Knowee
Questions
Features
Study Tools

Select the correct answer.A researcher for a travel company is looking into the prices people are willing to pay for airplane tickets. The company has communicated that the overall population mean is $265 with a standard deviation of $40.93. The researcher has a sample of 130 ticket purchases from one location. By the central limit theorem, which interval can the researcher be 95% certain that the sample mean will fall within? A. $261.41 and $268.59 B. $264.37 and $265.63 C. $257.82 and $272.18 D. $254.23 and $275.77

Question

Select the correct answer.A researcher for a travel company is looking into the prices people are willing to pay for airplane tickets. The company has communicated that the overall population mean is 265withastandarddeviationof265 with a standard deviation of 40.93. The researcher has a sample of 130 ticket purchases from one location. By the central limit theorem, which interval can the researcher be 95% certain that the sample mean will fall within? A. 261.41and261.41 and 268.59 B. 264.37and264.37 and 265.63 C. 257.82and257.82 and 272.18 D. 254.23and254.23 and 275.77

...expand
🧐 Not the exact question you are looking for?Go ask a question

Solution

To answer this question, we need to use the Central Limit Theorem and the concept of a confidence interval.

The Central Limit Theorem states that if you have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

This will allow us to create a confidence interval for the sample mean. The formula for a confidence interval is:

CI = x̄ ± Z * (σ/√n)

where:

  • x̄ is the sample mean
  • Z is the Z-score (for a 95% confidence interval, the Z-score is 1.96)
  • σ is the population standard deviation
  • n is the sample size

Given in the problem, we have:

  • μ = $265 (population mean)
  • σ = $40.93 (standard deviation)
  • n = 130 (sample size)

We are trying to find the interval (x̄ - E, x̄ + E) where E = Z * (σ/√n)

First, calculate E: E = 1.96 * (40.93/√130) ≈ 7.07

Then, calculate the interval:

  • Lower limit = μ - E = 265 - 7.07 = $257.93
  • Upper limit = μ + E = 265 + 7.07 = $272.07

So, the researcher can be 95% certain that the sample mean will fall within the interval 257.93and257.93 and 272.07.

Looking at the options, the closest one is C. 257.82and257.82 and 272.18. So, the correct answer is C.

This problem has been solved

Similar Questions

Suppose the population of all public Universities shows the annual parking fee per student is \$110 with a standard deviation of \$18. If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a sample mean between \$100 and \$115 is?Select one:a.0.9738.b.0.4738.c.0.0262.d.0.6103.e.0.1103.

The central limit theorem states that if a random sample of size n is drawn from a population, then the sampling distribution of the sample mean:Group of answer choicesis approximately normal if n ≥ 30.is approximately normal if the underlying population is normal.has the same variance as the population.is approximately normal if n < 30.

A report announced that the median sales price of new houses sold one year was $231,000,and the mean sales price was $271,600.Assume that the standard deviation of the prices is $90,000. Complete parts (a)through(d)below. (c) If you select a random sample of n=100,what is the probability that the sample mean will be less than $300,000? The probability that the sample mean will be less than $300,000 is (Round to four decimal places as needed.)

If all possible samples of size 16 are drawn from a normal population with mean equal to 50 and standard deviation equal to 5, what is the probability that a sample mean X¯ will fall in the interval from μX¯ −1.9σX¯ to μX¯ −0.4σX¯ ? Assume that the sample means can be measured to any degree of accuracy.

Suppose the commute times for employees of a large company follow a normal distribution. If the mean time is 22 minutes and the standard deviation is 5 minutes, 95% of the employees will have a travel time within which range?A.17.25 minutes to 26.75 minutesB.17 minutes to 27 minutesC.12 minutes to 32 minutesD.7 minutes to 37 minutesSUBMITarrow_backPREVIOUS

1/3

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.