Let p represent the population proportion of all female students who received a grade of A on this test. Use a 85% confidence interval to estimate p to four decimal places if possible. < p <
Q: Out of 100 people sampled, 47 had kids. Based on this, construct a 99% confidence interval for the…
A: X = 47 N = 100 99% confidence level
Q: If n = 100 and p (p-hat) = 0.9, construct a 95% confidence %3D interval. Give your answers to three…
A: Givensample size(n)=100p^=0.9α=1-0.95=0.05α2=0.025zc=zα2=z0.025=1.96 (from z table)
Q: Out of 500 people sampled, 390 had kids. Based on this, construct a 99% confidence interval for the…
A: Find z-critical value: For the two-tail test, Area on each tail. Area to the left of the normal…
Q: Out of 600 people sampled, 486 had kids. Based on this, construct a 95% confidence interval for the…
A: It is given that Favourable cases, X = 486 Sample size, n = 600
Q: construct a 99% confidence interval for the true population proportion of people with kids.
A: Given : n=600 , X=84 , α=0.01 Our aim is to construct a 99% confidence interval for the true…
Q: Out of 500 people sampled, 375 had kids. Based on this, construct a 99% confidence interval for the…
A: Number of people (n) = 500 Number of people had kids (n) = 375 Find z-critical value:…
Q: You recently sent out a survey to determine if the percentage of adults who use social media has…
A: It is given that sample size (n) is 2,953 and the number of respondents stated that they currently…
Q: Identify the t-score for a 90% confidence interval if the sample size is 31. t =
A: c1=90%n=31α=1-0.90=0.10df=n-1=31-1=30
Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 600…
A: Obtain the 95% confidence interval for the true proportion of adult residents who are parents in…
Q: Out of 500 people sampled, 280 had kids. Based on this, construct a 95% confidence interval for the…
A: Given Information:n=500x=280c=95%=0.95
Q: lculating the confidence interval.
A: x Freq 1 3 2 5 3 19 4 7 5 6 Total 40
Q: Out of 600 people sampled, 492 had kids. Based on this, construct a 99% confidence interval for the…
A:
Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 100…
A: According to the given information, we have Out of 100 adult residents sampled, 88 had kids. 99%…
Q: Out of 200 people sampled, 166 had kids. Based on this, construct a 90% confidence interval for the…
A:
Q: Out of 400 people sampled, 200 had kids. Based on this, construct a 99% confidence interval for the…
A: Number of peoples (n) =400 Number of peoples with kids (x) =200 Find z-critical value:…
Q: A confidence interval for the difference in the number of friends that men and women have was found…
A: The formula of the midpoint of the confidence interval formula is,
Q: Out of 300 people sampled, 135 had kids. Based on this, construct a 99% confidence interval for the…
A: According to the given information in this question We need to construct 99% confidence interval
Q: Out of 200 people sampled, 78 had kids. Based on this, construct a 99% confidence interval for the…
A:
Q: Out of 500 people sampled, 150 had kids. Based on this, construct a 99% confidence interval for the…
A: To construct a 99% confidence interval for the true population proportion of people with kids, we…
Q: t of 300 people sampled, 237 had kids. Based on this, construct a 90% confidence interval for the…
A: given data, n=300x=had kids=237p^=xn=237300=0.79CI=0.90α=1-0.90=0.10
Q: Out of 500 people sampled, 245 had kids. Based on this, construct a 95% confidence interval for the…
A: From the provided information, Out of 500 people sampled, 245 had kids. That is x = 245 and n = 500…
Q: Out of 100 people sampled, 14 had kids. Based on this, construct a 95% confidence interval for the…
A: Given data, x=14 n=100 p=x/n= 14/100= 0.14 The z value at 95% confidence is Z=1.96
Q: Out of 500 people sampled, 470 had kids. Based on this, construct a 95% confidence interval for the…
A: The sample size n= 500 The sample proportion is 470/500 = 0.94. The confidence level is 95%. The…
Q: Out of 500 people sampled, 415 had kids. Based on this, construct a 95% confidence interval for the…
A:
Q: Out of 500 people sampled, 280 had kids. Based on this, construct a 99% confidence interval for the…
A: Given information Total people sampled (n) = 500 No of people having kids(x) = 280 Confidence level…
Q: Out of 200 people sampled, 112 had kids. Based on this, construct a 95% confidence interval for the…
A:
Q: Out of 100 people sampled, 27 had kids. Based on this, construct a 95% confidence interval for the…
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Q: In survey of 7000 women, 4431 says they change their nail polish once a week. Construct a 95%…
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Q: In a sample of 580 adults, 348 had children. Construct a 95% confidence interval for the true…
A: Given that data Sample size n =580 Favorable cases x =348 Sample proportion p^=x/n =348/580 =0.6
Q: Out of 500 people sampled, 280 had kids. Based on this, construct a 90% confidence interval for the…
A: Given that n=500
Q: Out of 300 people sampled, 66 had kids. Based on this, construct a 99% confidence interval for the…
A:
Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 600…
A: The objective of this question is to construct a 90% confidence interval for the proportion of adult…
Q: Identify the t score for a 99% confidence interval if the sample size is 35 t=
A:
Q: If n=230 and p (p-hat) = 0.61, construct a 95% confidence interval. Give your answers to three…
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Q: Identify the t-score for a 99% confidence interval if the sample size is 70. t = The answer isnt…
A:
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- A simple random sample of 5 recently hospitalized migraine sufferers were given a new treatment; the results are listed below. Construct the 90% confidence estimate of the mean difference in the number of migraines per month before and after the treatment. Summary of # of migraines before and after treatment (5/3, 9/9, 7/4, 5/2, 3/4) Subject A Subject B Subject C Subject D Subject E # of migraines before 5 9 7 5 3 # of migraines after 3 9 4 2 4 **Let the first population be of the "before" results and the second population be of the "after" results.** The right hand critical value is . Round to 2 decimal places. The error is . Round to 2 decimal places. The mean difference in migraines per month is between and .How many pairs of shoes, on average, do female teens have? To find out, an AP®® Statistics class selected an SRS of 20 female students form their school. Then they recorded the number of pairs of shoes that each student reported having. This class also asked an SRS of 20 boys at their school how many pairs of shoes they have. A 95% confidence interval for ??−??=μG−μB= the true difference in the mean number of pairs of shoes for girls and boys is 10.9 to 26.5. Interpret the confidence interval. We are 95% confident that the interval from 10.9 to 26.5 captures the true difference (Girls – Boys) in the mean number of pairs of shoes owned by all girls and boys at this school. We are 95% confident that the interval from 10.9 to 26.5 captures the true mean difference (??????−????)(μGirls−Boys) in the number of pairs of shoes owned by all girls and boys at this school. Girls at this school have between 10.9 and 26.5 more pairs of shoes than boys 95% of the time. We are 95% confident…Find the critical t value to use in making an 80% confidence interval for a mean if the sample size is 35. Write your answer with 3 decimal places.
- This is a matched pairs design. Find a 99% confidence interval of the mean fuel mileage improvement uD from using premium gasoline rather than regular to two decimal places. Give you interval and a brief description of you calculations.In a recent year, a sample of the South American Redeye Piranha was collected by fishermen on a particular river. Wildlife biologists regard this sample as a random sample of all Redeye Piranha in that river. The mean length of the 60 fish in the sample was 12.88 cm, and a 99% confidence level for the true mean length of all Redeye Piranha in that river is 12.09 cm to 13.67 cm. Explain what would happen to the length of the interval if the confidence level were decreased to 95%.In a recent year, a sample of the South American Redeye Piranha was collected by fishermen on a particular river. Wildlife biologists regard this sample as a random sample of all Redeye Piranha in that river. The mean length of the 60 fish in the sample was 12.88 cm, and a 99% confidence level for the true mean length of all Redeye Piranha in that river is 12.09 cm to 13.67 cm. Explain what would happen to the length of the original interval if the sample size were decreased to 30 fish. The confidence interval will be (wider, narrower, the same) because decreasing the sample size (increases, decreases, does not affect) the margin of error.
- Using the data from the 2018 professional baseball season, the American League (AL) had 10 players bat over .300 out of 179 players who played at least 82 games. In the national League (NL), there were 10 players who batted over .300 out of 196 players who played at least 82 games. Find a 90% confidence interval for the difference in porportion of .300+ hitters in the AL and NL.What is the correct ans.? The change in confidence level (increase, decrease, does not change, none of the choices) as the size of the confidence interval containing the mean decreases.On average, adults spend approximately 5 hours per night in NREM (non-rapid eye movement) sleep. A sleep study selects a random sample of adults and gives them each a weighted blanket. They would like to know if using the weighted blanket changes the mean number of hours per night of NREM sleep. We want to test = 5 versus ≠ 5 where = the true mean hours of NREM sleep per night for adults who use a weighted blanket. A 99% confidence interval for the true mean hours of NREM sleep per night for adults who use a weighted blanket is (5.1, 5.8) hours. Based on the confidence interval, what conclusion would you make for a test of these hypotheses? The 99% confidence interval include as a plausible value, so we would H0. We convincing evidence that the true mean hours of NREM sleep per night for adults who use a weighted blanket 5 hours.
- A sample of 82 Oscar-winning actresses had a mean age of 35.9 years when they won the award. s = 11.1 years. Use a 0.01 significance level to test the claim that the mean age of actresses when they win an Oscar is greater than 33 years. Once you have your conclusion, find the 99% confidence interval of the mean age and compare your results.A repeated measures study is done to measure the change in IQ test scores taken on Monday versus those taken on Friday. There were n=9 participants in the study. The mean difference was MD =-5 points and the standard error for the mean difference was SMD=0.51. Construct a 95% confidence interval to estimate the size of the population mean difference.What proportion of students are willing to report cheating by other students? A student projectput this question to an SRS of 172 undergraduates at a large university: "You witness two students cheating on aquiz. Do you go to the professor?" The Minitab output below shows the results of a significance test and a 95%confidence interval based on the survey data.El SessionTest and Cl for One ProportionTeat of p = 0.5 vs p not = 0.5Sample1X 1 Sample P95: CI246 439 0.560364 (0.513935, 0.606794)