question archive The accompanying data represent the total travel tax? (in dollars) for a? 3-day business trip in 8 randomly selected cities
The accompanying data represent the total travel tax? (in dollars) for a? 3-day business trip in 8 randomly selected cities
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The accompanying data represent the total travel tax? (in dollars) for a? 3-day business trip in 8 randomly selected cities. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. Complete parts? (a) through? (c) below.
68.29
78.17
69.26
83.28
80.45
87.56
100.43
?(a) Determine a point estimate for the population mean travel tax.
A point estimate for the population mean travel tax is ?$____.
?(Round to two decimal places as? needed.)
?(b) Construct and interpret a 95?% confidence interval for the mean tax paid for a? three-day business trip. ?(Round to two decimal places as? needed.)
A. One can be nothing ?% confident that the all cities have a travel tax between ?$ ______ and ?$ ______.
B. The travel tax is between ?$_____ and ?$______ for ____% of all cities.
C. There is a nothing ?% probability that the mean travel tax for all cities is between ?$ ___ and ?$ _____.
D. One can be ____ ?% confident that the mean travel tax for all cities is between ?$ ______ and $ _____.
?(c) What would you recommend to a researcher who wants to increase the precision of the? interval, but does not have access to additional? data?
A. The researcher could increase the sample mean.
B. The researcher could increase the level of confidence.
C. The researcher could decrease the level of confidence.
D. The researcher could decrease the sample standard deviation.
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(a) point estimate = 81.06
(b) D.One can be 95?% confident that the mean travel tax for all cities is between ?$ 70.838 and $ 91.288
(c) B. The researcher could increase the level of confidence.
given that,
sample data: (68.29 , 78.17 , 69.26 , 83.28 , 80.45 , 87.56 , 100.43 )
sample mean, x =81.063
standard deviation, s =11.056
sample size, n =7
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 11.056/ sqrt ( 7) )
= 4.179
margin of error = t alpha/2 * (standard error)
where,
ta/2 = t-table value
level of significance, alpha = 0.05
from standard normal table, two tailed value of |t alpha/2| with n-1 = 6 d.f is 2.447
margin of error = 2.447 * 4.179
= 10.225
CI = x ± margin of error
confidence interval = [ 81.063 ± 10.225 ]
= [ 70.838 , 91.288 ]
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