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 normallydistributed. A boxplot indicates there are no outliers. Complete parts (a) through (c) below.Question Help68.3579.5170.9383.4379.6285.02Click the icon to view the table of critical t-values.100.3597.14(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.Select the correct choice below and fill in the answer boxes to complete your choice.(Round to two decimal places as needed.)O A.One can be% confident that the mean travel tax for all cities is between $and $B. There is a% probability that the mean travel tax for all cities is between $and $OC. The travel tax is between $and $% of all cities.forClick to select your answer(s).7:48 PMN19OEt11/10/2019Type here to searchBUSdeleteprt scSYSTGinsertpousebreak1119$7X1615f3boJEesc&76#432UYTREWfobKCOI 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 normallydistributed. A boxplot indicates there are no outliers. Complete parts (a) through (c) below.8.3579.5170.9383.4379.6285.02Click the icon to view the table of critical t-values.100.3597.14B. There is a% probability that the mean travel tax for all cities is between $and $OC. The travel tax is between $and $for% of all cities.D. One can be% confident that the all cities have a travel tax 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?O A. The researcher could decrease the level of confidence.B. The researcher could increase the level of confidenceC. The researcher could decrease the sample standard deviationD. The researcher could increase the sample mean.Click to select your answer(s).7:48 PMO EtN1911/10/2019Type here to searchSUSprt scinspausebreakhobisAsf91715JE%5esc7#3427UYTREWKHCO

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Asked Nov 11, 2019
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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.
Question Help
68.35
79.51
70.93
83.43
79.62
85.02
Click the icon to view the table of critical t-values.
100.35
97.14
(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.
Select the correct choice below and fill in the answer boxes to complete your choice.
(Round to two decimal places as needed.)
O A.
One can be
% confident that the mean travel tax for all cities is between $
and $
B. There is a
% probability that the mean travel tax for all cities is between $
and $
OC. The travel tax is between $
and $
% of all cities.
for
Click to select your answer(s).
7:48 PM
N
19
OEt
11/10/2019
Type here to search
BUS
delete
prt sc
SYSTG
insert
pouse
break
11
19
$7
X
16
15
f3
bo
JE
esc
&
7
6
#
4
3
2
U
Y
T
R
E
W
fob
K
CO
I
help_outline

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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. Question Help 68.35 79.51 70.93 83.43 79.62 85.02 Click the icon to view the table of critical t-values. 100.35 97.14 (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. Select the correct choice below and fill in the answer boxes to complete your choice. (Round to two decimal places as needed.) O A. One can be % confident that the mean travel tax for all cities is between $ and $ B. There is a % probability that the mean travel tax for all cities is between $ and $ OC. The travel tax is between $ and $ % of all cities. for Click to select your answer(s). 7:48 PM N 19 OEt 11/10/2019 Type here to search BUS delete prt sc SYSTG insert pouse break 11 19 $7 X 16 15 f3 bo JE esc & 7 6 # 4 3 2 U Y T R E W fob K CO I

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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.
8.35
79.51
70.93
83.43
79.62
85.02
Click the icon to view the table of critical t-values.
100.35
97.14
B. There is a
% probability that the mean travel tax for all cities is between $
and $
OC. The travel tax is between $
and $
for
% of all cities.
D. One can be
% confident that the all cities have a travel tax 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?
O A. The researcher could decrease the level of confidence.
B. The researcher could increase the level of confidence
C. The researcher could decrease the sample standard deviation
D. The researcher could increase the sample mean.
Click to select your answer(s).
7:48 PM
O Et
N
19
11/10/2019
Type here to search
SUS
prt sc
ins
pause
break
ho
bisAs
f9
17
15
JE
%
5
esc
7
#
3
4
2
7
U
Y
T
R
E
W
K
H
CO
help_outline

Image Transcriptionclose

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. 8.35 79.51 70.93 83.43 79.62 85.02 Click the icon to view the table of critical t-values. 100.35 97.14 B. There is a % probability that the mean travel tax for all cities is between $ and $ OC. The travel tax is between $ and $ for % of all cities. D. One can be % confident that the all cities have a travel tax 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? O A. The researcher could decrease the level of confidence. B. The researcher could increase the level of confidence C. The researcher could decrease the sample standard deviation D. The researcher could increase the sample mean. Click to select your answer(s). 7:48 PM O Et N 19 11/10/2019 Type here to search SUS prt sc ins pause break ho bisAs f9 17 15 JE % 5 esc 7 # 3 4 2 7 U Y T R E W K H CO

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Expert Answer

Step 1

Given data

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(x - 2 Data (x) 68.35 -14.69375 215.9062891 79.51 -3.53375 12.48738906 70.93 -12.11375 146.7429391 83.43 0.38625 0.149189063 79.62 -3.42375 11.72206406 85.02 1.97625 3.905564062 100.35 17.30625 299.5062891 97.14 14.09625 198.7042641 Sum 664.35 889.124 664.35 Mean = 83.04375 Standard deviation is given by Σ-1 (x - )2 889.124 11.27021 8 1 п — 1

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Step 2

b)

95% confidence interval is given by

The 100 (1 – α) % confidence interval for the population mean, μ, for given sample standard deviation, s is: ( – (/2; n – 1­) (s/√n),  + (/2; n – 1) (s/√n)).

Here, n is the sample size,  is the sample mean, and /2; n – 1 is the critical value of the t-distribution with (n – 1) degrees of freedom, above which, 100 (α/2) % or α/2 proportion of the observations lie.

The t-distribution is used, because the population standard deviation is unknown and the sample standard deviation is being used as a substitute.

Calculation:

Here, n = 8;  = 83.04375; s = 11.27021.

Therefore, degrees of freedom = 8 – 1 = 7.

Again, 100 (1 – α) % = 95% = 0.95.

Thus, α = 0.05.

From the Excel formula: =T.I...

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