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Rom4ik [11]
3 years ago
6

An experiment was conducted at a small supermarket for a period of 8 days on the sales of a single brand of dog​ food, involving

three levels of shelf​ height: knee​ level, waist​ level, and eye level. During each​ day, the shelf height of the dog food was randomly changed on three different occasions.​ Sales, in hundreds of​ dollars, of the dog food per day for the three shelf heights are given below. Based on the​ data, is there a significant difference in the average daily sales of this dog food based on shelf​ height? Use a 0.05 level of significance. Shelf Height Knee Level Waist Level Eye Level 77 88 85 82 94 85 86 93 87 78 90 81 81 91 80 86 94 79 77 90 87 81 87 93
Mathematics
1 answer:
sammy [17]3 years ago
4 0

Answer:

There is no significant difference in the average daily sales of this dog food based on shelf height

Step-by-step explanation:

Null hypothesis: There is no significant difference in the average sales

Alternate hypothesis: There is a significant difference in the average daily sales

Using the F-table, at 0.05 significance level and (2, 21) degrees of freedom (3-1, 24-3), the critical value is 3.47

Conclusion: Reject the null hypothesis if the computed F exceeds 3.47

Data values

Knee level: 77,88,85,82,94,85,86,93 (Mean is 86.25)

Waist level: 87,78,90,81,81,91,80,86 (Mean is 84.25)

Eye level: 94,79,77,90,87,81,87,93 (Mean is 86)

Grand mean = (86.25+84.25+86)/3 = 85.5

Sum of Squares (SS) for knee level= summation(sales - grand mean)^2 = 220

SS for waist level = 180

SS for eye level = 288

SStotal = 220+180+288 = 688

Sum of Squares due to Treatment (SST) for knee level = n(mean - grand mean)^2 = 8(86.25 - 85.5)^2 = 4.5

SST for waist level = 12.5

SST for eye level = 0.25

Total SST = 4.5+12.5+0.25 =

17.25

Mean Sum of Treatment (MST)= SST/(number of food types - 1) = 17.25/(3-1) = 17.25/2 = 8.625

Sum of Squares due to Error (SSE) = SStotal - SST = 688 - 17.25 = 670.75

Mean Sum of Error (MSE) = SSE/(24-3) = 670.75/21 = 31.940

F = MST/MSE = 8.625/31.940 = 0.27

The computed F 0.27 is less than the critical value 3.47, so we fail to reject the null hypothesis

Conclusion: There is no significant difference in the average daily sales of this dog food based on shelf height

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Find answers below

Step-by-step explanation:

H0: P <= 0.5

Ha: P > 0.5

, the number who prefer gut strings is <= a number or the test tends towards the  left-tailed.

{0,1,2,3,4,5} ;  

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b)

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Assume alpha = 0.05

z(alpha) = -1.645

Reject if (x/20 - 0.5) / sqrt[ (0.5)(0.5)/20 ] < -1.645

Reject if (x/20 - 0.5) <  < (-1.645) sqrt ( (0.5)(0.5)/20 )

Reject if x/20   < (-1.645) sqrt ( (0.5)(0.5)/20 ) + 0.5

Reject if x/20   < 0.316

Reject if x   < (0.316)(20) = 6.32

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According to (a),  reject H0 if x <= 5

P( Type II error) = P( do not reject H0/ when Ha is true)

P( Type II error) = P( x > 5/ P=0.6)

x ---p(x)

6  0.004854  0.998388  

7  0.014563    

8  0.035497  

9  0.070995  

10  0.117142  

11  0.159738  

12  0.179706  

13  0.165882  

14  0.124412  

15  0.074647  

16  0.034991  

17  0.012350  

18  0.003087  

19  0.000487  

20  0.000037  

add: 0.9984 --  proba bility of a type II error

Assuming P=0.8

P( Type II error) = P( x > 5/ P=0.8)

6  0.000002  1.000000  

7  0.000013  

8  0.000087  

9  0.000462  

10  0.002031  

11  0.007387  

12  0.022161  

13  0.054550  

14  0.109100  

15  0.174560  

16  0.218199  

17  0.205364  

18  0.136909  

19  0.057646  

20  0.011529  

add: 1.0000  probability of a type II error

d)

P( x <= 13) =  

0  0.000001  

1  0.000019  

2  0.000181  

3  0.001087  

4  0.004621  

5  0.014786  

6  0.036964  

7  0.073929  

8  0.120134  

9  0.160179  

10  0.176197  

11  0.160179  

12  0.120134  

13  0.073929  

add: 0.9423 < 0.10 ,  H0 cannot be rejected

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