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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 through below.
68.87 78.25 70.44 84.67 79.79 86.33 100.24 98.26
Click the icon to view the table of critical t-values.
a. Determine a point estimate for the population mean travel tax A point estimate for the population mean travel tax is $ 83.36. (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.)
A. The lower bound is $ and the upper bound is $. One can be % confident that all cities have a travel tax between these values.
B. The lower bound is $ and the upper bound is $ The travel tax is between these values for % of all cities.
C. The lower bound is $ and the upper bound is $ There is a % probability that the mean travel tax for all cities is between these values.
D. The lower bound is $ and the upper bound is One can be [95]% confident that the mean travel tax for all cities is between these values.
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 decrease the level of confidence.
B. The researcher could decrease the sample standard deviation.
C. The researcher could increase the level of confidence.
D. The researcher could increase the sample mean

Respuesta :

shallomisaiah19

Answer:

Step-by-step explanation:

Given that:

68.87, 78.25, 70.44, 84.67, 79.79, 86.33, 100.24, 98.26

we calculate sample mean and standard deviation from given data

Sample Mean

[tex]\bar x = \frac{\sum (x)}{n} =\frac{666.85}{8} \\\\=83.35625[/tex]

Sample Variance

[tex]s^2= \frac{\sum (x- \bar x )^2}{n-1} \\\\=\frac{933.224787}{7} =133.317827[/tex]

sample standard deviation

[tex]s=\sqrt{s^2} \\=\sqrt{133.317827} \\ =11.546334[/tex]

95% CI for [tex]\mu[/tex] using t - dist

Sample mean = 83.35625

Sample standard deviation = 11.546334

Sample size = n = 8

Significance level = α = 1 - 0.95 = 0.05

Degrees of freedom for t - distribution

d-f = n - 1 = 7

Critical value

[tex]t_{\alpha 12, df}= t_{0.025, df=7}=2.365[/tex] ( from t - table , two tails, d.f =7)

Margin of Error

[tex]E = t_{\alpha 12, df}\times \frac{s_x}{\sqrt{n} } \\\\=2.365 \times \frac{11.546334}{\sqrt{8} } \\\\=2.365 \times 4.082246\\\\E=9.654512[/tex]

Limits of 95% Confidence Interval are given by:

Lower limit

[tex]\bar x - E = 83.35625-9.654512\\\\=73.701738\approx 73.702[/tex]

Upper Limit

[tex]= \bar x + E\\=83.35625+ 9.654512\\=93.010762 \approx 93.011[/tex]

95% Confidence interval is

[tex]\bar x \pm E = 83.35625 \pm 9.654512\\\\=(73.701738,93.010762)[/tex]

95% CI using t - dist (73.70 < μ < 93.01)

D. The lower bound is $ and the upper bound is One can be [95]% confident that the mean travel tax for all cities is between these values.

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 decrease the level of confidence.