, we do not have enough evidence against H0 to reject it, so we fail to reject the null hypothesis.


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