If 32 students from the sample use Samsung smartphone and the accuracy level will be within + 3%. At 95% confidence level, determine the sample size n
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Q: A poll of 1995 U.S. adults found that 1818 U.S. adults, or 91.128% regularly used Facebook as a news…
A: Given that: Sample proportion, p^=91.128%=0.91128 The sample size, i.e., the sampled no. of US…
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A: Given Information: Sample size (n) = 300 Proportion of students who smoke (p) = 165/300 = 0.55
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A: Sample sizen=10 The 90% confidence interval for the mean differences is [-2, 5]. At the 90% level…
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
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Q: uppose a researcher collected data with n = 200 and came up with a sample mean of 83 and 95%…
A: Given n=200 Mean=83
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A: Given that, a random sample of 16 movies has been selected from a normally distributed population.…
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Q: If n = 320 and p (p-hat) = 0.7, construct a 90% confidence interval. Give your answers to three…
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Q: A report statesb that 33% of home owners had a vegetable garden. how large a sample is needed to…
A: p=33%=0.33, E=5%=0.05
Q: If n=190 and p' (p-prime) = 0.6, construct a 99% confidence interval. Give your answers to three…
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Q: A poll of 1785 U.S. adults found that 900 U.S. adults, or 50.42% regularly used Facebook as a news…
A: Given that: Sample proportion, p^=0.5042 sample size (n) =900
Q: A report statesb that 44% of home owners had a vegetable garden. how large a sample is needed to…
A: Answer: Using the given data, A report stated that 44% of home owners had a vegetable garden, P =…
Q: Out of 400 people sampled, 164 had kids. Based on this, construct a 95% confidence interval for the…
A: Given:Sample size = n = 400 peoplePeople having kids = x = 164Confidence level = 0.95So, α = 1 –…
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Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 600…
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A: Given information
Q: Out of 500 people sampled, 355 had kids. Based on this, construct a 90% confidence interval for the…
A: By using the given information, we have Out of 500 people sampled, 355 had kids. 90% confidence…
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Q: calculate t value if df is 24 and 95% confidence interval
A: Given Information: Degrees of freedom df=24 Confidence level = 95% = 0.95 Significance level…
Q: Out of 300 people sampled, 207 had kids. Based on this, construct a 90% confidence interval for the…
A: Givensample size(n)=300x=207sample proportion(p^)=xn=207300=0.69confidence level=90%
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: Type only the upper interval by using type your answer to the nearest hundredths place.
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Q: A sample of 175 U.S. adults is composed of 100 women and 75 men. What is the estimate of the true…
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Q: If n=580 and p' (p-prime) = 0.5, construct a 90% confidence interval. Give your answers to three…
A: The population proportion CI is, p^±Zp^ (1-p^)nwherep^=sample proportionn=sample size
Q: Out of 200 people sampled, 132 had kids. Based on this, construct a 90% confidence interval for the…
A: given data, x=132n=200CI=0.90α=1-0.90=0.10p^=xn=132200=0.66
Q: Suppose that a 92% Confidence Interval for estimating the average heights, in inches, of all college…
A: we use the following, margin of error = Z a/2 * (standard error)where, Za/2 = Z-table…
Q: Out of 400 people sampled, 332 had kids. Based on this, construct a 95% confidence interval for the…
A: sample size (n) = 400favorable outcomes (x) = 33295% ci p.
Q: Out of 400 people sampled, 304 had kids. Based on this, construct a 95% confidence interval for the…
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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…
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A: Given, Proportion of college students use their smartphones for study = 0.61Total number of students…
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A: We want to find the 95% CI for population proportion
Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 400…
A: Solution: Let X be the number of parents had kids and n be the sample number of adults. From the…
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A: Let p is defined as the population proportion of people with kids. A sample of 100 people considered…
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 300 people sampled, 258 had kids. Based on this, construct a 99% confidence interval for the…
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Q: A teacher gives a test to his class of 25 students. Since he wants to quickly obtain an estimate of…
A: The class of 25 students is considered as the population. Sample size n =12 Sample mean x¯=82 Sample…
Suppose Berty, chose from the population of students a sample of 64 students and he confirms that the 99%. Confidence interval of the mean time spent in the sample is between [ 250 ; 350 ] minutes.
If 32 students from the sample use Samsung smartphone and the accuracy level will be within + 3%.
At 95% confidence level, determine the
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- Out of 500 people sampled, 415 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 four places < p <A researcher comparing the means of dependent samples consisting of 30 matched pairs and obtained a mean difference of 5 and a standard error of the mean difference of 2. The confidence interval for the mean difference was (1.6, 8.4). What confidence level was used in constructing the interval?Solve it correctly please. I will rate accordingly with 3votes.
- Out of 400 people sampled, 212 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to three placesOut of 600 people sampled, 180 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.Nearly one-half of Americans ages 25 to 29 are unmarried. How large a sample is necessary to estimate the true proportion of unmarried Americans in this age group within 2 percentage points with 95% confidence?In a sample of 460 adults, 281 had children. Construct a 90% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three places < p <Out of 300 people sampled, 270 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to three places. < p <If n=550 and p' (p-prime) = 0.62, construct a 99% confidence interval.Give your answers to three decimals.Construct a 96% confidence interval if x=75 and n=200. Round your answers to 3 decimal places.We wish to estimate what percent of adult residents in a certain county are parents. Out of 200 adult residents sampled, 34 had kids. Based on this, construct a 90% confidence interval for the proportion, p, of adult residents who are parents in this county. Give your answers as decimals, to three places.We randomly sample 600 people in a specific town. 342 of them like cats. Based on this, construct a 95% confidence interval for the true population proportion of people in this town who like cats. Give your answers as decimals, rounded to four places.SEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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