Solution :
Given that,
n = 200
x = 42
Point estimate = sample proportion = = x / n = 42 / 200 = 0.210
1 - = 1 -0.210 = 0.790
At 95% confidence level
= 1-0.85% =1-0.85 =0.15
/2
=0.15/ 2= 0.075
Z/2
= Z0.075 = 1.44
Z/2 = 1.44
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.44 * ((0.210*(0.790) / 200)
= 0.041
A 85% confidence interval for population proportion p is ,
- E < p < + E
0.210 -0.041 < p < 0.210 +0.041
0.169 < p < 0.251
(0.169 , 0.251 )
A 85% confidence interval for population proportion p is=(0.169 , 0.251 )
A sociologist sampled 200 people who work in computer related jobs, and found that 42 of...
A sociologist sampled 200 people who work in computer-related jobs and found that 43 of them have changed jobs in the past 6 months. Can you conclude that more than 15% of computer-related workers have changed jobs in the past 6 months? Find the P-value and state a conclusion. The P-value is (Round the final answer to four decimal places.) We (Click to select) conclude that more than 15% of computer-related workers have changed jobs in the past 6 months.
Problem 4 (20 pts) A sociologist is interested in surveving workers in computer-related jobs to estimate the proportion of such workers who have changed the jobs within the past year. (a) (10 pts) In the absence of preliminary data, how large a sample must be taken to ensure that a 99% confidence interval will specify the proportion to within 20.05? (b) (5 pts) is the sample size obtnined in (a) large enough to ensure that confidence interval will specify the...
B: construct a 99% confidence interval for the proportion of people who performed volunteer work during the past year, round answers to at least three decimal C construct a 95 confidence interval for the proportion of people who performed volunteer work during the past year, round answers to at least three decimal D A sociologist states that 43% of Americans perform volunteer work in a given year. Does the confidence interval of 99 or 95 contradict this statement Volunteering: The...
Out of 200 people sampled, 68 had kids. Based on this, construct a 95% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to three places
Out of 200 people sampled, 120 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to four places _____ < p < ______
Out of 200 people sampled, 94 had children. Based on this, construct a 99% confidence interval for the true population proportion of people with children. Preliminary: Is it safe to assume that n≤5% of all people with children? No Yes Verify nˆp(1−ˆp)≥10. YOU MUST ROUND YOUR ANSWER TO ONE DECIMAL PLACE. nˆp(1−ˆp)= 4.Confidence Interval: What is the 99% confidence interval to estimate the population proportion? YOU MUST ROUND ANSWER TO THREE DECIMALS PLACES. 5. ?????? <p< ????
Out of 200 people sampled, 112 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to four places <p< Question Help: Message instructor Submit Question
An internet service provider (ISP) sampled 450 customers and found that 86 of them experienced an interruption in their service the previous month. (a) Construct a 95% confidence interval for the proportion of customers who experience an outage. (b) A quality assurance technician claims that no more than 16% of its customers experience an outage. Does your confidence interval support this claim?
(COMPUTE AND INTERPRET CONFIDENCE INTERVAL ESTIMATES) An Internet service provider sampled 540 customers and found that 75 of them experienced an interruption in high-speed service during the previous month. Construct a 90% confidence interval for the proportion of all customers who experienced an interruption. Find the critical values, find E , the margin of error, then compute and interpret your interval estimate with a full sentence. Critical value: (draw, label & shade) Margin of error, E : Confidence Interval:
Volunteering: The General Social Survey asked 1299 people whether they performed any volunteer work during the past year. A total of 528 people said they did. al. Part 1 of 3 (a) Find a point estimate for the proportion of people who performed volunteer work during the past year. Round the answer to at least three decimal places. The point estimate for the proportion of people who performed volunteer work during the past year is 0.4065 Part: 1/3 Part 2...