Question

A random sample of 800 teenagers revealed that in this sample, the mean number of hours per week of TV watching is 13.2 with a sample standard deviation of 1.6. Find and interpret a 95% confidence interval for the true mean weekly hours of TV watching for teenagers.

Answer #1

The mean number of hours of part-time work per week for a sample
of 317 teenagers is 29. If the margin of error for the population
mean with a 95% confidence interval is 2.1, construct a 95%
confidence interval for the mean number of hours of part-time work
per week for all teenagers.

In order to determine how many hours per week freshmen college
students watch television, a random sample of 25 students was
selected. It was determined that the students in the sample spent
an average of 19.5 hours with a sample standard deviation of 3.9
hours watching TV per week. Please answer the following questions:
(a) Provide a 95% confidence interval estimate for the average
number of hours that all college freshmen spend watching TV per
week. (b) Assume that a...

A random sample of 49 teenagers from a large population was
surveyed, and the average number of movies that they had rented in
the past week was x = 1.9, with a sample standard deviation of s =
0.5.
a) Construct a 95% confidence interval for the average number of
movies rented by teenagers in a week.
b) Construct a 99% confidence interval for the average number of
movies rented by teenagers in a week.

A random sample of 49 teenagers from a large population was
surveyed, and the average number of movies that they had rented in
the past week was x = 1.9, with a sample standard deviation of s =
0.5. a) Construct a 95% confidence interval for the average number
of movies rented by teenagers in a week. b) Construct a 99%
confidence interval for the average number of movies rented by
teenagers in a week.

A Gallup poll conducted January 17 - February 6, 2019, asked
1024 teenagers 13 years of age to 17 years of age , “Typically, how
many hours per week do you spend watching TV?’’ The teenagers who
participated in the survey watched, on average, 13.0 hours of
television per week with a standard deviation of 2.3 hours per
week. Construct a 95% confidence interval for the average number of
hours that teenagers watch TV each week

A random sample of 49 teenagers from a large population was
surveyed, and the average number of movies that they had rented in
the past week was x = 1.9, with a sample standard deviation of s =
0.5. a) Construct a 95% confidence interval for the average number
of movies rented by teenagers in a week. b) Construct a 99%
confidence interval for the average number of movies rented by
teenagers in a week.t v

A random sample of 90 UT business students revealed that on
average they spent 8 hours per week on Facebook. The population
standard deviation is assumed to be 2 hours per week.
If we want to develop a 95% confidence interval for the average
time spent per week on Facebook by all UT business students, the
margin of error of this interval is
The 95% confidence interval for the average time spent per week
on Facebook by all UT business...

Question 1 A random sample of 49 teenagers from a large
population was surveyed, and the average number of movies that they
had rented in the past week was x = 1.9, with a sample standard
deviation of s = 0.5.
a) Construct a 95% confidence interval for the average number of
movies rented by teenagers in a week
. b) Construct a 99% confidence interval for the average number
of movies rented by teenagers in a week.

A
study of 50 teenagers revealed that the average number of times I
thought about food in one day was 100 times with a standard
deviation of 10 develop a 95 percent confidence interval of the
mean time is teenagers think about food in one day

A random sample from 19 shirt manufacturers in Thailand was
sampled for number of hours each week they operated. The average
number of hours was 96.5 with a standard deviation of 13.2 hours.
At the 5% significance level (use table below), the null hypothesis
that the mean hours operated exceeds 95 will

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