/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 BusinessWeek surveyed MBA alumni... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

BusinessWeek surveyed MBA alumni 10 years after graduation \((\text {Business Week} \text { , September } 22,\) 2003 ). One finding was that alumni spend an average of \(\$ 115.50\) per week eating out socially. You have been asked to conduct a follow-up study by taking a sample of 40 of these MBA alumni. Assume the population standard deviation is \(\$ 35\) a. Show the sampling distribution of \(\bar{x}\), the sample mean weekly expenditure for the \(40 \mathrm{MBA}\) alumni. b. What is the probability that the sample mean will be within \(\$ 10\) of the population mean? c. Suppose you find a sample mean of \(\$ 100 .\) What is the probability of finding a sample mean of \(\$ 100\) or less? Would you consider this sample to be an unusually low spending group of alumni? Why or why not?

Short Answer

Expert verified
a. \( \bar{x} \sim N(115.50, 5.54) \). b. Probability is \( 0.9296 \). c. Probability is \( 0.0024 \); yes, it is unusually low.

Step by step solution

01

Determine the Sampling Distribution

Since the problem states that we are dealing with a sample mean, we assume that the sampling distribution of the sample mean \( \bar{x} \) follows a normal distribution due to the Central Limit Theorem, as the sample size is 40, which is sufficiently large. The mean of the sampling distribution \( \mu_{\bar{x}} \) is the same as the population mean \( \mu = 115.50 \). The standard deviation of the sampling distribution, known as the standard error, is calculated as \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{35}{\sqrt{40}} \approx 5.54 \). Therefore, \( \bar{x} \sim N(115.50, 5.54) \).
02

Compute Probability for Part B

We need to find the probability that the sample mean \( \bar{x} \) is within \( \$10 \) of the population mean \( \mu = 115.50 \). This means \( 105.5 < \bar{x} < 125.5 \). We standardize these values to find the corresponding \( z \)-scores: \( z_1 = \frac{105.5 - 115.5}{5.54} \approx -1.805 \) and \( z_2 = \frac{125.5 - 115.5}{5.54} \approx 1.805 \). Using the standard normal distribution table, we find the probability: \( P(z_1 < z < z_2) \approx P(-1.805 < z < 1.805) = P(z < 1.805) - P(z < -1.805) \approx 0.9648 - 0.0352 = 0.9296 \).
03

Compute Probability for Part C

We are given a sample mean \( \bar{x} = 100 \) and need to find the probability that \( \bar{x} \leq 100 \). We calculate the corresponding \( z \)-score: \( z = \frac{100 - 115.5}{5.54} \approx -2.82 \). Using the standard normal distribution table, \( P(z \leq -2.82) \approx 0.0024 \). Thus, there is a \( 0.24\% \) chance of observing a sample mean of \$100 or less. Since this probability is very low, this sample mean is considered unusually low.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sampling Distribution
Imagine you have a large population, like all MBA alumni. If you take multiple random samples from this group and calculate the mean of each sample, you will get a range of means. This range of means forms something known as the sampling distribution. It is essentially the probability distribution of a statistic—here, the sample mean—across multiple samplings.

According to the Central Limit Theorem, the sampling distribution of the sample mean tends to be normally distributed, especially if your sample size is larger than 30. This is why, even if the original data is not normal, the sampling means will approximate a normal distribution.

The center of the sampling distribution, or its mean, \( \mu_{\bar{x}} \), is equal to the population mean, which in this exercise is \(115.50\). The spread of the distribution is described by the standard error, which tells us how much the sample mean \( \bar{x} \) is expected to vary. The sampling distribution essentially gives us a way to understand how representative a sample mean is compared to the population mean.
Standard Error
The standard error is a crucial concept when dealing with samples. It measures the dispersion or variability of the sample mean from the true population mean. Specifically, it tells us how much we expect the sample mean to differ from the population mean.

You can calculate the standard error if you know the population standard deviation and the sample size:\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]
where \(\sigma\) is the population standard deviation and \(n\) is the sample size. In our exercise, the standard error is calculated as \( \frac{35}{\sqrt{40}} \), which is approximately \(5.54\). This means any sample mean we calculate will most likely differ by about \(5.54\) from the population mean.

Understanding the standard error helps in determining how accurate our sample mean is as an estimate of the population mean. A smaller standard error indicates a more precise estimate of the population mean.
Normal Distribution
The concept of normal distribution is central to statistical analysis and plays a significant role in the sampling distribution of means. In a normal distribution, data is symmetrically distributed around a mean value, forming a bell-shaped curve. Most of the data points (about 68%) lie within one standard deviation of the mean in either direction, and about 95% lie within two standard deviations.

In the context of our exercise, thanks to the Central Limit Theorem, the sampling distribution of the sample means follows a normal distribution. This is true despite the shape of the actual population distribution, provided we have a sufficiently large sample size—typically 30 or more.

This makes normal distribution very useful because it allows us to apply probability theory to make inferences about our data. For example, by converting sample mean values to \( z \)-scores, we can use standard normal distribution tables to find probabilities associated with specific mean values. In this way, the normal distribution aids in predicting the likelihood of particular outcomes in our sampling data.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The following data are from a simple random sample. \\[ 5 \quad 8 \quad 10 \quad 7 \quad 10 \quad 14 \\] a. What is the point estimate of the population mean? b. What is the point estimate of the population standard deviation?

The mean annual cost of automobile insurance is \(\$ 939\) (CNBC, February 23,2006 ). Assume that the standard deviation is \(\sigma=\$ 245\) a. What is the probability that a simple random sample of automobile insurance policies will have a sample mean within \(\$ 25\) of the population mean for each of the following sample sizes: \(30,50,100,\) and \(400 ?\) b. What is the advantage of a larger sample size when attempting to estimate the population mean?

The average score for male golfers is 95 and the average score for female golfers is 106 (Golf Digest, April 2006). Use these values as the population means for men and women and assume that the population standard deviation is \(\sigma=14\) strokes for both. A simple random sample of 30 male golfers and another simple random sample of 45 female golfers will be taken. a. Show the sampling distribution of \(\bar{x}\) for male golfers. b. What is the probability that the sample mean is within three strokes of the population mean for the sample of male golfers? c. What is the probability that the sample mean is within three strokes of the population mean for the sample of female golfers? d. In which case, part (b) or part (c), is the probability of obtaining a sample mean within three strokes of the population mean higher? Why?

The Grocery Manufacturers of America reported that \(76 \%\) of consumers read the ingredients listed on a product's label. Assume the population proportion is \(p=.76\) and a sample of 400 consumers is selected from the population. a. Show the sampling distribution of the sample proportion \(\bar{p}\), where \(\bar{p}\) is the proportion of the sampled consumers who read the ingredients listed on a product's label. b. What is the probability that the sample proportion will be within ±.03 of the population proportion? c. Answer part (b) for a sample of 750 consumers.

A population has a mean of 200 and a standard deviation of \(50 .\) Suppose a simple random sample of size 100 is selected and \(\bar{x}\) is used to estimate \(\mu\) a. What is the probability that the sample mean will be within ±5 of the population mean? b. What is the probability that the sample mean will be within ±10 of the population mean?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.