produced a mean of x¯¯¯=23.2 from a population with a normal distribution and a standard deviation σ=2.86 (a) Find a 95% confidence interval for μ ≤μ≤ (b) Find a 90%
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A random sample of 80 observations produced a mean of x¯¯¯=23.2 from a population with a
(a) Find a 95% confidence interval for μ
≤μ≤
(b) Find a 90% confidence interval for μ
≤μ≤
(c) Find a 99% confidence interval for μ
≤μ≤
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- A random sample of size 30 from a normal population yields x̅ = 69 and s = 4. The lower bound of a 95 percent confidence interval is (Round off upto 2 decimal places).What critical value of t* should be used for a 95% confidence interval for the population mean based on a random sample of 21 observations? Find the t-table here. t* = 1.721 t* = 1.725 t* = 2.080 t* = 2.086Use the t-distribution to find a confidence interval for a difference in means μ1−μ2μ1-μ2 given the relevant sample results. Give the best estimate for μ1−μ2μ1-μ2, the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed.A 95% confidence interval for μ1−μ2μ1-μ2 using the sample results x⎯⎯1=515x¯1=515, s1=136s1=136, n1=300n1=300 and x⎯⎯2=462x¯2=462, s2=94s2=94, n2=200n2=200Enter 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 = Enter your answer; Best estimateMargin of error = Enter your answer; Margin of errorConfidence interval : Enter your answer; Confidence interval, value 1 to Enter your answer; Confidence interval, value 2
- A random sample of 70 observations produced a mean of 27.5 and a standard deviation of 3.26. Find a 90% confidence interval for < U < (b) Find a 95% confidence interval for < U < (c) Find a 99% confidence interval for < U <The mean and the standard deviation of the sample of 100 bank customer waiting times are x¯ = 5.36 and s = 2.455. (1) Calculate a t-based 95 percent confidence interval for µ, the mean of all possible bank customer waiting times using the new system. (2) Are we 95 percent confident that µ is less than 6 minutes?.A random sample of n measurements was selected from a population with standard deviation = 19 and unknown meanu. Calculate a 95% confidence interval for μ for each of the following situations: (a) n = 35, z = 77.8 MFM (b) n = 60, = 77.8A sample of size n=79 Is drawn from a population whose standard deviation is o = 29, Part 1 of 2 (a) Find the margin of error for a 90% confidence interval for l. Round the answer to at least three decimal places. The margin of error for a 90% confidence interval for u is Part 2 of 2 (b) If the sample size weren=44, would the margin of error be larger or smaller? (Choose one) ▼ because the sample size is (Choose one) Larger SmallerA warehouse owner wants to know the average weight of books at his facility. He weighs 18 books and finds that the sample has a mean weight of 9.41 pounds and standard deviation of 2.89 pounds. Construct a 99%confidence interval for the population mean of book weights. Assume the population is normally distributed.Choose the correct confidence interval and critical value used in the calculation. (7.46,11.36)(tc=2.8609) (7.83,10.99)(Zc=2.326) (7.44,11.38) (tc=2.8982) (7.45,11.37) (tc=2.8784) (7.66,11.16)(Zc=2.576)The marginal error for confidence intervals concerning μ doesn't depend on the sample size. However, it depends on the critical value t*, as well as the standard deviation of the observed sample. Is this true or false? And why?An article reports that in a sample of 9 men, the average volume of femoral cartilage (located in the knee) was 18.7 cm3 with a standard deviation of 3.3 cm3 and the average volume in a sample of 9 women was 11.2 cm3 with a standard deviation of 2.4 cm2. Let μXμX represent the population mean for men and let μYμY represent the population mean for women. Find a 95% confidence interval for the difference μX−μYμX−μY . Round down the degrees of freedom to the nearest integer and round the answers to three decimal places. The 95% confidence interval isA simple random sample from a population with a nomal distribution of 100 body temperatures has x= 98.20°F and s 0.64°F. Construct a 98% confidence interval estimate of the standard deviation of body temperature of alI healthy humans.The lifetime of a certain type of battery is known to be normally distributed with standard deviation o = 22 hours. A sample of 50 batteries had a mean lifetime of 120.1 hours. It is desired to construct a 95% confidence interval for the mean lifetime for this type of battery. ▬▬ Part 1 of 6 (a) What is the point estimate? The point estimate is 120.1 Part: 1 / 6 Part 2 of 6 (b) Find the critical value. The critical value is Skip Part Check Type here to search X 11 S I 81 17 hp fa Es Save For Later Submit Assi © 2023 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Ac (?) 55°F 10:36 F 3/28/2 10
