College Statistics Quiz: Chi Square Distribution

20 questions · exam conditions

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Chi Square DistributionQuestion 1 of 20

In a chi-square goodness-of-fit test, a researcher compares observed frequencies to expected frequencies across 7 categories. Assuming the conditions for the test are met, what is the expected value of the resulting chi-square test statistic under the null hypothesis?

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College Statistics Quiz

College Statistics Quiz: Chi Square Distribution

Practice Chi Square Distribution in College Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Chi Square Distribution, giving you a quick way to practice the rules, question types, and explanations that matter most for College Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

5
6 (correct answer)
7
12

Explanation: For a chi-square goodness-of-fit test, the degrees of freedom (df) are calculated as

k-1

, where

k

is the number of categories. In this case,

k=7

, so

df = 7 - 1 = 6

. The test statistic follows a chi-square distribution with 6 degrees of freedom. The expected value (or mean) of a chi-square distribution is equal to its degrees of freedom. Therefore, the expected value of the test statistic is 6.

Question 2

Let the random variable $X$ follow a chi-square distribution with 15 degrees of freedom. What is the expected value of the random variable $Y = 3X + 5$ ?

50 (correct answer)
45
95
15

Explanation: This question tests your understanding of linear transformations of random variables and properties of the chi-square distribution. When you see a problem involving transformations like

Y = aX + b

, remember that expectation is a linear operator. First, recall that for a chi-square distribution with

\nu

degrees of freedom, the expected value is simply

\nu

. So

E[X] = 15

since

X

follows a chi-square distribution with 15 degrees of freedom. For the linear transformation

Y = 3X + 5

, we use the linearity property of expectation:

E[Y] = E[3X + 5] = 3E[X] + 5 = 3(15) + 5 = 45 + 5 = 50

Therefore, the correct answer is A) 50. Looking at the wrong answers: B) 45 represents the mistake of forgetting the constant term—this would be

3E[X] = 3(15) = 45

but ignores the

+5

. C) 95 might result from incorrectly thinking the degrees of freedom get transformed (like

3(15) + 5 = 50

, then somehow adding more) or other computational errors. D) 15 is simply the original expected value of

X

, ignoring the transformation entirely. Study tip: Always remember that

E[aX + b] = aE[X] + b

for any constants

a

and

b

. This linearity property works regardless of the underlying distribution. For chi-square distributions specifically, memorize that

E[X] = \nu

where

\nu

is the degrees of freedom—this appears frequently on exams.

Question 3

Let the random variable $X$ follow a chi-square distribution with 15 degrees of freedom. What is the expected value of the random variable $Y = 3X + 5$ ?

50 (correct answer)
45
95
15

Y = aX + b

, remember that expectation is a linear operator. First, recall that for a chi-square distribution with

\nu

degrees of freedom, the expected value is simply

\nu

. So

E[X] = 15

since

X

follows a chi-square distribution with 15 degrees of freedom. For the linear transformation

Y = 3X + 5

, we use the linearity property of expectation:

E[Y] = E[3X + 5] = 3E[X] + 5 = 3(15) + 5 = 45 + 5 = 50

Therefore, the correct answer is A) 50. Looking at the wrong answers: B) 45 represents the mistake of forgetting the constant term—this would be

3E[X] = 3(15) = 45

but ignores the

+5

. C) 95 might result from incorrectly thinking the degrees of freedom get transformed (like

3(15) + 5 = 50

, then somehow adding more) or other computational errors. D) 15 is simply the original expected value of

X

, ignoring the transformation entirely. Study tip: Always remember that

E[aX + b] = aE[X] + b

for any constants

a

and

b

. This linearity property works regardless of the underlying distribution. For chi-square distributions specifically, memorize that

E[X] = \nu

where

\nu

is the degrees of freedom—this appears frequently on exams.

Question 4

The chi-square distribution is a family of distributions, where each member is defined by a single parameter, the degrees of freedom (df). A primary reason for the importance of the degrees of freedom parameter is that it determines which of the following?

Only the mean of the distribution, while the shape and variance are constant for all members.
Only whether the distribution applies to categorical or continuous data.
The specific shape, center (mean), and spread (variance) of the distribution. (correct answer)
The probability that a chi-square random variable will be positive, which increases with df.

Explanation: The degrees of freedom (df) parameter is the sole parameter that defines a specific chi-square distribution within its family. It dictates all the key properties of the curve: its center (mean = df), its spread (variance = 2*df), and its shape (the degree of right-skewness, which decreases as df increases).

Question 5

An auditor takes a random sample of 25 accounts from a large company. The sample variance of the account balances is $s^2 = 400$ . The auditor assumes the account balances are normally distributed with a true population variance of $\sigma^2$ . Which of the following random variables, constructed from this data, would follow a chi-square distribution?

$\frac{25 s^2}{\sigma^2}$
$\frac{s^2}{400}$
$\frac{24 (s - \mu)}{\sigma}$
$\frac{24 s^2}{\sigma^2}$ (correct answer)

Explanation: When you encounter questions about chi-square distributions in statistics, you're typically dealing with sample variances from normally distributed populations. The key relationship to remember is that for a sample of size

n

from a normal population, the quantity

\frac{(n-1)s^2}{\sigma^2}

follows a chi-square distribution with

(n-1)

degrees of freedom. The correct answer is D because it properly applies this fundamental relationship. With a sample size of 25, we have

n-1 = 24

degrees of freedom. Therefore,

\frac{24s^2}{\sigma^2}

follows a chi-square distribution with 24 degrees of freedom. Let's examine why the other options are incorrect: Option A uses

\frac{25s^2}{\sigma^2}

, which incorrectly uses

n

instead of

(n-1)

. This is a common mistake that stems from confusing the sample size with the degrees of freedom. Option B,

\frac{s^2}{400}

, simply divides the sample variance by its observed value, yielding 1. This has no connection to the chi-square distribution and doesn't involve the population variance

\sigma^2

at all. Option C attempts to create a standardized variable using

(s-\mu)

, but this form doesn't follow any standard distribution. The chi-square distribution specifically relates to squared deviations and variances, not linear differences. Remember this pattern: for chi-square problems involving sample variance, always look for the form

\frac{(n-1)s^2}{\sigma^2}

where

(n-1)

represents the degrees of freedom. The degrees of freedom is always one less than the sample size.

Question 6

Let $U$ be a random variable following a chi-square distribution with 8 degrees of freedom, and let $V$ be a random variable following a chi-square distribution with 12 degrees of freedom. If $U$ and $V$ are independent, what is the variance of the random variable $W = U + V$ ?

20
40 (correct answer)
16
28

Explanation: One property of the chi-square distribution is that the sum of two independent chi-square random variables is also a chi-square random variable with degrees of freedom equal to the sum of the individual degrees of freedom. So,

W = U + V

follows a

\chi^2(8+12) = \chi^2(20)

distribution. The variance of a chi-square distribution is twice its degrees of freedom. Therefore, the variance of

W

2 \times 20 = 40

. Alternatively, because U and V are independent,

Var(U+V) = Var(U) + Var(V) = (2 \times 8) + (2 \times 12) = 16 + 24 = 40

Question 7

5
6 (correct answer)
7
12

Explanation: For a chi-square goodness-of-fit test, the degrees of freedom (df) are calculated as

k-1

, where

k

is the number of categories. In this case,

k=7

, so

df = 7 - 1 = 6

Question 8

Let $X_1, X_2, ..., X_{10}$ be a random sample from a normal distribution with mean $\mu = 5$ and standard deviation $\sigma = 2$ . Which of the following statistics would have a chi-square distribution with 10 degrees of freedom?

$\sum_{i=1}^{10} \left(\frac{X_i - 5}{2}\right)^2$ (correct answer)
$\sum_{i=1}^{10} X_i^2$
$\frac{\sum_{i=1}^{10} (X_i - \bar{X})^2}{4}$
$\sum_{i=1}^{10} (X_i - 5)^2$

Explanation: When you encounter questions about chi-square distributions, remember that they arise from sums of squared standard normal random variables. The key is identifying which statistic creates independent standard normal terms. Choice A is correct because it transforms each observation into standard normal form. Since

X_i \sim N(5, 2^2)

, we have

\frac{X_i - 5}{2} \sim N(0,1)

. When you square independent standard normal variables and sum them, you get a chi-square distribution. Here,

\sum_{i=1}^{10} \left(\frac{X_i - 5}{2}\right)^2

follows

\chi^2_{10}

because we're summing 10 independent squared standard normals. Choice B sums the squares of the original normal variables without standardizing them. Since

X_i^2

doesn't follow a standard normal distribution, this sum won't be chi-square. Choice C involves the sample mean

\bar{X}

, which creates dependence among the terms

(X_i - \bar{X})

. While this statistic is related to chi-square distributions, it actually follows

\chi^2_9

(with 9 degrees of freedom) because using the sample mean "costs" one degree of freedom. Choice D squares the deviations from the true mean but fails to divide by the variance (4). Without proper standardization, these terms aren't standard normal, so their sum isn't chi-square distributed. Study tip: For chi-square recognition, look for the pattern: sum of squared standardized terms. The formula is

\sum \left(\frac{X_i - \mu}{\sigma}\right)^2

. Always check that each term is properly standardized and that you're not introducing dependencies through sample statistics.

Question 9

A researcher collects a random sample of size $n=20$ from a population to construct a confidence interval for the population variance $\sigma^2$ . The statistic $T = \frac{19s^2}{\sigma^2}$ is used, where $s^2$ is the sample variance. For $T$ to follow a chi-square distribution with 19 degrees of freedom, which of the following is a necessary assumption?

The sample size must be greater than 30.
The population from which the sample is drawn must be approximately normal. (correct answer)
The population variance $\sigma^2$ must be known before sampling.
The sample variance $s^2$ must be an unbiased estimator of the population mean.

Explanation: The theorem stating that the statistic

\frac{(n-1)s^2}{\sigma^2}

follows a chi-square distribution with

n-1

degrees of freedom relies on the condition that the sample is drawn from a normally distributed population. The other options are incorrect: there is no

n>30

requirement for this procedure,

\sigma^2

is the unknown parameter being estimated, and

s^2

is an estimator for the population variance, not the mean.

Question 10

Let $\chi^2_{\alpha, k}$ denote the critical value for a chi-square distribution with $k$ degrees of freedom such that $P(\chi^2(k) > \chi^2_{\alpha, k}) = \alpha$ . How does the value of $\chi^2_{0.05, k}$ change as $k$ increases?

It decreases because the distribution becomes less skewed.
It remains constant because the tail probability is fixed at 0.05.
It increases because both the mean and variance of the distribution increase. (correct answer)
It may increase or decrease depending on the specific value of $k$ .

Explanation: As the degrees of freedom

k

increase, the mean (

k

) and variance ((2k)) of the chi-square distribution both increase. This means the entire distribution shifts to the right and becomes more spread out. Consequently, any specific upper-tail percentile point (like the 95th percentile, which is the critical value

\chi^2_{0.05, k}

) must also shift to the right to maintain the same 0.05 area in the tail. Therefore, the value of

\chi^2_{0.05, k}

increases as

k

increases.

Question 11

A random variable $X$ follows a chi-square distribution with 3 degrees of freedom. Let $\mu$ be the mean of this distribution. Which of the following statements about the probability $P(X < \mu)$ is correct?

$P(X < \mu) = 0.5$
$P(X < \mu) < 0.5$
$P(X < \mu) > 0.5$ (correct answer)
The value cannot be determined without a chi-square probability table.

Explanation: A chi-square distribution with a small number of degrees of freedom (like 3) is strongly skewed to the right. In any right-skewed distribution, the mean is pulled to the right by the long tail, so the mean is greater than the median. The median is the value

m

for which

P(X < m) = 0.5

. Since

\mu > \text{median}

, the area to the left of the mean must be greater than 0.5. Thus,

P(X < \mu) > 0.5

Question 12

A random variable follows a chi-square distribution with 18 degrees of freedom. What is the standard deviation of this distribution?

$\sqrt{18}$
6 (correct answer)
18
36

Explanation: For a chi-square distribution with

k

degrees of freedom, the variance is (2k). Here,

k=18

, so the variance is

2 \times 18 = 36

. The standard deviation is the square root of the variance. Therefore, the standard deviation is

\sqrt{36} = 6

Question 13

Consider a sequence of chi-square distributions, $\chi^2(k)$ , where the degrees of freedom $k$ increase towards infinity. Which distribution does the shape of the $\chi^2(k)$ distribution approach?

A standard normal distribution, $N(0,1)$ .
An F-distribution with increasing denominator degrees of freedom.
A uniform distribution over the positive real numbers.
A normal distribution with mean $k$ and variance (2k). (correct answer)

Explanation: This question tests your understanding of asymptotic behavior in probability distributions—how certain distributions behave as their parameters approach limiting values. As the degrees of freedom

k

in a chi-square distribution increase toward infinity, the Central Limit Theorem comes into play. A chi-square random variable with

k

degrees of freedom is actually the sum of

k

independent squared standard normal variables. The key properties you need to remember are that

\chi^2(k)

has mean

k

and variance

2k

. When

k

becomes large, this sum of many independent random variables (the squared normals) approaches a normal distribution by the CLT. The limiting normal distribution maintains the same mean and variance as the original chi-square: mean

k

and variance

2k

. This makes answer D correct. Now for the incorrect options: Answer A suggests convergence to a standard normal

N(0,1)

, but this ignores that both the mean and variance of

\chi^2(k)

grow with

k

—they don't stabilize at 0 and 1. Answer B mentions an F-distribution, which is completely unrelated to the asymptotic behavior of chi-square distributions. Answer C proposes a uniform distribution, which makes no sense since chi-square distributions are always right-skewed and bell-shaped as

k

increases, never flat like a uniform distribution. Remember this pattern: when a distribution involves a parameter that represents the sum of many independent components (like degrees of freedom in chi-square), increasing that parameter often leads to normal convergence via the Central Limit Theorem.

Question 14

For a chi-square distribution with 8 degrees of freedom, which of the following correctly orders the mode, median, and mean from least to greatest?

Mode < Median < Mean (correct answer)
Mean < Median < Mode
Mode < Mean < Median
Mean = Median = Mode

Explanation: A chi-square distribution with

k>2

degrees of freedom is unimodal and skewed to the right. For any right-skewed distribution, the relationship between the three measures of central tendency is Mode < Median < Mean. Specifically, for

\chi^2(8)

, the Mean is

k=8

, and the Mode is

k-2=6

. The Median is between these two values.

Question 15

A test statistic $T$ is known to follow a chi-square distribution under the null hypothesis. A researcher calculates $T$ for their data and finds that its value corresponds to the 99th percentile of its distribution. The variance of this chi-square distribution is 16. What is the mean of the distribution?

8 (correct answer)
4
16
It cannot be determined without knowing the value of the test statistic.

Explanation: When you encounter chi-square distributions in statistics, remember that they have a special relationship between their mean and variance that's determined entirely by the degrees of freedom parameter. For any chi-square distribution with

k

degrees of freedom, the mean equals

k

and the variance equals

2k

. This is a fundamental property you should memorize. Since you're told the variance is 16, you can work backwards: if variance =

2k = 16

, then

k = 8

. Therefore, the mean must also equal the degrees of freedom, which is 8. The correct answer is A) 8. The information about the 99th percentile is irrelevant red herring - you don't need the actual test statistic value or its percentile position to determine the distribution's parameters. Let's examine why the other options are wrong. Answer B) 4 would be correct if you mistakenly thought the mean equaled half the variance, but that's not how chi-square distributions work. Answer C) 16 incorrectly assumes the mean equals the variance, which confuses the two parameters. Answer D) It cannot be determined is wrong because chi-square distributions are completely characterized by their degrees of freedom - once you know the variance, you automatically know the mean. Study tip: Memorize this chi-square relationship: mean =

k

, variance =

2k

, where

k

is degrees of freedom. Many statistics problems will give you one parameter and ask for another, so knowing this connection saves time and prevents errors.

Question 16

Let $X \sim \chi^2(5)$ and $Y \sim \chi^2(15)$ . Which of the following statements most accurately describes the relationship between the distributions of $X$ and $Y$ ?

The distribution of $Y$ is more skewed to the right than the distribution of $X$ .
The distribution of $Y$ has a larger mean and is more symmetric than the distribution of $X$ . (correct answer)
The distribution of $Y$ has a smaller variance than the distribution of $X$ .
Both distributions are symmetric and centered at their respective degrees of freedom.

Explanation: As the degrees of freedom (df) of a chi-square distribution increase, the distribution becomes less skewed (more symmetric) and approaches a normal distribution. The mean (df) and variance (2*df) also increase. Since

Y

has more degrees of freedom (15) than

X

(5), its distribution has a larger mean (15 vs. 5) and is more symmetric (less skewed) than the distribution of

X

Question 17

$P(X < \mu) = 0.5$
$P(X < \mu) < 0.5$
$P(X < \mu) > 0.5$ (correct answer)
The value cannot be determined without a chi-square probability table.

m

for which

P(X < m) = 0.5

. Since

\mu > \text{median}

, the area to the left of the mean must be greater than 0.5. Thus,

P(X < \mu) > 0.5

Question 18

20
40 (correct answer)
16
28

W = U + V

follows a

\chi^2(8+12) = \chi^2(20)

distribution. The variance of a chi-square distribution is twice its degrees of freedom. Therefore, the variance of

W

2 \times 20 = 40

. Alternatively, because U and V are independent,

Var(U+V) = Var(U) + Var(V) = (2 \times 8) + (2 \times 12) = 16 + 24 = 40

Question 19

The sample size must be greater than 30.
The population from which the sample is drawn must be approximately normal. (correct answer)
The population variance $\sigma^2$ must be known before sampling.
The sample variance $s^2$ must be an unbiased estimator of the population mean.

Explanation: The theorem stating that the statistic

\frac{(n-1)s^2}{\sigma^2}

follows a chi-square distribution with

n-1

degrees of freedom relies on the condition that the sample is drawn from a normally distributed population. The other options are incorrect: there is no

n>30

requirement for this procedure,

\sigma^2

is the unknown parameter being estimated, and

s^2

is an estimator for the population variance, not the mean.

Question 20

8 (correct answer)
4
16
It cannot be determined without knowing the value of the test statistic.

k

degrees of freedom, the mean equals

k

and the variance equals

2k

. This is a fundamental property you should memorize. Since you're told the variance is 16, you can work backwards: if variance =

2k = 16

, then

k = 8

k

, variance =

2k

, where

k

is degrees of freedom. Many statistics problems will give you one parameter and ask for another, so knowing this connection saves time and prevents errors.