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i don’t need the process. i just need the answers.I questionary is in a weBWork system.if there is any wrong answer it wont accept it and the system would let me know which one is wrong, then i will let you know which one isnt correct.

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A random sample of 20 U.S. weddings yielded the following data on wedding costs in dollars:

8944, 23989, 18464, 28681, 30364, 18838, 10644, 32438, 32115, 5510,

15200, 11732, 31979, 29310, 11035, 12701, 27406, 20589, 19212, 5442

a) Use the data to obtain a point estimate for the population mean wedding cost,

weddings. Note: The sum of the data is 394593.

b) Is your point estimate in part (a) likely equal to

4) A random sample of

, of all recent U.S.

exactly (yes or no)?

measurements was selected from a population with standard

deviation

and unknown mean

of the following situations:

. Calculate a

% confidence interval for

for each

(a)

(b)

(c)

(d) In general, we can say that for the same confidence level, increasing the sample size the margin of

error (width) of the confidence interval. (Enter: ”DECREASES”, ”DOES NOT CHANGE” or

”INCREASES”, without the quotes.)

6) For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.45 million dollars.

Assuming a population standard deviation gross earnings of 0.46 million dollars, obtain a 99% confidence

interval for the mean gross earnings of all Rolling Stones concerts (in millions).

Confidence interval: ( , ).

b) Which of the following is the correct interpretation for your answer in part (a)?

A. We can be 99% confident that the mean gross earnings for this sample of 30 Rolling Stones concerts lies

in the interval

B. There is a 99% chance that the mean gross earnings of all Rolling Stones concerts lies in the interval

C. We can be 99% confident that the mean gross earnings of all Rolling Stones concerts lies in the interval

D. None of the above

7) A random sample of

observations produced a mean of

normal distribution and a standard deviation

(a) Find a

% confidence interval for

(b) Find a

% confidence interval for

from a population with a

.

Note: Round your answers to 4 decimal places.

8) A study of the career paths of hotel general managers sent questionnaires to an simple random sample

of 200 hotels belonging to major U.S. hotel chains. There were 92 responses. The average time these 92

general managers had spent with their current company was 10.82 years. (Take it as known that the

standard deviation of time with the company for all general managers is 3 years.)

Find the margin of error for an 85% confidence interval to estimate the mean time a general manager had

spent with their current company. Round your answer to 2 decimal places.

Answer: years

9) Starting salaries of 110 college graduates who have taken a statistics course have a mean of $44,260.

Suppose the distribution of this population is approximately normal and has a standard deviation of

$8,825.

Using an 95% confidence level, find both of the following:

(NOTE: Do not use commas nor dollar signs in your answers.)

(a) The margin of error E:

(b) The confidence interval for the mean

:

10) Suppose you have selected a random sample of

distribution. Compare the standard normal

the following confidence intervals.

(a)

% confidence interval

measurements from a normal

values with the corresponding

values if you were forming

(b)

% confidence interval

(c)

% confidence interval

11) Periodically, the county Water Department tests the drinking water of homeowners for contaminants

such as lead and copper. The lead levels in water specimens collected in 1998 for a sample of 5 residents of

a subdevelopement of the county is shown below.

lead (

g/L )

(a) Fill in the following chart to determine the sample mean

and sample standard deviation,

Then,

(b) Determine the critical value,

.

=

(c) Determine the maximal margin of error.

=

(d) Construct a

subdevelopment.

% confidence interval for the mean lead level in water specimans of the

.

12) Justin is interested in buying a digital phone. He visited 10 stores at random and recorded the price of

the particular phone he wants. The sample of prices had a mean of $389.19 and a standard deviation of

$10.33.

(a) What critical value

distribution?

should be used for a 95% confidence interval for the mean,

, of the

=

(b) Calculate a 95% confidence interval for the mean price of this model of digital phone:

(Enter the smaller value in the left answer box.)

$ to $

Note: Round your answer to 2 decimal places.

13) A random sample of 18 size AA batteries for toys yield a mean of 3.26 hours with standard deviation,

1.42 hours.

(a) Find the critical value,

, for a 99% confidence interval.

=

(b) Find the margin of error for a 99% confidence interval. hour(s).

Note: Round off your answer to 2 decimal places.

14) A study is conducted to determine if a newly designed text book is more helpful to learning the

material than the old edition. The mean score on the final exam for a course using the old edition is 75.

Ten randomly selected people who used the new text take the final exam. Their scores are shown in the

table below.

Person A B C D E F G H I J

Test Score 71 86 95 80 90 96 69 88 77 73

Use a

significance level to test the claim that people do better with the new edition. Assume the

standard deviation is 10.2. (Note: You may wish to use statistical software.)

(a) What kind of test should be used?

A. One-Tailed

B. Two-Tailed

C. It does not matter.

(b) The test statistic is (rounded to 2 decimals).

(c) The P-value is

(d) Is there sufficient evidence to support the claim that people do better than 75 on this exam?

A. Yes

B. No

(e) Construct a

% confidence interval for the mean score for students using the new text.

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Tags:

standard deviation

Margin of error

point estimate

gross earnings

correct interpretation

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