Question 2 Homework 6-Chapter 5 Due: Wednesday, October 18 1. The waiting time f

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Question 2

Homework 6-Chapter 5 Due: Wednesday, October 18 1. The waiting time for an oil change is distriuted with mean 1o minutes an standard deviation 10 minutes (a.) If they receive 50 customers, what is the probability that the average oil changs time is greater than 12 minutes? (b.) If they receive 100 customers, what is the probability that the average oil changs time is between 7 and 9 minutes? (c.) If they receive 100 customers, what is the probability that the average oil times is less than 12 minutes? (d.) If John helps 40 customers (back-to-back) with oil changes, what is the proba- bility that he will finish within 7 hours? 2. Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content (in percent) of each bottle is determined. The average alcohol content is found to be 8.9 percent. Assume that the true standard deviation of alcohol content among cough syrup is 1.2 percent. (a) Find a 95% confidence interval for the true average alcohol content. (b) Find a 99% upper confidence interval for the true average alcohol content. (c.) How large would your sample have to be in order for the sample average to be within 0.15 percent of the true mean (with 95% confidence)?

Explanation / Answer

PART A.

TRADITIONAL METHOD

given that,

standard deviation, =1.2

sample mean, x =8.9

population size (n)=50

I.

stanadard error = sd/ sqrt(n)

where,

sd = population standard deviation

n = population size

stanadard error = ( 1.2/ sqrt ( 50) )

= 0.17

II.

margin of error = Z a/2 * (stanadard error)

where,

Za/2 = Z-table value

level of significance, = 0.05

from standard normal table, two tailed z /2 =1.96

since our test is two-tailed

value of z table is 1.96

margin of error = 1.96 * 0.17

= 0.333

III.

CI = x ± margin of error

confidence interval = [ 8.9 ± 0.333 ]

= [ 8.567,9.233 ]

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DIRECT METHOD

given that,

standard deviation, =1.2

sample mean, x =8.9

population size (n)=50

level of significance, = 0.05

from standard normal table, two tailed z /2 =1.96

since our test is two-tailed

value of z table is 1.96

we use CI = x ± Z a/2 * (sd/ Sqrt(n))

where,

x = mean

sd = standard deviation

a = 1 - (confidence level/100)

Za/2 = Z-table value

CI = confidence interval

confidence interval = [ 8.9 ± Z a/2 ( 1.2/ Sqrt ( 50) ) ]

= [ 8.9 - 1.96 * (0.17) , 8.9 + 1.96 * (0.17) ]

= [ 8.567,9.233 ]

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interpretations:

1. we are 95% sure that the interval [8.567 , 9.233 ] contains the true population mean

2. if a large number of samples are collected, and a confidence interval is created

for each sample, 95% of these intervals will contains the true population mean

[ANSWERS]

best point of estimate = mean = 8.9

standard error =0.17

z table value = 1.96

margin of error = 0.333

confidence interval = [ 8.567 , 9.233 ]

PART B.

when alpha = 99%

confidence interval = [ 8.9 ± Z a/2 ( 1.2/ Sqrt ( 50) ) ]

= [ 8.9 - 2.576 * (0.17) , 8.9 + 2.576 * (0.17) ]

= [ 8.463,9.337 ]

PART C.

Compute Sample Size

n = (Z a/2 * S.D / ME ) ^2

Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )

Standard Deviation ( S.D) = 1.2

ME =0.15

n = ( 1.96*1.2/0.15) ^2

= (2.352/0.15 ) ^2

= 245.862 ~ 246

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