Problem 3
Question
Is it believable that the numbers 65,30 , and 55 are a random sample of size 3 from a normal distribution with \(\mu=50\) and \(\sigma=10\) ? Answer the question by using a chi square distribution. (Hint: Let \(Z_{i}=\left(Y_{i}-50\right) / 10\) and use Theorem 7.3.1.)
Step-by-Step Solution
Verified Answer
Whether the numbers 65, 30, 55 are a believable random sample from the given normal distribution with \(\mu = 50\) and \(\sigma = 10\) will depend on the comparison result of the calculated chi square statistic with the chi square distribution table for 2 degrees of freedom. If the calculated statistic is within the acceptable region, then the numbers are believable as a random sample from the normal distribution. Otherwise, they are not believable as a random sample from the given normal distribution.
1Step 1 - Standardization
Calculate the standardized values \(Z_i\) using the formula \(Z_i = (Y_i - \mu) / \sigma\). For \(Y_i = 65, 30, 55\), \(\mu = 50\), \(\sigma = 10\) calculate the corresponding \(Z_i\).
2Step 2 - Calculate Chi Square Statistic
Next, calculate the squared values of each standardized value, \(Z_i^2\). Finally, get the sum of this squared standardized values. This will give the chi square statistic, say signified as X.
3Step 3 - Validate the Sample
Compare the calculated chi square statistic, X, with a chi square distribution table with degrees of freedom equal to the sample size minus 1, which is 3-1=2. If X is within the acceptable region of the chi square distribution for the given degrees of freedom, then it is believable that the numbers are from the specified normal distribution.
Key Concepts
Normal DistributionStandardization in StatisticsChi Square TestDegrees of Freedom
Normal Distribution
The normal distribution is a bell-shaped curve that is ubiquitous in statistics and represents the spread of a set of data. Its perfect symmetry suggests that for a given mean \(\mu\), the probability of observing a value above \(\mu\) is equal to the probability of observing a value below it. The normal distribution is defined by two parameters: the mean \(\mu\) and the standard deviation \(\sigma\).
In our context, the assumption is that sample values are from a normal distribution with a mean of 50 and a standard deviation of 10. If this is true, the vast majority of the samples should fall within a certain range of the mean, and observing extreme values becomes increasingly unlikely.
In our context, the assumption is that sample values are from a normal distribution with a mean of 50 and a standard deviation of 10. If this is true, the vast majority of the samples should fall within a certain range of the mean, and observing extreme values becomes increasingly unlikely.
Standardization in Statistics
Standardization is a technique that transforms your data into a common scale, allowing you to compare scores from different types of distributions. This is particularly useful when dealing with the normal distribution.
The formula for standardization is given by \( Z_i = \frac{Y_i - \mu}{\sigma} \), where \( Z_i \) is the standardized value, \( Y_i \) is the original value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
In the given exercise, the values 65, 30, and 55 are standardized using a mean (\( \mu \) ) of 50 and a standard deviation (\( \sigma \) ) of 10. This process adjusts the numbers so that they can be directly compared to a standard normal distribution, which has a mean of 0 and a standard deviation of 1.
The formula for standardization is given by \( Z_i = \frac{Y_i - \mu}{\sigma} \), where \( Z_i \) is the standardized value, \( Y_i \) is the original value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
In the given exercise, the values 65, 30, and 55 are standardized using a mean (\( \mu \) ) of 50 and a standard deviation (\( \sigma \) ) of 10. This process adjusts the numbers so that they can be directly compared to a standard normal distribution, which has a mean of 0 and a standard deviation of 1.
Chi Square Test
The chi square test is a statistical method used to compare observed data with data we expect to obtain according to a specific hypothesis. It's very useful when you want to assess whether the differences between your observed and expected data are due to chance.
The test involves calculating a statistic, represented by the symbol \( X^2 \), which summarizes the aggregated discrepancies between the observed and expected frequencies. In the given exercise, by squaring the standardized values and summing them up, we get our chi square statistic. Under the null hypothesis - that the observed data comes from a given distribution - this statistic follows a chi square distribution.
The test involves calculating a statistic, represented by the symbol \( X^2 \), which summarizes the aggregated discrepancies between the observed and expected frequencies. In the given exercise, by squaring the standardized values and summing them up, we get our chi square statistic. Under the null hypothesis - that the observed data comes from a given distribution - this statistic follows a chi square distribution.
Degrees of Freedom
The concept of degrees of freedom in statistics is a measure of the number of independent ways by which a dynamic system can move, without violating any constraints imposed on it. In the context of the chi square test, degrees of freedom roughly correspond to the number of categories reduced by the number of parameters estimated from the data.
For the chi square test explanation provided in the solution, degrees of freedom are computed as the size of the sample minus one (n-1). In the exercise, for a sample size of 3, you would have 2 degrees of freedom (3-1=2). The degrees of freedom are essential when determining the critical value from the chi square distribution to assess the believability of the sample being from a normal distribution with specified mean and variance.
For the chi square test explanation provided in the solution, degrees of freedom are computed as the size of the sample minus one (n-1). In the exercise, for a sample size of 3, you would have 2 degrees of freedom (3-1=2). The degrees of freedom are essential when determining the critical value from the chi square distribution to assess the believability of the sample being from a normal distribution with specified mean and variance.
Other exercises in this chapter
Problem 2
Find the moment-generating function for a chi square random variable and use it to show that \(E\left(\chi_{n}^{2}\right)=n\) and \(\operatorname{Var}\left(\chi
View solution Problem 4
Use the fact that \((n-1) S^{2} / \sigma^{2}\) is a chi square random variable with \(n-1\) df to prove that $$ \operatorname{Var}\left(S^{2}\right)=\frac{2 \si
View solution Problem 6
If \(Y\) is a chi square random variable with \(n\) degrees of freedom, the pdf of \((Y-n) / \sqrt{2 n}\) converges to \(f_{Z}(z)\) as \(n\) goes to infinity (r
View solution Problem 8
Let \(V\) and \(U\) be independent chi square random variables with 7 and 9 degrees of freedom, respectively. Is it more likely that \(\frac{V / 7}{U / 9}\) wil
View solution