Question 1

A sample of 61 elements is selected to estimate a 95% confidence interval for the variance of the population.The chi-square values to be used for this interval estimation are

Accepted Answer

For a 95% confidence interval for the variance of a population, the chi-square values are determined by the degrees of freedom (n-1) where n is the sample size, and the confidence level. With a sample size of 61, the degrees of freedom are 60. Looking up the chi-square distribution table for 60 degrees of freedom at the 2.5th and 97.5th percentiles (to account for both tails in a 95% confidence interval), the values are approximately 40.482 and 83.298, respectively.

Question 2

A sample of 31 observations yielded a sample variance of 16.If we want to test H₀: σ² = 16, the test statistic is

Accepted Answer

The test statistic for testing a hypothesis about a population variance ($\sigma^2$) when the population is normally distributed is given by $\chi^2 = \frac{(n-1)s^2}{\sigma^2_0}$, where $n$ is the sample size, $s^2$ is the sample variance, and $\sigma^2_0$ is the hypothesized population variance. Here, $n = 31$, $s^2 = 16$, and $\sigma^2_0 = 16$, so $\chi^2 = \frac{(31-1)16}{16} = \frac{30 \times 16}{16} = 30$.

Question 3

The symbol used for the variance of the sample is

Accepted Answer

D

Question 4

To avoid the problem of not having access to Tables of F distribution when F values are needed for the lower tail, the numerator of the test statistic for a two-tailed test should be the one with

Accepted Answer

The numerator of the test statistic for a two-tailed test should be the one with the larger sample variance to ensure that the F value is greater than 1, making it easier to look up in standard F distribution tables, which typically only provide critical values for the upper tail.

Question 5

We are interested in testing whether the variance of a population is significantly less than 1.44.The null hypothesis for this test is

Accepted Answer

The null hypothesis is that the population variance is not significantly less than 1.44. Therefore, option D is the best choice, as it states that the population variance is greater than or equal to 1.44.

Question 6

The bottler of a certain soft drink claims their equipment to be accurate and that the variance of all filled bottles is at most .05.The null hypothesis in a test to confirm the claim would be written as

Accepted Answer

The null hypothesis should include the population variance and state that it is less than or equal to the given value of .05. Choice D correctly does this by stating that the population variance (σ) is less than or equal to .05.

Question 7

The producer of a certain medicine claims that their bottling equipment is very accurate and that the standard deviation of all their filled bottles is .2 ounces or less.A sample of 20 bottles showed a standard deviation of .12 ounces.The test statistic to test the claim is

Accepted Answer

We need to perform a one-sample t-test since the population standard deviation is not known. The test statistic can be calculated as: 
t = (x̄ - μ) / (s / √n) 
where x̄ is the sample mean, μ is the claimed mean, s is the sample standard deviation, and n is the sample size.

In this case, the claimed mean is not given, but we can assume it to be the mean of the sample, which is reasonable if the sample is representative of the population. Therefore, we have: 
t = (x̄ - μ) / (s / √n) = (0 - x̄) / (0.2 / √20) = -√20 * x̄ / 0.2828

The critical value for a one-tailed t-test with 19 degrees of freedom at a significance level of 0.05 is 1.734. Since the alternative hypothesis is that the true standard deviation is greater than 0.2, we only need to check if t > 1.734.

Substituting the sample values, we get: 
t = -√20 * 0 / 0.2828 = 0 
Since t < 1.734, we cannot reject the null hypothesis that the true standard deviation is less than or equal to 0.2.

The calculated test statistic is 0, which corresponds to a p-value of 0.5. Alternatively, we can calculate the test statistic as t = (0.12 - 0) / (0.2 / √20) = 6.84, which gives the same conclusion. Therefore, the answer is C.

Question 8

A sample of n observations is taken from a normal population.The appropriate chi-square distribution has

Accepted Answer

The appropriate chi-square distribution for a sample of n observations taken from a normal population, when estimating variance, has n - 1 degrees of freedom. This accounts for the loss of one degree of freedom due to the estimation of the sample mean.

Question 9

For an F distribution, the number of degrees of freedom for the numerator

Accepted Answer

The F distribution does not have a requirement for the relationship between the numerator and denominator degrees of freedom; they can be larger, smaller, or equal to each other.

Question 10

Which of the following has a chi-square distribution?

Accepted Answer

The expression (n - 1)s^2/σ^2 has a chi-square distribution. This is because it is the formula for the chi-square statistic when testing the variance of a normally distributed sample, where s^2 is the sample variance, σ^2 is the population variance, and n is the sample size.

Question 11

The manager of the service department of a local car dealership has noted that the service times of a sample of 16 new automobiles has a standard deviation of 5 minutes.A 95% confidence interval estimate for the variance of service times for all their new automobiles is

Accepted Answer

To estimate the confidence interval for the variance of service times, we use the chi-square distribution because the sample standard deviation is known, and we're dealing with variance. The formula for the confidence interval for variance ($\sigma^2$) is given by:$$\left( \frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}} \right)$$where $n$ is the sample size, $s^2$ is the sample variance, and $\chi^2_{\alpha/2, n-1}$ and $\chi^2_{1-\alpha/2, n-1}$ are the chi-square values for $\alpha/2$ and $1-\alpha/2$ degrees of freedom, respectively.Given:- $n = 16$- Standard deviation $s = 5$ minutes, so the variance $s^2 = 25$- Confidence level = 95%, so $\alpha = 0.05$The degrees of freedom ($df$) is $n - 1 = 15$. For a 95% confidence interval and 15 degrees of freedom, the chi-square values (from a chi-square distribution table) are approximately:- $\chi^2_{0.025, 15} \approx 27.488$- $\chi^2_{0.975, 15} \approx 6.262$Plugging these values into the formula gives:$$\left( \frac{(16-1) \cdot 25}{27.488}, \frac{(16-1) \cdot 25}{6.262} \right) \approx (13.642, 59.885)$$Therefore, the correct answer is A) 13.642 to 59.885.

Question 12

The producer of a certain bottling equipment claims that the variance of all their filled bottles is .027 or less.A sample of 30 bottles showed a standard deviation of .2 ounces.The p-value for the test is

Accepted Answer

The p-value is determined by conducting a chi-square test for variance. Given the sample size (n=30), the sample standard deviation (s=.2), and the claimed population variance (σ^2=.027), we calculate the test statistic and then find the corresponding p-value. The fact that the p-value falls between .025 to .05 suggests that the sample variance significantly differs from the claimed variance, but without the exact calculations or the chi-square distribution table, we can't pinpoint the precise p-value, only that it falls within this range based on the options provided.

Question 13

The sampling distribution used when making inferences about a single population's variance is

Accepted Answer

When making inferences about a single population's variance, we use a chi-square distribution for the sampling distribution. This is because the variance is a measure of the spread of the data and is therefore always non-negative. The chi-square distribution is also non-negative and is used specifically for testing hypotheses about variances.

Question 14

A sample of 20 cans of tomato juice showed a standard deviation of .4 ounces.A 95% confidence interval estimate of the variance for the population is

Accepted Answer

To find the confidence interval for the population variance, we use the chi-square distribution.

The formula for the chi-square distribution is:

((n-1)*s^2)/chi-square critical value =< sigma^2 =< ((n-1)*s^2)/chi-square critical value

where:
n = sample size
s = sample standard deviation
sigma^2 = population variance
chi-square critical value = value from chi-square distribution table based on the degrees of freedom (df) and confidence level

In this case, n=20, s=0.4, and we want a 95% confidence interval, which corresponds to a chi-square critical value of 31.4104 (based on df=19). Substituting these values into the formula, we get:

((20-1)*0.4^2)/31.4104 =< sigma^2 =< ((20-1)*0.4^2)/31.4104
0.0925 =< sigma^2 =< 0.3413

Therefore, the 95% confidence interval for the population variance is 0.0925 to 0.3413, which corresponds to choice D.

Question 15

The value of F_.05 with 8 numerator and 19 denominator degrees of freedom is

Accepted Answer

The answer of The value of F_.05 with 8 numerator...

Question 16

The random variable for a chi-square distribution may assume

Accepted Answer

A chi-square distribution is a probability distribution in which the random variable takes only non-negative values, i.e., any value greater than zero. Thus, only choice D is correct.

Question 17

A sample of 51 observations yielded a sample standard deviation of 5.If we want to test H₀: σ² = 20, the test statistic is

Accepted Answer

The test statistic for testing a hypothesis about a population variance (σ^2) is calculated using the formula χ^2 = (n-1)s^2 / σ^2, where n is the sample size, s is the sample standard deviation, and σ is the population standard deviation under the null hypothesis. Here, n = 51, s = 5, and σ = 20. Plugging these values into the formula gives χ^2 = (51-1)(5^2) / 20^2 = 50*25 / 400 = 1250 / 400 = 62.5.

Question 18

The manager of the service department of a local car dealership has noted that the service times of a sample of 30 new automobiles has a standard deviation of 5 minutes.A 95% confidence interval estimate for the standard deviation of the service times (in minutes) for all their new automobiles is

Accepted Answer

Since we are estimating the standard deviation of the population, we will use the chi-square distribution. The formula for the confidence interval is:

[ (n-1)s^2) / chi-square(alpha/2,n-1), (n-1)s^2) / chi-square(1-alpha/2,n-1) ]

where s is the sample standard deviation, n is the sample size, and chi-square is the chi-square distribution value with alpha/2 (lower tail) and 1-alpha/2 (upper tail) degrees of freedom.

Using a chi-square table or calculator, we find that the chi-square values for alpha/2 = 0.025 and degrees of freedom = 29 and 1-alpha/2 = 0.975 and degrees of freedom = 29 are 16.047 and 45.722, respectively.

Plugging in the given values, we get:

[ (30-1)(5^2) / 45.722, (30-1)(5^2) / 16.047 ]

Simplifying the expression, we get:

[ 3.982, 6.722 ]

Therefore, the 95% confidence interval estimate for the standard deviation of the service times is 3.982 to 6.722 minutes. Therefore, the best choice is C.

Question 19

The symbol used for the variance of the population is

Accepted Answer

The symbol for the variance of the population is represented by σ^2, which is option B (σ²), where the notation indicates sigma squared, denoting population variance.

Question 20

Which of the following has an F distribution?

Accepted Answer

The answer of Which of the following has an F...

Quiz 11: Inferences About Population Variances