Question

PLEASE SHOW ALL WORK IN EXCEL.

Use Bilingual sheet to answer this question.

A national survey of companies included a question that asked whether the company had at least one bilingual telephone operator. The sample results of 90 companies follow (Y denotes that the company does have at least one bilingual operator; N denotes that it does not).

N	N	N	N	Y	N	Y	N	N
Y	N	N	N	Y	Y	N	N	N
N	N	Y	N	Y	N	Y	N	Y
Y	Y	N	Y	N	N	N	Y	N
N	Y	N	N	N	N	N	N	N
Y	N	Y	Y	N	N	Y	N	Y
N	N	Y	Y	N	N	N	N	N
Y	N	N	N	N	Y	N	N	N
Y	Y	Y	N	N	Y	N	N	N
N	N	N	Y	Y	N	N	Y	N

The file Dataset_HW-3, sheet named Bilingual also contains the above survey results. Use this information to estimate with 80% confidence the proportion of the population that does have at least one bilingual operator.

[3 points]

Use Part-2 sheet to answer this question.

You are trying to estimate the average amount a family spends on food during a year. In the past the standard deviation of the amount a family has spent on food during a year has been approximately $1000. If you want to be 99% sure that you have estimated average family food expenditures within $50, how many families do you need to survey?

[2.5 points]

Use Part-3 sheet to answer this question.

You have been assigned to determine whether more people prefer Coke or Pepsi. Assume that roughly half the population prefers Coke and half prefers Pepsi. How large a sample do you need to take to ensure that you can estimate, with 95% confidence, the proportion of people preferring Coke within 3% of the actual value? [Hint: proportion est. = 0.5]

N

N

N

N

Y

N

Y

N

N

Y

N

N

N

Y

Y

N

N

N

N

N

Y

N

Y

N

Y

N

Y

Y

Y

N

Y

N

N

N

Y

N

N

Y

N

N

N

N

N

N

N

Y

N

Y

Y

N

N

Y

N

Y

N

N

Y

Y

N

N

N

N

N

Y

N

N

N

N

Y

N

N

N

Y

Y

Y

N

N

Y

N

N

N

N

N

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Y

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Y

N

Answer 1

#1) we are asked to find 80% confidence interval for proportion of the population that does have at least one bilingual operator.

So first we need to find proportion of Y in the data set.There are 30 Y's in the given chart and total sample of 90 companies are provided.

So the proportion of Y's ( ) = 30/ 90 = 0.3333

Confidence interval for proportion is given by,

Lower limit = - E and Upper limit = + E

E is margin of error =  

z is the critical value follows standard normal distribution and n is the sample size.

We can find value of z using z score table for given confidence level.

We are given confidence level (c) = 0.80 , So = 1- c = 1 - 0.80 = 0.20

Then /2 = 0.1 , 1 - ( /2) = 1 - 0.1 = 0.9

so we need to find z score corresponding to area 0.9 on z score table.

So first we need to find number 0.9 or nearest on the table..0.8997 is the nearest one to 0.9

So we have to add corresponding row and column of 0.8997

So 1.2+0.08 = 1.28

z =1.28 is the required z value ..

We have = 0.3333 and n = 90

Therefore E = = 1.28* = 0.0636

E = 0.0636

So lower limit = 0.3333 - 0.0636 = 0.2697

lower limit = 0.3333 + 0.0636 = 0.3969

So 80% confidence the proportion of the population that does have at least one bilingual operator is (26.97%,39.69%)

#2) We are given  standard deviation of the amount a family has spent on food = $1000.

E = 50

We are asked to find n for 99% confidence level to estimate the mean. it is given by ,

n =

We are given = $1000 and E = 50 , we have to find z for given value of confidence level 0.99 using z score table.

confidence level (c) = 0.99 , So = 1- c = 1 - 0.99 = 0.01

Then /2 =0.005 , 1 - ( /2) = 1 - 0.005 = 0.995

so we need to find z score corresponding to area 0.995 on z score table.

So first we need to find number 0.995 or nearest on the table..0.9951 is the nearest one to 0.995

So we have to add corresponding row and column of 0.995

So 2.5+0.08 = 2.58

z =2.58 is the required z value .

n = = = = 2662.56 2663

Therefore we need to survey 2663 families .

#3)

We are given = 0.5 , E = 0.03

We are asked to find n for 99% confidence level to estimate the proportion. it is given by ,

n =

we have to find z for given value of confidence level 0.95 using z score table.

confidence level (c) = 0.95 , So = 1- c = 1 - 0.95 = 0.05

Then /2 =0.025 , 1 - ( /2) = 1 - 0.025 = 0.975

so we need to find z score corresponding to area 0.975 on z score table.

So first we need to find number 0.975 or nearest on the table..0.9750 is on the table.

So we have to add corresponding row and column of 0.9750

1.9+0.06 = 1.96

z =1.96 is the required z value .

n = =  

n = 1067.11 1068

n = 1068