PLEASE RESPONSE QUESTION 2 AND 3. THANK YOU PLEASE RESPONSE QUESTION 2 AND 3. TH

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PLEASE RESPONSE QUESTION 2 AND 3. THANK YOU

PLEASE RESPONSE QUESTION 2 AND 3. THANK YOU

Assignment 4: Confidence Intervals - Submit Files Hide Submission Folder Information Submission Folder Assignment 4: Confidence Intervals Instructions Complete the problems below. Show some of your steps and explain the answer, too. If you use StatCrunch to compute a confidence interval, shot a screen shot of your output, or type out your input (so your work can be error-checked) Question #1 A 95% confidence interval for a population proportion yielded the interval (.345, .455). (This is enough information to answer the following questions. Don't over-think these questions!) 1. Compute the margin of error 2. Compute the sample proportion. 3, Will a 90% confidence interval be wider or narrower? Explain. Question #2 Many companies are interested in knowing the percentage of adults who buy clothing online. How many adults must be surveyed in order to be 95% confident that the sample percentage is in error by no more than three percentage points? 1. Use a recent result from the Census Bureau: 70% of adults buy clothing online 2. Assume that we have no prior information suggesting a possible value of the proportion. Question #3 A Pew Research Center poll of 1007 randomly selected adults showed that 42% of those survey feel the Internet is a bad influence on morality 1. What is the value of n? 2. What is the value of p? 3. Compute the margin of error, E, that corresponds to a 95% confidence level 4. Compute the 95% confidence interval estimate of the population proportion p 5. Based on the results, can we safely conclude that more than half of adults feel the Internet is a bad influence on morality? 6. Assuming that you are a newspaper reporter, write a brief statement that accurately describes the results and includes all of the relevant information.

Explanation / Answer

2.
TRADITIONAL METHOD
given that,
possibile chances (x)=70
sample size(n)=100
success rate ( p )= x/n = 0.7
I.
sample proportion = 0.7
standard error = Sqrt ( (0.7*0.3) /100) )
= 0.046
II.B9
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.046
= 0.09
III.
CI = [ p ± margin of error ]
confidence interval = [0.7 ± 0.09]
= [ 0.61 , 0.79]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possibile chances (x)=70
sample size(n)=100
success rate ( p )= x/n = 0.7
CI = confidence interval
confidence interval = [ 0.7 ± 1.96 * Sqrt ( (0.7*0.3) /100) ) ]
= [0.7 - 1.96 * Sqrt ( (0.7*0.3) /100) , 0.7 + 1.96 * Sqrt ( (0.7*0.3) /100) ]
= [0.61 , 0.79]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.61 , 0.79] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion

3.

TRADITIONAL METHOD
given that,
possibile chances (x)=422.94
sample size(n)=1007
success rate ( p )= x/n = 0.42
I.
sample proportion = 0.42
standard error = Sqrt ( (0.42*0.58) /1007) )
= 0.016
II.B9
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.016
= 0.03
III.
CI = [ p ± margin of error ]
confidence interval = [0.42 ± 0.03]
= [ 0.39 , 0.45]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possibile chances (x)=422.94
sample size(n)=1007
success rate ( p )= x/n = 0.42
CI = confidence interval
confidence interval = [ 0.42 ± 1.96 * Sqrt ( (0.42*0.58) /1007) ) ]
= [0.42 - 1.96 * Sqrt ( (0.42*0.58) /1007) , 0.42 + 1.96 * Sqrt ( (0.42*0.58) /1007) ]
= [0.39 , 0.45]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 95% sure that the interval [ 0.39 , 0.45] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion

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