It was observed that 4 out of 10 randomly selected students went to the cinema. Find the 95% confidence interval estimate for the proportion of moviegoers in the population, \pi. P( 0,20 <1< 0,30 ) = 0,95 OA) P( 0,30
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A: Given information- Sample size (n) = 59 Mean x̅ = 29.5 Standard deviation (s) = 5.2 Significance…
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A: Given Data: x¯=21.74n=400σ=45.65 As per 68 – 95 – 99.7 Rule 68% of the data falls within 1…
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Q: A study was done to determine whether there is a relationship between snoring and the risk of heart…
A: a. From the given information, x1=89, n1=1105, x2=22 and n2=1379.
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Q: It was observed that 30 out of 100 randomly selected students smoked. Estimate the 95% confidence…
A: Given,sample size(n)=100no.of students smoked(x) =30sample proportion(p^)=xnsample…
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A: Sample sizen=10 The 90% confidence interval for the mean differences is [-2, 5]. At the 90% level…
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Q: A sample of 5 water speciments selected for treatment found a sample mean of arsenic concentration…
A: Solution: Let X be the arsenic concentration. From the given information, x-bar=24.3, s=4.1 and n=5.…
Q: A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2…
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Q: A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older…
A: Given information Sample size (n) = 19 Sample mean x̅ = 77.7 min Standard deviation (s) = 21.1 min…
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Q: An SRS of 350 high school seniors gained an average of ?¯=20.58 points in their second attempt at…
A: It is given that,Sample mean = 20.58Standard deviation, sigma is 48.19.The sample size, n is 350.
Q: Calculate the 95% confidence interval for the standard deviation. b) Can it be said at the 5%…
A: Given that sample size n = 30 Sample SD , s = 0.37 The critical values of Chi square for 95%…
Q: Data on 4100 college graduates show that the mean time required to graduate with a bachelor's…
A: The sample mean = 5.39. Standard deviation is 1.59. The sample size, n is 4100. From the…
Q: has a sample mean foot length of 27.5 cm. Assuming that the standard deviation of foot lengths for…
A: We have given that Sample size n = 64 Sample mean x = 27.5 Standard deviation σ= 2 95% confidence…
Q: A random sample of 10 one year old Rainbow trout in a fish hatchery showed the average weight was…
A: From the given information Sample mean =1.2 Standard deviation = 0.8 We want to find 99% confidence…
Q: A recent study of 1000 college students in UAE and Lebanon showed that 55% of them sleep less than 7…
A: Given data, p=55%=0.55 n=1000 Construct a 95% confidence interval?
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Q: An SRS of 450 high school seniors gained an average of x¯=21.72 points in their second attempt at…
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Q: Blood platelet counts of women have a bell-shaped distribution with a mean of 255.1 and a standard…
A: The 68-95-99.7% (Empirical) Rule: For a normal distribution, the empirical rule (68-95-99.7% Rule)…
Q: Sludge samples are collected with an automatic sampling device from lake sediment. The device takes…
A: Confidence interval : Confidence interval is a range of values it contains the true population…
Q: A random sample of 358 graduating high school seniors was polled across a particular tristate area,…
A: No. of high school graduates in the sample(n)=358 No. of high school graduates who had taken the…
Q: A researcher collected a random sample of n = 46 squirrels and found 6 with a broken tail. Find a…
A: From the provided information Sample size (n) = 46, x = 10 The sample proportion can be obtained as:
Q: A random sample of 64 student workers showed that they called in sick from work an average of 10…
A: Given Sample size n=64, sample mean x̄=10, standard deviations σ=3 Construct 95% CI for population…
Q: A random sample of hybrid vehicle fuel consumptions has a sample mean of x=53.2 mpg and sample…
A: From the provided information, According to the empirical rule: Approximately 68% of the data lie…
Q: Data on 4400 college graduates show that the mean time required to graduate with a bachelor's…
A: The sample mean = 5.65 Standard deviation is 1.15 The sample size, n is 4400. From the…
Q: A random sample of 64 students at a university showed an average age of 25 years and a sample…
A: Let be the population mean age of all students in the university.Given that,Sample size Average age…
Q: A sample of 36 different payroll departments found that the employees worked an average of 242.5…
A: given data sample size (n) = 36sample mean ( x¯ ) = 242.5sample standard deviation (s) =21.295% ci…
Q: A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2…
A: Given data n=59 mean=29.5 standard deviation=5.2 confidence interval is 90%
Q: A researcher comparing the means of dependent samples consisting of 30 matched pairs and obtained a…
A: Given that: Sample size(n) =30 Mean difference =5 Standard error of mean difference =2 The…
Q: a data set includes 109 body temperatures of healthy adult humans having a mean of 98.3 f and a…
A: Given, sample size n=109, mean(x-)=98.3F and standard deviation(s) =0.54F.Formula for confidence…
Q: An SRS of 450 high school seniors gained an average of x = 20.80 points in their second attempt at…
A: The sample size is defined as: n=450. The sample average change in SAT score, x¯=20.80 And the…
Q: 2) (9.2) Suppose a random sample of 1001 Americans age 15 or older were asked "How much time do you…
A: We have given that, Sample mean (x̄) = 1.22, standard deviation (s) = 0.65 and sample size (n) =…
Q: A study was done to determine whether there is a relationship between snoring and the risk of heart…
A: Given: Consider Snoring to be data set 1 and Risk of heart disease to be data set 2.…
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Q: 11th - It was observed that 30 out of 100 randomly selected students smoked. Estimate the 95%…
A: we associate with the interval estimate is called confidence interval. This probability indicates…
Q: An SRS of 29 students at HCC gave an average height of 5.5 feet and a standard deviation of .1 feet.…
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- A simple random sample from a population with a normal distribution of 97 body temperatures has x=98.20°F and s=0.63°F. Construct a 99% confidence interval estimate of the standard deviation of body temperature of all healthy humans.A sample of size 65 gave a mean 7.4 and standard deviation as 1.6 then 95% confidence interval for population mean will be Select one: O (17.008, 17.792) O [27.008, 30.008] O (7.008, 7.792) O (9,265, 10.735)Estimate the average number of feathers that a mature tom turkey has with 90% confidence interval by using a sample of 32 turkeys that had a mean of 3450 feathers and a standard deviation of 25 feathers.
- A statistical study is made to compare the % of smokers in small rural towns to the overall smoking rate of the US. Of 1500 asults selected randomly from small towns it is found that 285 of them smoke. With a cofidence interval of (017, 0.21). The CDC reports that 15.1% of all American adults smoke. Does the confidence interval suggest that smoking rate is higher in small towns than in the country overall?A research company has carried out a poll to find out about people’s economic outlook. 894 of 1854 young adults (aged between 20–29 years) said that the economic situation is likely to improve over the next year. By contrast, 1292 of 2243 mature adults (aged between 30–39 years) agreed with this statement.) Calculate the upper limit of an 80% confidence interval for the difference between the proportions of people holding this optimistic view in the two age groups in the population. Form the confidence interval around a positive point estimate for the difference between the proportions. In order to obtain a positive point estimate, make sure you subtract the lower value from the higher value when calculating the difference between the sample proportions. Provide your answer as a number between 0 and 1 rounded to 4 decimal places.The oxide thickness of semiconductors for a random sample of n = 16 yielded a mean of 424 with a standard deviation of 8. It is assumed that oxide thickness is normally distributed. Calculate the 90% confidence interval for the mean oxide thickness of semiconductors. Round off final calculations to 1 decimal place. 1. What are the Lower limit and Upper limit
- An SRS of 400 high school seniors gained an average of x = 21.87 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation o = 48.66. We want to estimate the mean change in score µ in the population of all high school seniors. (a) Using the 68-95-99.7 Rule or the z-table (Table A), give a 95% confidence interval (a, b) for u based on this sample. (Enter your answers rounded to three decimal places. If you are using CrunchIt, adjust the default precision under Preferences as necessary. See the instructional video on how to adjust precision settings.) a: b: (b) Based on your confidence interval in part (a), how certain are you that the mean change in score u in the population of all high school seniors is greater than 0? O The upper endpoint of the interval is larger than 0, so we are 95% certain that the mean change in score in the population of all high school seniors is greater than 0. O We cannot be…A random sample of 100 accounts from a branch of certain bank, has mean balance of Rs. 18,600 with S.D.= 1160. A random sample of 80 accounts from another branch of same bank has mean balance of Rs. 15,200; with S.D.= Rs. 920. Find 90% confidence interval of difference between Mean balance of all accounts in both branches.Blood platelet counts of women have a bell-shaped distribution with a mean of 255.1 and a standard deviation of 65.4 (all units are 1000 cells/µL). Using the empirical-rule, what is the approximate percentage of women with platelet counts Within 2 standard deviation of the mean, or between 124.3 and 385.9 Between 189.7 and 320
- Milan n.nA study was done to determine whether there is a relationship between snoring and the risk of heart disease. Among 1,105 snorers in the study, 85 had heart disease, while only 21 of 1,379 nonsnorers had heart disease. In USE SALT (a) Determine a 95% confidence interval that estimates p, - p2 difference in proportions of snorers and nonsnorers who have heart disease. (Let p, be the proportion of snorers and p, be the proportion of nonsnorers. Round your answers to three decimal places.) |-0.079 X to -0.045A random sample of 64 students at a university showed an average age of 25 years and a sample standard deviation of 2 years. The 98 % confidence interval for the true average age of all students in the university is * 20.5 to 29.5. O24.4 to 25.6. 23.0 to 27.0. O 20.0 to 30.0.