BUAD310 Chapter 8

14.88 to 16.72 tons. in excel: =T.INV(0.95,48) =1.677242 Interval 1: =15.8-1.677242*3.85/SQRT(49) =14.88 Interval 2: =15.8+1.677242*3.85/SQRT(49) =16.72

A highway inspector needs an estimate of the mean weight of trucks crossing a bridge on the interstate highway system. She selects a random sample of 49 trucks and finds a mean of 15.8 tons with a sample standard deviation of 3.85 tons. The 90 percent confidence interval for the population mean is

±0.639 The width is ts/(n1/2) = (2.131)(1.2)/(16)1/2 = 0.639.

A random sample of 16 ATM transactions at the Last National Bank of Flat Rock revealed a mean transaction time of 2.8 minutes with a standard deviation of 1.2 minutes. The width (in minutes) of the 95 percent confidence interval for the true mean transaction time is

a statistic why: A statistic is a random variable. its sampling distribution describes its behavior

A sampling distribution describes the distribution of

(a) In excel: z-multiplier =NORM.S.INV(0.995) Interval 1 =0.937-2.5758293*0.009/SQRT(32) =.9329 Interval 2 =0.937+2.5758293*0.009/SQRT(32) =.9411 (b) 4. Approximately normal with μ = .9370 and σ(xbar)=0.0090/√(32). (c) a. Central Limit Theorem

Concerns about climate change and CO2 reduction have initiated the commercial production of blends of biodiesel (e.g., from renewable sources) and petrodiesel (from fossil fuel). Random samples of 32 blended fuels are tested in a lab to ascertain the bio/total carbon ratio. (a) If the true mean is .9370 with a standard deviation of 0.0090, within what interval will 99 percent of the sample means fall? (Round your answers to 4 decimal places.) b) If the true mean is .9370 with a standard deviation of 0.0090, what is the sampling distribution of Xbar? 1. Exactly normal with μ = .9370 and σ = 0.0090. 2. Approximately normal with μ = .9370 and σ = 0.0090. 3. Exactly normal with μ = .9370 and σ(xbar)=0.0090/√(32). 4. Approximately normal with μ = .9370 and σ(xbar)=0.0090/√(32). (c) What theorem did you use to answer part (b)? a. Central Limit Theorem b. Chebyshev's Theorem c. Pythagorean Theorem d. Law of Large Numbers

(a) In excel: step 1: z-multiplier =NORM.S.INV(0.975) Interval 1 =183-1.95996398*20/SQRT(50) = 177.46 Interval 2 =183+1.95996398*20/SQRT(50) = 188.54 (b) In excel: step 1: z-multiplier =NORM.S.INV(0.975) = 1.95996398 Interval 1 =860-1.95996398*12/SQRT(6) = 850.40 Interval 2 =860+1.95996398*12/SQRT(6) = 869.60 (c) In excel: step 1: z-multiplier =NORM.S.INV(0.975) = 1.95996398 Interval 1 =77-1.95996398*4/SQRT(27) = 75.49 Interval 2 =77+1.95996398*4/SQRT(27) = 78.51

Find the interval [μ−zσ/√(n),μ+zσ/√(n)] within which 95 percent of the sample means would be expected to fall, assuming that each sample is from a normal population. Item2 9 points eBook Print References Check my work Check My Work button is now enabledItem 2 Item 2 9 points Section Exercise 8-2 Find the interval [ μ−zσn√,μ+zσn√ ] within which 95 percent of the sample means would be expected to fall, assuming that each sample is from a normal population. (a) μ = 183, σ = 20, n = 50. (Round your answers to 2 decimal places.) b) μ = 860, σ = 12, n = 6. (Round your answers to 2 decimal places.) (c) μ = 77, σ = 4, n = 27. (Round your answers to 3 decimal places.)

No Why: z is always smaller than t (ceteris paribus), so the interval would be narrower than is justified.

In constructing a confidence interval for a mean with unknown variance with a sample of 25 items, Beth used z instead of t. "Well, at least my interval will be wider than necessary, so it was a conservative error," said she. Is Beth's statement correct?

the distribution of the mean is approximately normal for large n. why: The sampling distribution of the mean is asymptotically normal for any population.

The Central Limit Theorem implies that

(a) Standard Error of Xbar= stdev/sqrt(n) 2.5/√(16)=.6250 (b) In excel: Z-multiplier =NORM.S.INV(0.95) =1.64485363 Interval 1: =26-1.64485363*2.5/SQRT(16) Interval 2: =26+1.64485363*2.5/SQRT(16)

The highway fuel economy of a 2016 Lexus RX 350 FWD 6-cylinder 3.5-L automatic 5-speed using premium fuel is a normally distributed random variable with a mean of μ = 26.00 mpg and a standard deviation of σ = 2.50 mpg. (a) What is the standard error of Xbar , the mean from a random sample of 16 fill-ups by one driver? (Round your answer to 4 decimal places.) (b) Within what interval would you expect the sample mean to fall, with 90 percent probability? (Round your answers to 4 decimal places.)

the sample mean why: the mean is not used in calculating the width of the confidence interval zσ/(n1/2).

The width of a confidence interval for μ is not affected by

step 1: 100%-72% = 28% step 2: 28%/2 = 14% step 3: 72%+14%=86% or .86 step 4: in excel enter =NORM.S.INV(0.86) = 1.08

To determine a 72 percent level of confidence for a proportion, the value of z is approximately

$825.48 to $874.52 in excel: =T.INV(0.995,35) =2.72380559 Interval 1: =850-2.72380559*54/SQRT(36) =825.48 Interval 2: =840+2.72380559*54/SQRT(36) =874.52

To estimate the average annual expenses of students on books and class materials, a sample of size 36 is taken. The sample mean is $850 and the sample standard deviation is $54. A 99 percent confidence interval for the population mean is