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I don't know how to use Minitab software. and I actually need a**More:**

**I don't know how to use Minitab software. and I actually need a picture that I can copy paste in the word file.It is the Problem number 3. **

**3. Using Minitab, repeat problems 1A and 2C. Copy and paste the output to your assignment.**

Confidence Intervals (three problems)

Confidence Intervals (three problems)

1. A college admissions officer for the school?s online undergraduate program wants to estimate the

mean age of its graduating students. From a previous study the standard deviation was approximately

two (2) years.

A. The administrator took a random sample of 40 from which the mean was 24 years and the standard

deviation was 1.7 years. From this sample, what is the 95 confidence interval for the population mean of

interest?

N = 40 24 ± 2.042× 1.7

? 40 ( ) = 24 ± 2.042× 0.269 = 24 ± 0.549 = 23.451 ? ? ? 24.549 B. What is a proper statistical interpretation of the confidence interval you calculated in Part B?

95% confidence that the school?s online undergraduate program?s mean age of its graduating students is

between 23.451 to 24.549 years.

C. Would a 95% confidence interval using a larger sample size, assuming the same sample mean and

standard deviation, be wider or narrower than the interval using the administrator?s sample size?

Explain why it would be wider or narrower.

Narrower, because sample size in part A is larger than in part B.

D. Typically, undergraduate resident students graduate by the time they are 23 years of age. Does the

interval you calculate in Part B reflect that the average graduating student age of the online student is

older than that of the graduating resident student? Explain.

Yes, because the interval doesn?t include 23. So, average graduating of online students is older than that

of the graduating resident student.

2. You run the advertising department for a large hotel chain. You want to estimate with 98% confidence

the proportion of your guests who make reservations using your hotel?s website. You wish for the

estimate to be within 5% of the true proportion.

A. With no known prior proportion to go by, calculate minimum sample size required. 0.05

(¿¿ 2)

n = (2.332 )(0.5)( 0.5) = 543

¿ B. Assuming the data was gathered via survey methods that had approximately a 40% response rate,

what would be the actual sample size needed to produce the minimum size required in Part A? 543

N = 0.4 = 1358 C. A study was done that included responses from 600 randomly sampled quests, of which 55% said they

register using the hotel website. Calculate the 98% confidence interval for the population of interest.

p-hat = 0.55

n= 600 ± ( 600 )( 0.55 ) = 330 ? ? ? 270 (0.55)±(2.33) ×( ( 0.55 ) (0.45) 0.5

)

= 0.503 ? p ? 0.597

600 D. What is a proper statistical interpretation of the confidence interval you calculated in Part B?

98% confidence that proportion of your guests who make reservations using hotel?s website is between

50.3% to 59.7%.

E. If the more common 95% confidence were calculated, assuming the same sample proportion and

sample size from Part C, would the resulting interval be wider or narrower? Do not recalculate the

interval, but instead explain why it would be wider or narrower.

Narrower, because 95% confidence interval is narrower than 98% confidence interval. So, the result from

95% is less than that of 98%, which makes small margin of error. Thus, narrows the interval.

F. Using the 98% confidence interval you calculated in Part C, would it be reasonable to conclude that

more than 50% of the hotel guests register using the hotel website? Using your answer to Part E, what

would be your answer if the interval were at 95% confidence?

Yes, because interval reaches just above 50%. Since it?s narrower for 95%, it shows range as well that

reaches above 50%.

3. Using Minitab, repeat problems 1A and 2C. Copy and paste the output to your assignment.

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This question was answered on: * Feb 21, 2020 *

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