What is the formula to use when solving for the confidence interval estimate under T-Distribution? a. x̄ ± t(sn√)(sn) b. x̄ ± z(σn√)(σn) c. μ ± z(σn√)(σn) d. p̂ ± z (p̂ q̂ n−−√)
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What is the formula to use when solving for the confidence
x̄ ± t(sn√)(sn)
x̄ ± z(σn√)(σn)
μ ± z(σn√)(σn)
p̂ ± z (p̂ q̂ n−−√)
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- For a 98% confidence interval of the population mean based on a sample of n = 20 with s = 0.05, the critical value of t is:Q: Using the data posted below this question, determine if the company can claim, at 90% confidence, that the proportion of those that want to pay a lower premium for a higher deductible (i.e. Premium/Deductible = “Y”) is less than 0.30. Insurance Survey Age Gender Education Marital Status Years Employed Satisfaction* Premium/Deductible** 36 F Some college Divorced 4 4 N 55 F Some college Divorced 2 1 N 61 M Graduate degree Widowed 26 3 N 65 F Some college Married 9 4 N 53 F Graduate degree Married 6 4 N 50 F Graduate degree Married 10 5 N 28 F College graduate Married 4 5 N 62 F College graduate Divorced 9 3 N 48 M Graduate degree Married 6 5 N 31 M Graduate degree Married 1 5 N 57 F College graduate Married 4 5 N 44 M College graduate Married 2 3 N 38 M Some college Married 3 2 N 27 M Some college Married 2 3 N 56 M Graduate degree Married 4 4 Y 43 F College graduate Married 5 3 Y 45 M College graduate Married 15 3 Y…With α = .05, what is the critical t value for a one-tailed test with n = 25? t = 2.064 t = 1.711 t = 2.060 t = 2.708
- A simple random sample of men is obtained and the elbow-to-fingertip length of each man is measured. The population of those lengths has a distribution that is normal. The sample statistics are n=31, x=14.1, s=0.7. Determine the critical value of t and the margin of error, and then construct the 95% confidence interval estimate for the population mean. the critical value is t =?Use the given information to find the number of degrees of freedom, the critical values χ2L and χ2R, and the confidence interval estimate of σ. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. White Blood Counts of Women 98% confidence; n=146, s=1.96 (1000 cells/μL). df = x2L= x2r=…For parts a and b, use technology to estimate the following. a) The critical value of t for a 98% confidence interval with df=4. b) The critical value of t for a 90% confidence interval with df=111.
- In testing the hypotheses H 0 : p = 0.50 H0:p=0.50 versus H A : p ≠ 0.50 HA:p≠0.50 , a sample of 100 observations produces a test statistic z = 0 z=0 . What is the p p -value for this test? a1.0000 b 0.6915 c 0.5000 d 0.0000 cNone of the above.Use a t-distribution to find a confidence interval for the difference in means μd=μ1-μ2 using the relevant sample results from paired data. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using d=x1-x2.A 90% confidence interval for μd using the paired difference sample results x¯d=551.7, sd=141.4, nd=100.Give the best estimate for μd, the margin of error, and the confidence interval.Enter the exact answer for the best estimate, and round your answers for the margin of error and the confidence interval to two decimal places.Use the given information to find the number of degrees of freedom, the critical values χ2L and χ2R, and the confidence interval estimate of σ.It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Nicotine in menthol cigarettes 90% confidence; n=21, s=0.29 mg. a) df = b) X2/L = c) X 2/R = d) lower limit = _______ < p < upper limit = ________
- Let Y ∼N(μ,1) for μ unknown. Use the fact that P(|Y −μ|< 1.96σ) ≈0.95 toconstruct a random interval (a(Y ),b(Y )) (that is, an interval whose endpoints are r.v.s), suchthat the probability that μ is in the interval is approximately 0.95. This interval is calleda confidence interval for μ; such intervals are often desired in statistics when estimatingunknown parameters based on data.Use a t-distribution to find a confidence interval for the difference in means μd=μ1-μ2 using the relevant sample results from paired data. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using d=x1-x2. A 95% confidence interval for μd using the paired difference sample results x¯d=2.2, sd=2.2, nd=30.Give the best estimate for μd, the margin of error, and the confidence interval.Enter the exact answer for the best estimate, and round your answers for the margin of error and the confidence interval to two decimal places. Best estimate = _______Margin of error = _________The 95% confidence interval is _______ to _______T has an exponential distribution such that P(T≤2)=2P(T>4) What is Var(T)=?