Variable Nor95% CIStDev SE MeanMeanBuyItNow 9 630.78 20.47(615.04 646.52)6.82c. Interpret the 95 % confidence interval in context.d. Compared to the confidence interval (569, 599) forprices from auctions obtained in Example 6, is thereevidence that the mean price is higher when purchasedthrough Buy It Now? Explain.e. Excluding the observation of $675, the confidence in-terval equals (615, 636). Does this change your answerto part d?Time spent on e-mail When the GSS askedn =people in 2012, "About how many hours per week doyou spend sending and answering e-mail?" (EMAILHR),the summary statistics were = 6.89 and s 13.05. TIoutput with these data (available on the book's website)is shown in the screen shot.10508.36afcta. What is the margin of error at the 95% confidence levelfor the point estimate of the mean number of hours?b. Interpret the 95% confidence interval shown in context.c. The distribution of hours spent is right-skewed (e.g.,the minimum of zero hours is only 0.5 standard de-viations below the mean, but the largest value of 168hours has a z-score larger than 12). Is this a concern forthe validity of the confidence interval?rn.m-NORMAL FIX3 AUTO REAL RADIAN CLa?TInterval(6.100.7.680)X-6.890Sx-13.050n=1050.00012Grandmas using e-mail For the question about e-mail inthe previous exercise, the 14 females in the GSS ofage at8.37

Question
Asked Nov 8, 2019
10 views

8.36

Variable N
or
95% CI
StDev SE Mean
Mean
BuyItNow 9 630.78 20.47
(615.04 646.52)
6.82
c. Interpret the 95 % confidence interval in context.
d. Compared to the confidence interval (569, 599) for
prices from auctions obtained in Example 6, is there
evidence that the mean price is higher when purchased
through Buy It Now? Explain.
e. Excluding the observation of $675, the confidence in-
terval equals (615, 636). Does this change your answer
to part d?
Time spent on e-mail When the GSS askedn =
people in 2012, "About how many hours per week do
you spend sending and answering e-mail?" (EMAILHR),
the summary statistics were = 6.89 and s 13.05. TI
output with these data (available on the book's website)
is shown in the screen shot.
1050
8.36
af
ct
a. What is the margin of error at the 95% confidence level
for the point estimate of the mean number of hours?
b. Interpret the 95% confidence interval shown in context.
c. The distribution of hours spent is right-skewed (e.g.,
the minimum of zero hours is only 0.5 standard de-
viations below the mean, but the largest value of 168
hours has a z-score larger than 12). Is this a concern for
the validity of the confidence interval?
r
n.
m-
NORMAL FIX3 AUTO REAL RADIAN CL
a?
TInterval
(6.100.7.680)
X-6.890
Sx-13.050
n=1050.000
12
Grandmas using e-mail For the question about e-mail in
the previous exercise, the 14 females in the GSS ofage at
8.37
help_outline

Image Transcriptionclose

Variable N or 95% CI StDev SE Mean Mean BuyItNow 9 630.78 20.47 (615.04 646.52) 6.82 c. Interpret the 95 % confidence interval in context. d. Compared to the confidence interval (569, 599) for prices from auctions obtained in Example 6, is there evidence that the mean price is higher when purchased through Buy It Now? Explain. e. Excluding the observation of $675, the confidence in- terval equals (615, 636). Does this change your answer to part d? Time spent on e-mail When the GSS askedn = people in 2012, "About how many hours per week do you spend sending and answering e-mail?" (EMAILHR), the summary statistics were = 6.89 and s 13.05. TI output with these data (available on the book's website) is shown in the screen shot. 1050 8.36 af ct a. What is the margin of error at the 95% confidence level for the point estimate of the mean number of hours? b. Interpret the 95% confidence interval shown in context. c. The distribution of hours spent is right-skewed (e.g., the minimum of zero hours is only 0.5 standard de- viations below the mean, but the largest value of 168 hours has a z-score larger than 12). Is this a concern for the validity of the confidence interval? r n. m- NORMAL FIX3 AUTO REAL RADIAN CL a? TInterval (6.100.7.680) X-6.890 Sx-13.050 n=1050.000 12 Grandmas using e-mail For the question about e-mail in the previous exercise, the 14 females in the GSS ofage at 8.37

fullscreen
check_circle

Expert Answer

Step 1

a)

 

The margin of error at the 95% confidence level is 0.79 and it is obtained below:

 

From the given information,

 

The lower limit is 6.100 and upper limit is 7.680.

help_outline

Image Transcriptionclose

Upper limit -lower limit МЕ- 2 7.680-6.100 2 0.79

fullscreen
Step 2

b)

 

Interpret the 95% confidence interval.

 

The 95% confidence interval is (6.100, 7.680).

There is a 95% cha...

Want to see the full answer?

See Solution

Check out a sample Q&A here.

Want to see this answer and more?

Solutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*

See Solution
*Response times may vary by subject and question.
Tagged in

Math

Statistics

Related Statistics Q&A

Find answers to questions asked by student like you
Show more Q&A
add
question_answer

Q: Company XYZ know that replacement times for the portable MP3 players it produces are normally distri...

A: Consider,X be a random variable that shows the replacement times for the portable MP3 which follows ...

question_answer

Q: A sample of 230 observations is selected from a normal population with a population standard deviati...

A: Given:

question_answer

Q: 2. 2. Provided that a large group of students has ages that are normally "distributed with a mean of...

A: Given dataMean = 22Standard deviation = 2a)Since the data is normally distributed about mean 22. So,...

question_answer

Q: The lengths of pregnancies are normally distributed with a mean of 273 days and a standard deviation...

A: We are given,μ = 273σ = 20n = 50

question_answer

Q: The average number of pounds of red meat a person consumes each year is 196 with a standard deviatio...

A: We are given,μ = 196σ = 22n = 60

question_answer

Q: Use the given information to find the P-value.   The test statistic in a right-tailed test is z = 1....

A: It is given that the z-score is 1.43.

question_answer

Q: Suppose 36 participants complete an experiment where ads are presented subliminally during a task (e...

A: a). From the provided information, the null and alternative hypothesis can be stated as:H0:µ=0.50, t...

question_answer

Q: (d) what proportion of 18-ounce bags of chips ahoy! contains fewer than 1125 chocolate chips? (e) Wh...

A: Given information:

question_answer

Q: Find the area of the shaded region. The graph depicts the standard normal distribution of bone densi...

A: The area of the shaded region is 0.7277 or 72.77% and it is calculated below: