Question

two-question survey of 80 of its customers, who the management of the chain believe are representative of all of their shoppers. The first question on the survey asked the shoppers to estimate their average weekly expenditures ,The resulting responses were normally distributed with a mean of $40 and a standard deviation of $8. Because the chain is considering implementing a grocery delivery service, the second question on the survey asked whether or not the respondent would be interested in using such a service. 28 of the 80 respondents said that they would like to use a delivery service and the rest said that they would not. Customers' interest in using a delivery service and their weekly spending appear to be independent of one another.

Q1. : Provide the upper bound of a 90% confidence interval for the average weekly expenditure for the chain's entire population of shoppers.

Q2.: Provide the upper bound of a 90% confidence interval for the proportion of the chain's entire population of shoppers who would be likely to use a grocery delivery service

Q3.what is the estimated proportion of the chain's customers who are both interested in home deliveries and spend at least $50 per week at the stores?

Q4.: If we know that a customer who was not surveyed spends less than $40 per week at the chain, what is the estimated probability that this customer will be interested in home delivery, based on the estimates provided by this sample

Q5. Using the estimates provided by this sample, what is our best guess as to the percentage of the chain's customers who spend between $40 and $60 per week at the chain?

Answer 1

1.
TRADITIONAL METHOD
given that,
sample mean, x =40
standard deviation, s =8
sample size, n =80
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 8/ sqrt ( 80) )
= 0.894
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 79 d.f is 1.664
margin of error = 1.664 * 0.894
= 1.488
III.
CI = x ± margin of error
confidence interval = [ 40 ± 1.488 ]
= [ 38.512 , 41.488 ]
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DIRECT METHOD
given that,
sample mean, x =40
standard deviation, s =8
sample size, n =80
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 79 d.f is 1.664
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 40 ± t a/2 ( 8/ Sqrt ( 80) ]
= [ 40-(1.664 * 0.894) , 40+(1.664 * 0.894) ]
= [ 38.512 , 41.488 ]
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interpretations:
1) we are 90% sure that the interval [ 38.512 , 41.488 ]$ contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean

2.
TRADITIONAL METHOD
given that,
possible chances (x)=28
sample size(n)=80
success rate ( p )= x/n = 0.35
I.
sample proportion = 0.35
standard error = Sqrt ( (0.35*0.65) /80) )
= 0.0533
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
margin of error = 1.645 * 0.0533
= 0.0877
III.
CI = [ p ± margin of error ]
confidence interval = [0.35 ± 0.0877]
= [ 0.2623 , 0.4377]
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DIRECT METHOD
given that,
possible chances (x)=28
sample size(n)=80
success rate ( p )= x/n = 0.35
CI = confidence interval
confidence interval = [ 0.35 ± 1.645 * Sqrt ( (0.35*0.65) /80) ) ]
= [0.35 - 1.645 * Sqrt ( (0.35*0.65) /80) , 0.35 + 1.645 * Sqrt ( (0.35*0.65) /80) ]
= [0.2623 , 0.4377]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 90% sure that the interval [ 0.2623 , 0.4377] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion

3.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 40
standard Deviation ( sd )= 8
he estimated proportion of the chain's customers who are both interested in home deliveries and
spend at least $50 per week at the stores
P(X < 50) = (50-40)/8
= 10/8= 1.25
= P ( Z <1.25) From Standard Normal Table
= 0.8944
P(X > = 50) = (1 - P(X < 50)
= 1 - 0.8944 = 0.1056

4.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 40
standard Deviation ( sd )= 8/ Sqrt ( 80 ) =0.8944
sample size (n) = 80
the estimated probability that this customer will be interested in home delivery,
based on the estimates provided by this sample.
we know that a customer who was not surveyed spends less than $40 per week at the chain
P(X < 40) = (40-40)/8/ Sqrt ( 80 )
= 0/0.8944= 0
= P ( Z <0) From Standard NOrmal Table
= 0.5
P(X > = 40) = 1 - P(X < 40)
= 1 - 0.5 = 0.5

5.
our best guess as to the percentage of the chain's customers
who spend between $40 and $60 per week at the chain
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 40) = (40-40)/8/ Sqrt ( 80 )
= 0/0.894427
= 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(X < 60) = (60-40)/8/ Sqrt ( 80 )
= 20/0.894427 = 22.36068
= P ( Z <22.36068) From Standard Normal Table
= 1
P(40 < X < 60) = 1-0.5 = 0.5 =50%