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iren [92.7K]
2 years ago
15

Customers at your store select their groceries from shelves and then pay. Sales records and records of restocks for Flaming Roas

t Coffee (1 lb size) are as follows:
Number of bags Tue Wed Thu Fri Sat Sun Mon

Sales  5  8  2 19 22 15  3

Restock  6 12 12 18 24  0  0

What is the net change in how many bags are on the shelf, from the beginning of Tuesday to the end of Monday?
Mathematics
1 answer:
ioda2 years ago
5 0

Answer:

  • <em>The net change in how many bags are on the shelf, from the beginning of Tuesday to the end of Monday is -</em><u>2.</u>

Explanation:

The change in the number of bags any day is the number of bags is equal to the number of bags purchased to restock less the number of bags sold that day.

  • Change = bags purchased to restock - bags sold

At the end of <em>Tuesday</em>, the change is:

  • Restock:    6
  • Sales:        5
  • Change:    6 - 5 = 1    (note that this means that the number of bags increases by 1)

At the end of <em>Wednesday</em>, the change is:

  • Restock:    12
  • Sales:        8
  • Change:    12 - 8 = 4    (the number of bags increases by 4)

At the end of <em>Thursday</em>, the change is:

  • Restock:    12
  • Sales:        2
  • Change:    12 - 2 = 10    (the number of bags increases by 10)

At the end of <em>Friday</em>, the change is:

  • Restock:    18
  • Sales:        19
  • Change:    18 - 19 = - 1    (the number of bags decreases by 1).

At the end of <em>Saturday</em>, the change is:

  • Restock:    24
  • Sales:        22
  • Change:    24 - 22 = 2    (the number of bags increases by 2).

At the end of <em>Sunday</em>, the change is:

  • Restock:    0
  • Sales:        15
  • Change:    0 - 15 = - 15    (the number of bags decreases by 15).

At the end of <u>Monday</u>, the change is:

  • Restock:    0
  • Sales:        3
  • Change:    0 - 3 = - 3    (the number of bags decreases by 3).

The net change in how many bags are on the shelf, from the beginning of Tuesday to the end of Monday equals the algebraic sum of every change:

  • Net change = 1 + 4 + 10 + (-1) + 2 + (-15) + (-3)
  • Using associative property: (1 + 4 + 10 + 2) - (1 + 15 +3)
  • Simplifying: 17 - 19 = -2

<u>Conclusion</u>: the net change in how many bags are on the shelf, from the beginning of Tuesday to the end of Monday is -2, meaning that the number of bags, after taking into account all sales and restock, decreases by 2.

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Answer:

a) 0.68 - 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.630  

0.68 + 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.730  

And the 99% confidence interval would be given (0.630;0.730).  

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c) For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval  are satisfied, so then the results can be assumed for all the population

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

p represent the real population proportion of interest  

\hat p =0.68 represent the estimated proportion

n=575 is the sample size required (variable of interest)  

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

Part a

The confidence interval would be given by this formula  

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=2.58  

And replacing into the confidence interval formula we got:  

0.68 - 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.630  

0.68 + 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.730  

And the 99% confidence interval would be given (0.630;0.730).  

Part b

We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)

Part c

For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval  are satisfied, so then the results can be assumed for all the population

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