Testing for Normality Using a chi-square goodness-of-fit test, you can decide, with some degree of certainty, whether a variable is normally distributed. In all chi-square tests for normality, the null and alternative hypotheses are as listed below.


H₀: The variable has a normal distribution.


Hₐ: The variable does not have a normal distribution.


To determine the expected frequencies when performing a chi-square test for normality, first estimate the mean and standard deviation of the frequency distribution. Then, use the mean and standard deviation to compute the z-score for each class boundary. Then, use the z-scores to calculate the area under the standard normal curve for each class. Multiplying the resulting class areas by the sample size yields the expected frequency for each class.In Exercises 17 and 18, (a) find the expected frequencies, (b) find the critical value and identify the rejection region, (c) find the chi-square test statistic, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim.


In Exercises 17 and 18, (a) find the expected frequencies, (b) find the critical value and identify the rejection region, (c) find the chi-square test statistic, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim.


Test Scores At α=0.01, test the claim that the 200 test scores shown in the frequency distribution are normally distributed.

Verified step by step guidance

1. Step 1: Calculate the sample mean (\( \bar{x} \)) and sample standard deviation (\( s \)) using the midpoints of each class interval and their corresponding frequencies. The midpoint for each class is the average of the class boundaries. Use the formulas: \( \bar{x} = \frac{\sum f x}{n} \) and \( s = \sqrt{\frac{\sum f (x - \bar{x})^2}{n - 1}} \), where \( x \) is the midpoint and \( f \) is the frequency.
2. Step 2: Convert the class boundaries to z-scores using the formula \( z = \frac{x - \bar{x}}{s} \) for each boundary. This standardizes the class boundaries relative to the estimated normal distribution.
3. Step 3: Use the z-scores to find the area under the standard normal curve for each class interval. This is done by finding the cumulative probabilities for the z-scores at the boundaries and subtracting to get the area for each class.
4. Step 4: Multiply each class area by the total sample size (\( n = 200 \)) to find the expected frequencies for each class. These expected frequencies represent the counts we would expect if the data were normally distributed.
5. Step 5: Perform the chi-square goodness-of-fit test by calculating the test statistic \( \chi^2 = \sum \frac{(f - E)^2}{E} \), where \( f \) is the observed frequency and \( E \) is the expected frequency for each class. Then, determine the degrees of freedom (number of classes minus 1 minus the number of estimated parameters, usually 2 for mean and standard deviation), find the critical value from the chi-square distribution table at \( \alpha = 0.01 \), and compare the test statistic to the critical value to decide whether to reject or fail to reject the null hypothesis.

Key Concepts

Chi-Square Goodness-of-Fit Test

The chi-square goodness-of-fit test assesses whether observed frequency data fit a specific distribution, such as the normal distribution. It compares observed frequencies with expected frequencies calculated under the null hypothesis. The test statistic sums the squared differences between observed and expected frequencies, divided by expected frequencies.

Calculating Expected Frequencies Using the Normal Distribution

To find expected frequencies for each class, estimate the mean and standard deviation from the data. Convert class boundaries to z-scores, then find the area under the standard normal curve between these z-scores. Multiply these areas by the total sample size to get expected frequencies for each class.

Hypothesis Testing and Decision Making

In hypothesis testing, the null hypothesis (H₀) assumes the data follow a normal distribution, while the alternative (Hₐ) assumes they do not. Using the chi-square statistic and critical value at a chosen significance level (α=0.01), decide to reject or fail to reject H₀. This decision interprets whether the data support normality.