Question 2) A survey by Visa Spending reported that the average amount spent on

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Question 2) A survey by Visa Spending reported that the average amount spent on a wedding gift for a close family member is $166. A random sample of 45 people who purchased wedding gifts for a close family member spent an average of S160.50. Assume that the population standard deviation is $38.Use a 95% confidence interval to test the validity of this report and choose the one statement that is correct. A) Because this confidence interval does not include $16, the report by Visa Spending is B) Because this confidence interval does not include $166, the report by Visa Spending is C) Because this confidence interval does include $166, the report by Visa Spending is D) Because this confidence interval does include $166, the report by Visa Spending is not not validated validated validated validated

Explanation / Answer

TRADITIONAL METHOD
given that,
standard deviation, =38
sample mean, x =160.5
population size (n)=45
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 38/ sqrt ( 45) )
= 5.665
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 * 5.665
= 11.103
III.
CI = x ± margin of error
confidence interval = [ 160.5 ± 11.103 ]
= [ 149.397,171.603 ]
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DIRECT METHOD
given that,
standard deviation, =38
sample mean, x =160.5
population size (n)=45
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 = [ 160.5 ± Z a/2 ( 38/ Sqrt ( 45) ) ]
= [ 160.5 - 1.96 * (5.665) , 160.5 + 1.96 * (5.665) ]
= [ 149.397,171.603 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [149.397 , 171.603 ] 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

C) Because this confidence interval does include $166, the report by Visa Spending is
validated.

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