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Chapter 7: Problem 40

PDI The scores on the Psychomotor Development Index (PDI), a scale of infant development, have a normal population distribution with mean 100 and standard deviation 15. An infant is selected at random. a. Find the \(z\) -score for a PDI value of 90 . b. A study uses a random sample of 225 infants. Using the sampling distribution of the sample mean PDI, find the \(z\) -score corresponding to a sample mean of 90 . c. Explain why a PDI value of 90 is not surprising, but a sample mean PDI score of 90 for 225 infants would be surprising.

Short Answer

Expert verified
The z-score for a single PDI value of 90 is -0.67, while for a sample mean of 90, it is -10. A single PDI value of 90 is not surprising, but a sample mean of 90 among 225 infants is surprising.

Step by step solution

01

Understanding the Problem

We are given a normally distributed scale called the Psychomotor Development Index (PDI) with a mean \( \mu = 100 \) and a standard deviation \( \sigma = 15 \). We need to solve three parts: finding the \( z \)-score for a single PDI value of 90, finding the \( z \)-score for a sample mean of 90 for a sample of 225 infants, and explaining why a single PDI score might differ in significance from a sample mean.
02

Calculate the z-score for a single PDI value of 90

The formula for finding the \( z \)-score of an individual data point \( x \) is: \( z = \frac{x - \mu}{\sigma} \). Substitute \( x = 90 \), \( \mu = 100 \), and \( \sigma = 15 \) into the formula:\[z = \frac{90 - 100}{15} = \frac{-10}{15} = -0.67\]Thus, the \( z \)-score for a PDI value of 90 is \( -0.67 \).
03

Calculate the standard error of the mean

When dealing with a sample mean, we must first determine the standard error of the mean \( \sigma_{\bar{x}} \), which is calculated by \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \). Here, \( n = 225 \).\[\sigma_{\bar{x}} = \frac{15}{\sqrt{225}} = \frac{15}{15} = 1\]So, the standard error of the mean is 1.
04

Calculate the z-score for sample mean of 90

For the sample mean, the \( z \)-score formula is: \( z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} \), where \( \bar{x} = 90 \). We use \( \mu = 100 \) and \( \sigma_{\bar{x}} = 1 \) from Step 3.\[z = \frac{90 - 100}{1} = \frac{-10}{1} = -10\]The \( z \)-score for a sample mean of 90 is \( -10 \).
05

Compare individual PDI value to sample mean

An individual PDI score of 90 has a \( z \)-score of \( -0.67 \), which is within typical range (not far from zero) and thus not surprising in a normal distribution. However, a sample mean \( z \)-score of \( -10 \) is extremely low and suggests that a mean score of 90 for 225 infants is very unlikely, indicating it would be surprising or unusual.

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Key Concepts

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

Understanding Z-score
The z-score is a statistical measure that describes a data point's position in relation to the mean of a group of points.
It tells us how many standard deviations a data point is from the mean. This concept is crucial in statistics, particularly in inferential statistics, where it helps compare individual points to a group.
  • Formula: The z-score is calculated using the formula \[ z = \frac{x - \mu}{\sigma} \]where:
    • \( x \) is the data point.
    • \( \mu \) is the mean of the dataset.
    • \( \sigma \) is the standard deviation.
  • Interpretation: A z-score of 0 means the data point is precisely at the mean, while a positive score indicates it is above the mean, and a negative score signifies it is below.^
The concept of the z-score can help identify outliers or understand whether a particular data point is typical within a set.By using the formula, we found that a single Psychomotor Development Index (PDI) score of 90 corresponds to a z-score of -0.67.
This value shows the PDI score is two-thirds of a standard deviation below the mean, making it relatively common in a normal distribution.
The Role of Sampling Distribution
A sampling distribution arises from taking numerous samples of a certain size from a population and calculating a particular statistic, like the mean, for each sample.
This allows for analyzing the expected distribution of that statistic if different samples were taken.

Sampling Distribution of the Sample Mean

When we talk about the sampling distribution of the sample mean, we refer to the distribution of mean values from different samples of the same size from the same population.
  • This distribution will have the same mean as the population mean.
  • The standard deviation of this distribution, called the standard error, is calculated by \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \).
For a sample of 225 infants, the standard error is calculated as 1.With this knowledge, the z-score for a sample mean PDI value of 90 was calculated as -10.
This indicates that obtaining such a sample mean is extremely unlikely and considered surprising given the population distribution of mean 100 and standard deviation of 15.
The Basics of Normal Distribution
The normal distribution, also known as the Gaussian distribution, is a probability distribution that is symmetric about the mean.
It is well known for its bell-shaped curve and critical in statistics because it accurately models numerous real-world variables and data.

Understanding the Curve

The curve is peaked at the mean, and most observations fall within a few standard deviations from the mean:
  • 68% of the data falls within one standard deviation.
  • 95% within two standard deviations.
  • 99.7% within three standard deviations.
This property makes it useful for predicting outcomes and probabilistic calculations. In our case with the PDI scores: - A single score of 90 is only -0.67
standard deviations away from the mean (100), indicating it is quite typical in the distribution. - However, a sample mean of 90 for 225 infantscorresponding to a z-score of -10becomes very improbable in this distribution.
Thus, understanding the normal distribution helps contextualize statistical results and assess their likelihood.

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Most popular questions from this chapter

What good is a standard deviation? Explain how the standard deviation of the sampling distribution of a sample proportion gives you useful information to help gauge how close a sample proportion falls to the unknown population proportion.

Physicians' assistants The 2006 AAPA survey of the population of physicians' assistants who were working full time reported a mean annual income of \(\$ 84,396\) and standard deviation of \(\$ 21,975 .\) (Source: Data from 2006 AAPA survey [www.apa.org].) a. Suppose the AAPA had randomly sampled 100 physicians' assistants instead of collecting data for all of them. Describe the mean, standard deviation, and shape of the sampling distribution of the sample mean. b. Using this sampling distribution, find the \(z\) -score for a sample mean of \(\$ 80,000\) c. Using parts a and b, find the probability that the sample mean would fall within approximately \(\$ 4000\) of the population mean.

Multiple choice: CLT The central limit theorem implies a. All variables have approximately bell-shaped data distributions if a random sample contains at least about 30 observations. b. Population distributions are normal whenever the population size is large. c. For sufficiently large random samples, the sampling distribution of \(\bar{x}\) is approximately normal, regardless of the shape of the population distribution. d. The sampling distribution of the sample mean looks more like the population distribution as the sample size increases.

Canada lottery In one lottery option in Canada (Source: Lottery Canada), you bet on a six-digit number between 000000 and \(999999 .\) For a \(\$ 1\) bet, you win \(\$ 100,000\) if you are correct. The mean and standard deviation of the probability distribution for the lottery winnings are \(\mu=0.10\) (that is, 10 cents) and \(\sigma=100.00\). Joe figures that if he plays enough times every day, eventually he will strike it rich, by the law of large numbers. Over the course of several years, he plays 1 million times. Let \(\bar{x}\) denote his average winnings. a. Find the mean and standard deviation of the sampling distribution of \(\bar{x}\). b. About how likely is it that Joe's average winnings exceed \(\$ 1,\) the amount he paid to play each time? Use the central limit theorem to find an approximate answer.

Comparing pizza brands \(\quad\) The owners of Aunt Erma's Restaurant plan an advertising campaign with the claim that more people prefer the taste of their pizza (which we'll denote by A) than the current leading fast-food chain selling pizza (which we'll denote by \(\mathrm{D}\) ). To support their claim, they plan to randomly sample three people in Boston. Each person is asked to taste a slice of pizza \(A\) and a slice of pizza \(D\). Subjects are blindfolded so they cannot see the pizza when they taste it, and the order of giving them the two slices is randomized. They are then asked which pizza tastes better. Use a symbol with three letters to represent the responses for each possible sample. For instance, ADD represents a sample in which the first subject sampled preferred pizza \(A\) and the second and third subjects preferred pizza \(\mathrm{D}\) a. Identify the eight possible samples of size \(3,\) and for each sample report the proportion that preferred pizza \(A\). b. In the entire Boston population, suppose that exactly half would prefer pizza \(\mathrm{A}\) and half would prefer pizza D. Explain why the sampling distribution of the sample proportion who prefer Aunt Erma's pizza, when \(n=3,\) is \begin{tabular}{cc} \hline Sample Proportion & Probability \\ \hline 0 & \(1 / 8\) \\ \(1 / 3\) & \(3 / 8\) \\ \(2 / 3\) & \(3 / 8\) \\ 1 & \(1 / 8\) \\ \hline \end{tabular} c. In part b, we can also find the probabilities for each possible sample proportion value using the binomial distribution. Use the binomial with \(n=3\) and \(p=0.50\) to show that the probability of a sample proportion of \(1 / 3\) equals \(3 / 8 .\) (Hint: This equals the probability that \(x=1\) person out of \(n=3\) prefer pizza A. It's especially helpful to use the binomial formula when \(p\) differs from \(0.50,\) since then the eight possible samples listed in part a would not be equally likely.)
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