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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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A TV station claims that 38% of the 6:00 - 7:00 pm viewing audience watches its evening news program. A consumer group believes
Elena L [17]

Answer:

z=\frac{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374  

p_v =P(Z  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that regularly watch the TV station’s news program is significantly less than 0.38 .  

Step-by-step explanation:

1) Data given and notation

n=830 represent the random sample taken

X=282 represent the people that regularly watch the TV station’s news program

\hat p=\frac{282}{830}=0.340 estimated proportion of people that regularly watch the TV station’s news program

p_o=0.38 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is less than 0.38:  

Null hypothesis:p\geq 0.38  

Alternative hypothesis:p < 0.38  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(Z  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that regularly watch the TV station’s news program is significantly less than 0.38 .  

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hi

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