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strojnjashka [21]
3 years ago
9

A researcher wanted to determine the mean number of hours per week (Sunday through Saturday) the typical person watches televisi

on. Results from the Sullivan Statistics Survey I indicate that s = 7.5 hours.
(a) How many people are needed to estimate the number of hours people watch television per week within 2 hours with 95% confidence?


(b) How many people are needed to estimate the number of hours people watch television per week within 1 hour with 95% confidence?


(c) What effects does doubling the required accuracy have on the sample size?


(d) How many people are needed to estimate the number of hours people watch television per week within 2 hours with 90% confidence? Compare this result to part (a). How does decreasing the level of confidence in the estimate affect sample size? Why is this reasonable?
Business
1 answer:
GarryVolchara [31]3 years ago
4 0

Answer:

A.) 54 subjects

B.) 216 subjects

C.) doubling the accuracy results in 4 times the sample size

D.) 38 subjects

Decreasing confidence level decreases sample size. For fixed error margin, the lower the confidence level, the lower the sample size

Explanation:

Standard deviation (s) = 7.5 hours

A.)

Error margin 'E' = 2

Confidence level = 0.95

α = 1 - 0.95 = 0.05, α/2 = 0.025

Z - value at α/2 = 0.025 = 1.96

Sample size = [(1.96 × 7.5)/2]^2

Sample size = 7.35^2 = 54.022

54 subjects

B.) E = 1

Sample size = [(1.96 × 7.5)/1]^2

Sample size = 14.7^2 = 216.09

216 subjects

C.) from the above, doubling the accuracy results in 4 times the sample size.

D.) Using a confidence interval of 90%

Error margin 'E' = 2

Confidence level = 0.90

α = 1 - 0.90 = 0.1, α/2 = 0.05

Z - value at α/2 = 0.05 = 1.645

Sample size = [(1.645 × 7.5)/2]^2

Sample size = 6.16875^2 = 38.05

=38 subjects

Decreasing confidence level decreases sample size. For fixed error margin, the lower the confidence level, the lower the sample size

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c. innovators.

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The minimum cost will be "$214085".

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D = 1700 units \\\\S =  \$ 50 \\\\H=  20%\\

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ii) When quantity = 1501 -10,000,  price = $ 12.45 , and holding price is $12.45 * 20 %= $2.49.

iii) When quantity = 10,0001- and more,  price = $ 12.40 , and holding price is $12.40 * 20 %= $2.48.

EOQ= \sqrt{\frac{2DS}{H}} \\\\EOQ1= \sqrt{\frac{2\times 17000\times 50}{2.50}} \\\\EOQ1=824.62 \ \ \ or \ \ \ 825\\

EOQ2= \sqrt{\frac{2\times 17000\times 50}{2.49}} \\\\EOQ1=826.2T \ \ \ or \ \ \ 826\\

EOQ3= \sqrt{\frac{2\times 17000\times 50}{2.48}} \\\\EOQ3=827.93 \ \ \ or \ \ \ 828\\

know we should calculate the total cost of EOQ1 and break ever points (1501 to 10,000)units

total \ cost = odering \ cost + holding \ cost + \ Annual \ product \ cost\\\\total_c  = \frac{D}{Q} \times S +  \frac{Q}{2} \times H + (p \times D) \\\\T_c  = \frac{17000}{825} \times 50+  \frac{825}{2} \times 2.50 + (12.50 \times 17000)\\\\T_c = 1030 .30 +1031.25+212500\\\\T_c =$ 214561.55\\\\

T_c  = \frac{17000}{1501} \times 50+  \frac{1501}{2} \times 2.49 + (12.45 \times 17000)\\\\T_c = 566.28 +1868.74+211650\\\\T_c =$ 214085.02 \ \ \ or \ \ \  $ 214085\\\\

T_c  = \frac{17000}{10001} \times 50+  \frac{10001}{2} \times 2.48 + (12.40 \times 17000)\\\\T_c = 84.99+ 12401.24+210800\\\\T_c =$ 223286.23 \\

The total cost is less then 15001. So, optimal order quantity is 1501, that's why cost is = $214085.

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Note: See the attached xlsx file for the effect of each transaction on the individual accounts of the expanded accounting equation and the report of the total of each element.

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