Question 1

The contents of a sample of 26 cans of apple juice showed a standard deviation of .06 ounces. We are interested in testing whether the variance of the population is significantly more than .003. The null hypothesis is

Accepted Answer

The null hypothesis in this case is that the variance of the population is not significantly more than .003. This can be written as σ^2 ≤ .003. Therefore, the correct choice is D) σ² ≤ .003.

Question 2

The contents of a sample of 26 cans of apple juice showed a standard deviation of .06 ounces. We are interested in testing whether the variance of the population is significantly more than .003. The test statistic is

Accepted Answer

The test statistic for testing the hypothesis about the population variance is calculated as:

χ^2 = (n-1)*s^2/σ^2

where n is the sample size, s is the sample standard deviation, and σ^2 is the hypothesized population variance.

Substituting the given values, we get:

χ^2 = (26-1)*0.06^2/0.003 = 30

Therefore, the test statistic is 30.

Question 3

The contents of a sample of 26 cans of apple juice showed a standard deviation of .06 ounces. We are interested in testing whether the variance of the population is significantly more than .003. At the .05 level of significance, the null hypothesis

Accepted Answer

The answer of The contents of a sample of 26...

Question 4

The value of F_.01 with 9 numerator and 20 denominator degrees of freedom is

Accepted Answer

The F-value at the 0.01 significance level with 9 numerator and 20 denominator degrees of freedom is 3.46, based on F-distribution tables.

Question 5

The chi-square value for a one-tailed (upper tail) hypothesis test at a 5% level of significance and a sample size of 25 is

Accepted Answer

The chi-square value for a hypothesis test is determined by the degrees of freedom (df) and the level of significance. For a sample size of 25, the degrees of freedom would be 25 - 1 = 24. At a 5% level of significance for the upper tail, the chi-square critical value for 24 degrees of freedom is approximately 36.415.

Question 6

A sample of 21 elements is selected to estimate a 90% confidence interval for the variance of the population. The chi-square value(s) to be used for this interval estimation is(are)

Accepted Answer

To estimate the confidence interval for the variance of the population, we use the chi-square distribution. The degrees of freedom for this distribution is equal to the sample size minus one, which for this case is 21 - 1 = 20.

To find the chi-square value(s) to be used for a 90% confidence interval, we need to look up the values of chi-square with 20 degrees of freedom in a chi-square distribution table. The value for the upper tail probability of 0.05 (since it's a two-tailed test and 0.1 is divided among two tails) is 31.410, and the value for the lower tail probability of 0.05 is 10.851.

Therefore, the chi-square values to be used for this interval estimation are 10.851 and 31.410. Choice C is the only option that includes both values.

Question 7

The sampling distribution of the ratio of independent sample variances extracted from two normal populations with equal variances is the

Accepted Answer

The ratio of independent sample variances follows an F distribution. This is because the ratio of two independent chi-square distributions divided by their degrees of freedom follows an F distribution. In this case, the chi-square distributions come from the sample variances of two normal populations with equal variances, hence the resulting sampling distribution is an F distribution.

Question 8

For a sample size of 21 at 95% confidence, the chi-square values needed for interval estimation are

Accepted Answer

For a sample size of 21, the degrees of freedom are 20. At 95% confidence, the chi-square critical values for 20 degrees of freedom are approximately 9.591 (lower) and 34.170 (upper).

Question 9

We are interested in testing to see if the variance of a population is less than 7. The correct null hypothesis is

Accepted Answer

The correct null hypothesis for testing if the variance of a population is less than a specific value (in this case, 7) should state that the population variance is greater than or equal to this value. This is because null hypotheses typically represent a statement of no effect or no difference, which in this context means assuming the variance is not less than 7 until evidence suggests otherwise. Therefore, the correct null hypothesis is $ \sigma^2 \geq 7 $.

Question 10

A sample of 60 items from population 1 has a sample variance of 12 while a sample of 50 items from population 2 has a sample variance of 15. If we want to test whether the variances of the two populations are equal, the test statistic will have a value of

Accepted Answer

The answer of A sample of 60 items from population...

Question 11

The contents of a sample of 26 cans of apple juice showed a standard deviation of .06 ounces. We are interested in testing whether the variance of the population is significantly more than .003. The p-value for this test is

Accepted Answer

The p-value is greater than .10 because the calculated chi-square statistic for the sample variance compared to the hypothesized variance does not fall in the critical region for a significance level of 0.10.

Question 12

The sampling distribution of the quantity (n - 1)s²/σ² is the

Accepted Answer

The sampling distribution of the quantity $(n - 1)s^2/\sigma^2$ follows a chi-square distribution, where $s^2$ is the sample variance, $\sigma^2$ is the population variance, and $n$ is the sample size. This is a fundamental result in statistics related to the properties of variances.

Question 13

The chi-square value for a one-tailed (lower tail) test when the level of significance is .1 and the sample size is 15 is

Accepted Answer

To find the chi-square value for a one-tailed (lower tail) test, we need to use a chi-square distribution table. With a level of significance of .1 and 15 degrees of freedom, the chi-square value is 7.790. This is because the one-tailed test is only looking at one end of the distribution, which reduces the area under the curve and decreases the critical value.

Question 14

The 90% confidence interval estimate of a population standard deviation when a sample variance of 50 is obtained from a sample of 15 items is

Accepted Answer

The formula to calculate the confidence interval of the population standard deviation is:
(√[ (n-1) s² / χ²_(α/2)(n-1)] , √[ (n-1) s² / χ²_(1-α/2)(n-1)])
where 
n = sample size, 
s² = sample variance, 
χ²_(α/2)(n-1) and χ²_(1-α/2)(n-1) are the chi-square values for the given level of confidence and degree of freedom (n-1).

Substituting the values in the formula, we get
(√[(15-1) 50 / χ²_(0.05/2)(14)], √[(15-1) 50 / χ²_(1-0.05/2)(14)])
= (√[14*50 / 6.572], √[14*50 / 26.119])
= (5.436, 10.321)

Therefore, the 90% confidence interval estimate of a population standard deviation is 5.436 to 10.321. Option C is the correct answer.

Question 15

The 99% confidence interval estimate for a population variance when a sample standard deviation of 12 is obtained from a sample of 10 items is

Accepted Answer

The formula for calculating a confidence interval for population variance when sample standard deviation is known is:

((n-1)s^2) / chi-square value
where:
n = sample size
s = sample standard deviation
chi-square value = critical value from chi-square distribution table at (1- confidence level) and (n-1) degrees of freedom.

Substituting the given values, we get:
((10-1) * 12^2) / 3.247 = 54.941
((10-1) * 12^2) / 0.064 = 746.974

Therefore, the 99% confidence interval estimate for a population variance is 54.941 to 746.974. 
Hence, the correct choice is C.

Question 16

The critical value of F using α = .05 when there is a sample size of 21 for the sample with the smaller variance, and there is a sample size of 9 for the sample with the larger sample variance is

Accepted Answer

The critical value for F with α = .05, a numerator degrees of freedom of 8 and a denominator degrees of freedom of 20 is 2.45. Since the question specifies that the sample with the smaller variance has a sample size of 21, we know that this sample corresponds to the denominator degrees of freedom. Therefore, the critical value is 2.45.

Question 17

We are interested in testing whether the variance of a population is significantly more than 625. The null hypothesis for this test is

Accepted Answer

The null hypothesis for testing whether the variance of a population is significantly more than 625 is H₀: σ² ≤ 625. This is because the alternative hypothesis is that the variance is greater than 625. A one-tailed test with the alternative hypothesis of σ² > 625 would require a different null hypothesis. Option A incorrectly uses a greater than sign in the null hypothesis, Option C uses a greater than or equal to sign which is inconsistent with the alternative hypothesis, and Option D uses a much smaller value for the variance in the null hypothesis.

Question 18

The 95% confidence interval estimate of a population variance when a sample variance of 324 is obtained from a sample of 81 items is

Accepted Answer

The 95% confidence interval for a population variance is calculated using the chi-square distribution with the formula: $$ \left( \frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}} \right) $$ 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 critical values for $n-1$ degrees of freedom. Given $n=81$, $s^2=324$, and a 95% confidence level (so $\alpha=0.05$), we can find the chi-square critical values and calculate the interval. The correct calculation leads to the interval (243.086, 453.520), which matches option B.

Question 19

A sample of 61 observations yielded a sample standard deviation of 6. If we want to test H₀: σ² = 40, the test statistic is

Accepted Answer

The answer of A sample of 61 observations yielded a...

Question 20

A random sample of 31 sales charge showed a sample standard deviation of $50. A 90% confidence interval estimate of the population standard deviation is

Accepted Answer

The formula for a confidence interval estimate of the population standard deviation is:
Confidence interval = (n-1)S^2 / χ^2(df, α/2) to (n-1)S^2 / χ^2(df, 1- α/2)
where S = sample standard deviation, n = sample size, df = degrees of freedom (df = n-1), and α is the level of significance.

Plugging in the given values, we get the confidence interval as:
(30*50^2 / χ^2(30, 0.05/2)) to (30*50^2 / χ^2(30, 1- 0.05/2))
= (48035.2 / 45.72) to (48035.2 / 17.71)
= 1050.2 to 2701.7

Therefore, the 90% confidence interval estimate of the population standard deviation is (1050.2, 2701.7). Option D is the only choice that falls within the range of this interval.

Quiz 11: Inferences About Population Variances

The contents of a sample of 26 cans of apple juice showed a standard deviation of .06 ounces. We are interested in testing whether the variance of the population is significantly more than .003. The null hypothesis is

The contents of a sample of 26 cans of apple juice showed a standard deviation of .06 ounces. We are interested in testing whether the variance of the population is significantly more than .003. The test statistic is

The contents of a sample of 26 cans of apple juice showed a standard deviation of .06 ounces. We are interested in testing whether the variance of the population is significantly more than .003. At the .05 level of significance, the null hypothesis

The value of F_.01 with 9 numerator and 20 denominator degrees of freedom is

The chi-square value for a one-tailed (upper tail) hypothesis test at a 5% level of significance and a sample size of 25 is

A sample of 21 elements is selected to estimate a 90% confidence interval for the variance of the population. The chi-square value(s) to be used for this interval estimation is(are)

The sampling distribution of the ratio of independent sample variances extracted from two normal populations with equal variances is the

For a sample size of 21 at 95% confidence, the chi-square values needed for interval estimation are

We are interested in testing to see if the variance of a population is less than 7. The correct null hypothesis is

A sample of 60 items from population 1 has a sample variance of 12 while a sample of 50 items from population 2 has a sample variance of 15. If we want to test whether the variances of the two populations are equal, the test statistic will have a value of

The contents of a sample of 26 cans of apple juice showed a standard deviation of .06 ounces. We are interested in testing whether the variance of the population is significantly more than .003. The p-value for this test is

The sampling distribution of the quantity (n - 1) s²/σ² is the

The chi-square value for a one-tailed (lower tail) test when the level of significance is .1 and the sample size is 15 is

The 90% confidence interval estimate of a population standard deviation when a sample variance of 50 is obtained from a sample of 15 items is

The 99% confidence interval estimate for a population variance when a sample standard deviation of 12 is obtained from a sample of 10 items is

The critical value of F using α = .05 when there is a sample size of 21 for the sample with the smaller variance, and there is a sample size of 9 for the sample with the larger sample variance is

We are interested in testing whether the variance of a population is significantly more than 625. The null hypothesis for this test is

The 95% confidence interval estimate of a population variance when a sample variance of 324 is obtained from a sample of 81 items is

A sample of 61 observations yielded a sample standard deviation of 6. If we want to test H₀: σ² = 40, the test statistic is

A random sample of 31 sales charge showed a sample standard deviation of $50. A 90% confidence interval estimate of the population standard deviation is

Quiz 11: Inferences About Population Variances

The contents of a sample of 26 cans of apple juice showed a standard deviation of .06 ounces. We are interested in testing whether the variance of the population is significantly more than .003. The null hypothesis is

The contents of a sample of 26 cans of apple juice showed a standard deviation of .06 ounces. We are interested in testing whether the variance of the population is significantly more than .003. The test statistic is

The contents of a sample of 26 cans of apple juice showed a standard deviation of .06 ounces. We are interested in testing whether the variance of the population is significantly more than .003. At the .05 level of significance, the null hypothesis

The value of F.01 with 9 numerator and 20 denominator degrees of freedom is

The chi-square value for a one-tailed (upper tail) hypothesis test at a 5% level of significance and a sample size of 25 is

A sample of 21 elements is selected to estimate a 90% confidence interval for the variance of the population. The chi-square value(s) to be used for this interval estimation is(are)

The sampling distribution of the ratio of independent sample variances extracted from two normal populations with equal variances is the

For a sample size of 21 at 95% confidence, the chi-square values needed for interval estimation are

We are interested in testing to see if the variance of a population is less than 7. The correct null hypothesis is

A sample of 60 items from population 1 has a sample variance of 12 while a sample of 50 items from population 2 has a sample variance of 15. If we want to test whether the variances of the two populations are equal, the test statistic will have a value of

The contents of a sample of 26 cans of apple juice showed a standard deviation of .06 ounces. We are interested in testing whether the variance of the population is significantly more than .003. The p-value for this test is

The sampling distribution of the quantity (n - 1) s2/σ2 is the

The chi-square value for a one-tailed (lower tail) test when the level of significance is .1 and the sample size is 15 is

The 90% confidence interval estimate of a population standard deviation when a sample variance of 50 is obtained from a sample of 15 items is

The 99% confidence interval estimate for a population variance when a sample standard deviation of 12 is obtained from a sample of 10 items is

The critical value of F using α = .05 when there is a sample size of 21 for the sample with the smaller variance, and there is a sample size of 9 for the sample with the larger sample variance is

We are interested in testing whether the variance of a population is significantly more than 625. The null hypothesis for this test is

The 95% confidence interval estimate of a population variance when a sample variance of 324 is obtained from a sample of 81 items is

A sample of 61 observations yielded a sample standard deviation of 6. If we want to test H0: σ2 = 40, the test statistic is

A random sample of 31 sales charge showed a sample standard deviation of $50. A 90% confidence interval estimate of the population standard deviation is

The value of F_.01 with 9 numerator and 20 denominator degrees of freedom is

The sampling distribution of the quantity (n - 1) s²/σ² is the

A sample of 61 observations yielded a sample standard deviation of 6. If we want to test H₀: σ² = 40, the test statistic is