Question

Do Georgians (population 1) spend more time online compared to Floridians (population 2)? Last year data: Georgia: sample size = 100, sample mean = 35 hours, sample standard deviation 14.5 hours. Florida: sample size = 100, sample mean = 32 hours, sample standard deviation 10.5 hours. Formulate the hypothesis.

Calculate the p-value

What is your conclusion at alpha = 0.05?

Obtain a 95% confidence interval for the difference between the population of Georgia (population 1) and Florida (population 2).

What is your conclusion based on the interval?

Answer #1

Do women tend to spend more time on housework than men? If so,
how much more? A study reported the results shown in the table to
the right for the number of hours spent on housework per week.
Complete parts a through d below.
Housework Hours
Gender
Sample Size
Mean
Standard
Deviation
Women
474
30.5
13.7
Men
493
15.0
12.6
A. Based on this study, calculate how
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housework than...

You want to estimate the mean time
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It has been found that the average time Internet users spend
online per week is 18.3 hours. A random sample of LaTeX: n=48 n =
48 teenagers indicated that their mean amount of Internet time per
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deviation of LaTeX: \sigma=5.7 σ = 5.7 hours. At the LaTeX:
\alpha=0.02 α = 0.02 level of significance, can it be concluded
that the mean time differs from 18.3 hours per...

Consider the following data (Atlanta = population 1, Chicago =
population 2):
City
Atlanta (1)
Chicago (2)
Sample size
100 individuals
100 individuals
Own a Toyota
29 individuals
33 individuals
At population level, is the proportion of Toyota owners in
Atlanta smaller than the proportion of Toyota owners in Chicago?
Formulate the hypothesis.
A. Ho: p1-p2 = 0 . Ha: p1-p2≠0
B. Ho: p1-p2 ≥ 0 . Ha: p1-p2 < 0
C. Ho: p1-p2 ≤ 0 . Ha: p1-p2 >...

Do shoppers at the mall spend more money on average the day
after Thanksgiving compared to the day after Christmas? The 60
randomly surveyed shoppers on the day after Thanksgiving spent an
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$133. Their standard deviation was $32. What can be concluded at
the α = 0.10 level of significance? For this study, we should use
t-test for...

1. A nutritionist wants to determine how much time nationally
people spend eating and drinking. Suppose for a random sample of
933 people age 15 or older, the mean amount of time spent eating
or drinking per day is 1.92 hours with a standard deviation of 0.58
hour.
Determine and interpret a 99% confidence interval for the mean
amount of time Americans age 15 or older spend eating and drinking
each day.
2. A simple random sample of size n...

You are interested in the time students usually spend studying
for the final exams. Assume that the total study time during the
final week in the semester among students is approximately Normally
distributed with mean= 12 hours and a standard deviation = 4.4
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n=50 and compute the average total study time.
What is the standard deviation for the average studying time in
hours?
Imagine you sample 50...

1)Given a sample mean is82, the sample size is 100and the
population standard deviation is 20. Calculate the margin of error
to 2 decimalsfor a 90% confidence level.
2)Given a sample mean is 82, the sample size is 100 and the
population standard deviation is 20. Calculate the confidence
interval for 90% confidence level. What is the
lower limit value to 2 decimals?
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A graduate student believed that, on the average, college
students spend more time on the Internet compared to the rest of
the population. She conducted a study to determine if her
hypothesis was correct. The student randomly surveyed 25 students
and found that the average amount of time spent on the Internet was
11.75 hours per week with a SD = 1.5 hours. The last census found
that, on the average, people spent 11 hour per week on the
Internet....

Colleen knows the population mean she is working with is 100 and
the standard deviation of this population is 10. She predicts she
will obtain a sample mean of 101 with a sample size of 100. She
sets her alpha level to 0.05 and her research hypothesis is
one-tailed. What is the power of her study?
Question 15 options:
73.89%
55.34%
26.11%
94.89%

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