An interval estimate with a 99% confidence means that there is 0.99 probability that the population parameter and the estimator are always equal.
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A: Given that, σx=1μx=10X¯observed=12
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Q: A leading magazine reported at one time that the average number of weeks an individual is unemployed…
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Q: leading magazine, (like Barron's) reported at one time that the avera nemployed is 31 weeks. Assume…
A: Given, mean (µ)= 31 standard deviation (σ) = 4 sample size(n) = 34
Q: Suppose that memory for unrelated words is distributed normally with a mean of µ = 50. A random…
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Q: (b) A new sample of 200 employed adults is chosen. Find the probability that less than 15% of the…
A: It is given that n =200 and p =0.12.
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A: If the population variance is known then the width of the confidence interval is given by 2zσn,…
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A: It is given that n=65 and p=0.12.
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A: The objective of this question is to construct a confidence interval for the population mean using…
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A: The confidence level is 99%. The level of significance is 1 – confidence level = 1 – 0.99 = 0.01.…
Q: which one/s are true? 1. a 95% degree if confidence means that the probability that true value of…
A: which one/s are true? 1. a 95% degree if confidence means that the probability that true value of…
Q: A researcher administers a treatment to a sample from a population with a mean of m=50. If the…
A: Claim : the treatment is expected to increase scores.
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A: question: Suppose the mean of the pilot sample follows a normal distribution with a mean of 7 (from…
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Q: If a variable has a very small standard deviation, and the outcomes of the variable will... a.…
A: Standard Deviation: Standard deviation measures how dispersed your data in from the mean.
Q: Which of the following is the minimum sample size necessary to construct a 99% confidence interval…
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Q: Suppose a random sample of adult women has a sample mean height of x¯=64.3 inches, with a sample…
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A: GIVEN DATA 90% ci for meanσ = 12E = 4 n = ?
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Q: Find the probability that a single randomly selected value is less than 18. P(X < 18) = Find the…
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A: Hii ! Thanks for posting the question . Since your question has more than three subparts , we have…
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- A sample of size 115 will be drawn from a population with mean 48 and standard deviation 12. Use the TI-83 Plus/TI-84 Plus calculator. Part 1 of 2 (a) Find the probability that x will be less than 45. Round the answer to at least four decimal places. The probability that x will be less than 45 is 0.0037 Part: 1/ 2 Part 2 of 2 (b) Find the 25 th percentile of x. Round the answer to at least two decimal places. The 25 th percentile is Skip Part Check Save For Later Subm © 2022 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Cente ......... .........a ... MacBook Air DII 80 F10 F7 F9 esc F4 F6 F2 F3 F1 @ 23 $ % & 1 2 3 4 6. 8.Suppose that the height of Asian males aged 35-44 in the United States is normally distributed with a standard deviation of 3 in. Jeremy takes a simple random sample of 16 such men and notices that none of them are unusually tall or unusually short. He then calculates the average of the sample to be 67 in. and is planning on using this to find a 90% z-confidence interval for the true mean height. Can Jeremy use this information to find the z-confidence interval? Complete the following sentences. The population distribution is and the population standard deviation is Therefore, the sample to find a 90% z-confidence interval.A researcher is interested in studying the use of hypnosis to relieve pain. She took a random sample of 16 subjects and measured their sensory ratings. The mean of the sensory ratings was found to be 7.5 and standard deviation to be 2.3. It is known that the sensory ratings are normally distributed. Construct a 99% confidence interval for the mean sensory ratings of the population. a) (5.81, 9.19) b) (6.27, 8.73) c) (6.36, 8.64) d) (6.02, 8.98)
- The width of the confidence interval estimate of the population mean μ is a function of only two quantities: the population standard deviation σ and the sample size nIf the random variable of X is normally distributed with a mean of 50 and a standard deviation of 10, approximately what % of all possible observed values of X will lie between 30 and 70.Body Mass Index: In a survey of adults with diabetes, the average body mass index (BMI) in a sample of 1928 women was 31.30 with a standard deviation of 0.3. The BMI in a sample of 1559 men was 30.50, with a standard deviation of 0.6. Part 1 of 2 (a) Construct a 99% confidence interval for the difference in the mean BMI between women and men with diabetes. Let H, denote the mean BMI for women with diabetes. Use the TI-84 calculator and round the answers to two decimal places. A 99% confidence interval for the difference in the mean BMI between women and men with diabetes is Part: 1 / 2 Part 2 of 2 (b) Does the confidence interval contradict the claim that the mean BMI is the same for both men and women with diabetes? The confidence interval (Choose one) this claim.
- An IQ test is designed so that the mean is 100 and the standard deviation is 23 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 90% confidence that the sample mean is within 8 IQ points of the true mean. Assume that o = 23 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation. The required sample size is (Round up to the nearest integer.)Agricultural scientists are testing a new chicken feed to see whether it increases the number of eggs laid. The scientists divided a flock in half and gave half of the chickens the new feed, while the other half were given a regular feed. The eggs laid in a one-year period for a random sample of 10 chickens using the new feed and 10 chickens using the regular feed are recorded. Assume that the population standard deviation of number of eggs laid is 12 eggs per year for both groups and that the number of eggs laid per year is normally distributed. Let the chickens given the new feed be the first sample, and let the chickens given the regular feed be the second sample. The scientists conduct a two-mean hypothesis test at the 0.05 level of significance to test if there is evidence that the new feed increases the number of eggs laid. (a) Which answer choice shows the correct null and alternative hypotheses for this test? Select the correct answer below: H0:μ1=μ2;…Time to Complete the Course Right 45 47 50 49 50 50 48 44 Left | 45 | 43 48 47 50 52 | 44 42 Assume a Normal distribution. What can be concluded at the the a = 0.01 level of significance level of significance? For this study, we should use t-test for the difference between two dependent population means a. The null and alternative hypotheses would be: p1 p2 Ні: p1 p2 b. The test statistic t v v = 1.199 (please show your answer to 3 decimal places.) c. The p-value = [1338 your answer to 4 decimal places.) d. The p-value is > va e. Based on this, we should fail to reject f. Thus, the final conclusion is that ... * (Please show v the null hypothesis. O The results are statistically insignificant at a = 0.01, so there is insufficient evidence to conclude that the population mean time to complete the obstacle course with a patch over
- A leading magazine, (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 31 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 31 weeks and that the population standard deviation is 4 weeks. Suppose you would like to select a random sample of 34 unemployed individuals for a follow-up study. Find the probability that a single randomly selected value is less than 30. P(X < 30) = Find the probability that a sample of size n = 34 is randomly selected with a mean less than 30. P(M < 30) = Enter your answers as numbers accurate to 4 decimal places. Submit Question % & 3 6 7 y 60 00 %24 4Construct a 99% confidence interval for u, - H, with the sample statistics for mean cholesterol content of a hamburger from two fast food chains and confidence interval construction formula below. Assume the populations are approximately normal with unequal variances Stats X, = 98 mg, s,= 3.81 mg, n, = 14 x = 65 mg, s, =222 mg, n, =20 Confidence interval when variances are not equal +t. n2 d.f. is the smaller of n, - 1 or n, - 1 Enter the endpoints of the interval
