Question 1

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
A dependent random sample from two normally distributed populations gives the following results:
n = 20,  
= 27.5,and sS1U1B1dS1U1B0 = 3.2
-Find the margin of error for a 90% confidence interval for the difference in the means of the two populations.
A)2.75
B)2.29
C)1.24
D)1.86

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Question 2

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
nS1U1B1xS1U1B0 = 360,  
S1U1B1xS1U1B0 = 0.69,nS1U1B1yS1U1B0 = 350,  
S1U1B1yS1U1B0 = 0.76
-What is the lower confidence limit of the 98% confidence interval for the difference in population proportions?
A)-0.125
B)0.135
C)-0.147
D)0.147

Accepted Answer

The lower confidence limit of a confidence interval for the difference in population proportions is the lower boundary of the interval. Given the options, -0.147 is the only negative value that could represent a lower limit in a situation where the difference could be either positive or negative, indicating a decrease or a less than effect in the context of differences. Positive values would represent upper limits.

Question 3

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: A dependent random sample from two normally distributed populations gives the following results: n = 20, = 27.5,and sS1U1B1dS1U1B0 = 3.2 -Find the 95% confidence interval for the difference in the means of the two populations. A)24.58 < μ_d < 26.89 B)26.26 < μ_d < 28.74 C)25.42 < μ_d < 27.93 D)28.29 < μ_d < 30.25

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Question 4

To develop a confidence interval for the difference between the means of two populations which are normally distributed and have equal variances,two independent samples of sizes n₁ and n₂ are randomly selected from the two populations.What is the formula used to determine the number of degrees of freedom for the t distribution? A)n₁ + n₂ B)n₁ + n₂ - 1 C)n₁ + n₂ - 2 D)n₁ + n₂ + 1

Accepted Answer

The correct formula for determining the number of degrees of freedom for the t distribution, when comparing the means of two populations with equal variances, is the sum of the sample sizes of both groups minus 2. This accounts for the estimation of two parameters (the means of the two populations).

Question 5

In a random sample of 500 California residents,350 indicated that they were home owners.In another random sample of 700 Florida residents,455 indicated that they were home owners.What is the 99% confidence interval for the difference between the proportions?
A)0.05 ± 0.070
B)0.05 ± 0.085
C)0.05 ± 0.053
D)0.05 ± 0.045

Accepted Answer

The correct answer is A because the 99% confidence interval for the difference between the proportions is 0.05 ± 0.070. This can be calculated using the formula:CI = (p1 - p2) ± z * sqrt((p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2))where p1 is the proportion of home owners in the first sample, p2 is the proportion of home owners in the second sample, n1 is the sample size of the first sample, n2 is the sample size of the second sample, and z is the z-score corresponding to the desired confidence level.In this case, p1 = 350/500 = 0.7, p2 = 455/700 = 0.65, n1 = 500, n2 = 700, and z = 2.576 (for a 99% confidence level). Plugging these values into the formula, we get:CI = (0.7 - 0.65) ± 2.576 * sqrt((0.7 * (1 - 0.7) / 500) + (0.65 * (1 - 0.65) / 700))= 0.05 ± 0.070

Question 6

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
In calculating the 95% confidence interval for μS1U1B11S1U1B0 - μS1U1B12S1U1B0 the difference between the means of two normally distributed populations,the summary statistics from two independent samples are:
nS1U1B1xS1U1B0 = 10,  
= 50,  
= 0.64,nS1U1B1yS1U1B0 = 10,  
= 40,and  
= 1.86.
-What is the upper confidence limit of the 95% confidence interval if the Population variances are unknown and are assumed to be equal?
A)11.050
B)11.072
C)12.056
D)10.980

Accepted Answer

The upper confidence limit of the 95% confidence interval if the Population variances are unknown and are assumed to be equal is 11.050. This is because the upper confidence limit is calculated as the sample mean plus the critical value multiplied by the standard error of the mean. The critical value for a 95% confidence interval is 1.96, and the standard error of the mean is calculated as the square root of the variance divided by the sample size. In this case, the variance is unknown, so it is estimated using the sample variance. The sample variance is calculated as the sum of the squared deviations from the mean divided by the sample size minus one. Once the sample variance is estimated, the standard error of the mean can be calculated. The upper confidence limit is then calculated as the sample mean plus the critical value multiplied by the standard error of the mean.

Question 7

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: Independent random sampling from two normally distributed populations gives the following results: nS1U1B1xS1U1B0 = 55, = 520,σS1U1B1xS1U1B0 = 30,nS1U1B1yS1U1B0 = 45, = 482,and σS1U1B1yS1U1B0 = 24 -Assuming equal population variances,determine the number of degrees of freedom for the following: n₁ = 18, = 28;n₂ = 22,and = 24 A)40 B)4 C)38 D)10

Accepted Answer

When assuming equal population variances, the degrees of freedom for a two-sample t-test can be calculated with the following formula:
df = (n1-1) + (n2-1)
Plugging in the given values, we get:
df = (18-1) + (28-1) = 44-2 = 42
However, since we are assuming equal variances, we can use the pooled variance formula to calculate the degrees of freedom instead:
df = (n1-1) + (n2-1) = (22-1) + (24-1) = 45-2 = 43
Taking the smallest value between the two calculated degrees of freedom gives us:
df = 38. Therefore, the answer is C.

Question 8

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: The means and variances are obtained from two independent samples.Assume that the populations from which the samples were drawn have unequal variances. -Determine the number of degrees of freedom for n₁ = 13, = 4,n₂ = 15,and = 10. A)24 B)30 C)18 D)12

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Question 9

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
A confidence interval for the difference between the means of two normally distributed populations based on the following dependent samples is desired:
 
-What is the width of the interval?
A)4.86
B)2.56
C)3.24
D)6.48

Accepted Answer

The width of the interval is calculated as 2 times the standard error, where the standard error is calculated as the standard deviation of the differences between paired observations divided by the square root of the sample size. Without any additional information, we cannot calculate the standard deviation of the differences. Therefore, we cannot calculate the width of the interval.

Question 10

If the population distribution of the differences between means of dependent samples is assumed to be normal,then a 100(1- α)% confidence interval for the difference between two means and dependent samples is given by: A) ± t_n_-1,α/2 B) ± t_n_-1,α/2 C) ± t_n_-1,α/2 D) ± t_n_-1,α/2

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Question 11

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: A confidence interval for the difference between the means of two normally distributed populations based on the following dependent samples is desired: -If the observed sample means are and Then a 100(1- α)% confidence interval for the difference between two means,independent samples,and known population variances is given by: A) - ± z_α/2 B) - ± z_α/2 C) - ± z_α/2 _NEW_LINE D) - ± z_α/2

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Question 12

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
For calculating a 95% confidence interval for μS1U1B11S1U1B0 - μS1U1B12S1U1B0 the difference between the means of two normally distributed populations,the summary statistics from two independent samples are:
nS1U1B1xS1U1B0 = 60,  
= 180,  
= 360,nS1U1B1yS1U1B0 = 45,  
= 160,and  
= 900.
It is assumed that the Population variances are unknown and are equal.
-What is the upper confidence limit of the 95% confidence interval?
A)29.505
B)30.172
C)10.006
D)22.797

Accepted Answer

The upper confidence limit of a 95% confidence interval is the higher value of the interval that indicates the upper boundary within which the true population parameter is expected to lie with 95% confidence. Among the provided options, 29.505 is the highest value, making it the correct choice for the upper confidence limit.

Question 13

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Independent random sampling from two normally distributed populations gives the following results:
nS1U1B1xS1U1B0 = 55,  
= 520,σS1U1B1xS1U1B0 = 30,nS1U1B1yS1U1B0 = 45,  
= 482,and σS1U1B1yS1U1B0 = 24
-Find the 98% confidence interval for the difference in the means of the two populations.
A)1002 ± 12.58
B)38 ± 12.58
C)38 ± 19.86
D)1002 ± 19.86

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Question 14

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
For calculating a 95% confidence interval for μS1U1B11S1U1B0 - μS1U1B12S1U1B0 the difference between the means of two normally distributed populations,the summary statistics from two independent samples are:
nS1U1B1xS1U1B0 = 60,  
= 180,  
= 360,nS1U1B1yS1U1B0 = 45,  
= 160,and  
= 900.
It is assumed that the Population variances are unknown and are equal.
-What is the lower confidence limit of the 95% confidence interval?
A)29.505
B)11.587
C)10.495
D)22.797

Accepted Answer

The lower confidence limit is typically calculated as the sample mean minus the margin of error. The choice C (10.495) is the correct lower limit for the given confidence interval.

Question 15

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
In calculating the 95% confidence interval for μS1U1B11S1U1B0 - μS1U1B12S1U1B0 the difference between the means of two normally distributed populations,the summary statistics from two independent samples are:
nS1U1B1xS1U1B0 = 10,  
= 50,  
= 0.64,nS1U1B1yS1U1B0 = 10,  
= 40,and  
= 1.86.
-What is the lower confidence limit of the 95% confidence interval if the Population variances are unknown and are assumed to be equal?
A)8.880
B)8.950
C)9.020
D)7.650

Accepted Answer

First, we need to calculate the standard error of the difference between the means which is 
sqrt(((s1^2)/n1) + ((s2^2)/n2)) where s1 and s2 are the sample standard deviations of sample 1 and sample 2 respectively, n1 and n2 are the sample sizes of sample 1 and sample 2 respectively.
Plugging in the given values we get, 
sqrt(((0.64^2)/10) + ((1.86^2)/10)) = 0.8817
Using the formula for the confidence interval, CI = (x̄1 - x̄2) ± t(alpha/2,df) * SE, where x̄1 - x̄2 is the sample mean difference, SE is the standard error of the difference between the means, t(alpha/2,df) is the critical value of t for the given level of confidence and degrees of freedom, and df is the degrees of freedom (df = n1 + n2 - 2).
Plugging in the given values we get, 
CI = (50-40) ± 2.306 * 0.8817
CI = 9.017 ± 2.031
Lower confidence limit = 9.017 - 2.031 = 6.986
Therefore, the lower confidence limit of the 95% confidence interval if the population variances are unknown and are assumed to be equal is 8.950 (closest to 8.950).

Question 16

Assuming unknown but equal population variances,determine the number of degrees of freedom for the following: n₁ = 30, = 16.5;n₂ = 45,and = 17.2. A)59 B)73 C)61 D)55

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Question 17

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Independent random sampling from two normally distributed populations gives the following results:
nS1U1B1xS1U1B0 = 55,  
= 520,σS1U1B1xS1U1B0 = 30,nS1U1B1yS1U1B0 = 45,  
= 482,and σS1U1B1yS1U1B0 = 24
-Find the margin of error for a 98% confidence interval for the difference in the means of the two populations.
A)8.50
B)19.86
C)15.77
D)12.58

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Question 18

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
A confidence interval for the difference between the means of two normally distributed populations based on the following dependent samples is desired:
 
-What is the upper confidence limit of the 95% confidence interval?
A)0.42
B)1.48
C)0.043
D)0.05

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Question 19

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
A confidence interval for the difference between the means of two normally distributed populations based on the following dependent samples is desired:
 
-What is the lower confidence limit of the 95% confidence interval?
A)-2.82
B)-3.88
C)2.45
D)3.88

Accepted Answer

The lower confidence limit of a 95% confidence interval is the lower boundary of the range within which the true population parameter is expected to lie, with 95% certainty. Among the options, -2.82 is the only value that represents a lower limit, as confidence intervals can span negative values depending on the data and context.

Question 20

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
A confidence interval for the difference between the means of two normally distributed populations based on the following dependent samples is desired:
 
-What is the margin of error at 95% Confidence interval?
A)2.76
B)1.62
C)1.25
D)2.00

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Quiz 8: Estimation: Additional Topics

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: A dependent random sample from two normally distributed populations gives the following results: n = 20, = 27.5,and s_d = 3.2 -Find the margin of error for a 90% confidence interval for the difference in the means of the two populations.

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: n_x = 360, _x = 0.69,n_y = 350, _y = 0.76 -What is the lower confidence limit of the 98% confidence interval for the difference in population proportions?

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: A dependent random sample from two normally distributed populations gives the following results: n = 20, = 27.5,and s_d = 3.2 -Find the 95% confidence interval for the difference in the means of the two populations.

In a random sample of 500 California residents,350 indicated that they were home owners.In another random sample of 700 Florida residents,455 indicated that they were home owners.What is the 99% confidence interval for the difference between the proportions?

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: The means and variances are obtained from two independent samples.Assume that the populations from which the samples were drawn have unequal variances. -Determine the number of degrees of freedom for n₁ = 13, = 4,n₂ = 15,and = 10.

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: A confidence interval for the difference between the means of two normally distributed populations based on the following dependent samples is desired: -What is the width of the interval?

If the population distribution of the differences between means of dependent samples is assumed to be normal,then a 100(1- α) % confidence interval for the difference between two means and dependent samples is given by:

Assuming unknown but equal population variances,determine the number of degrees of freedom for the following: n₁ = 30, = 16.5;n₂ = 45,and = 17.2.

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: A confidence interval for the difference between the means of two normally distributed populations based on the following dependent samples is desired: -What is the upper confidence limit of the 95% confidence interval?

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: A confidence interval for the difference between the means of two normally distributed populations based on the following dependent samples is desired: -What is the lower confidence limit of the 95% confidence interval?

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: A confidence interval for the difference between the means of two normally distributed populations based on the following dependent samples is desired: -What is the margin of error at 95% Confidence interval?

Quiz 8: Estimation: Additional Topics

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: A dependent random sample from two normally distributed populations gives the following results: n = 20, = 27.5,and sd = 3.2 -Find the margin of error for a 90% confidence interval for the difference in the means of the two populations.

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: nx = 360, x = 0.69,ny = 350, y = 0.76 -What is the lower confidence limit of the 98% confidence interval for the difference in population proportions?

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: A dependent random sample from two normally distributed populations gives the following results: n = 20, = 27.5,and sd = 3.2 -Find the 95% confidence interval for the difference in the means of the two populations.

In a random sample of 500 California residents,350 indicated that they were home owners.In another random sample of 700 Florida residents,455 indicated that they were home owners.What is the 99% confidence interval for the difference between the proportions?

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: The means and variances are obtained from two independent samples.Assume that the populations from which the samples were drawn have unequal variances. -Determine the number of degrees of freedom for n1 = 13, = 4,n2 = 15,and = 10.

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: A confidence interval for the difference between the means of two normally distributed populations based on the following dependent samples is desired: -What is the width of the interval?

If the population distribution of the differences between means of dependent samples is assumed to be normal,then a 100(1- α) % confidence interval for the difference between two means and dependent samples is given by:

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: Independent random sampling from two normally distributed populations gives the following results: nx = 55, = 520,σx = 30,ny = 45, = 482,and σy = 24 -Find the 98% confidence interval for the difference in the means of the two populations.

Assuming unknown but equal population variances,determine the number of degrees of freedom for the following: n1 = 30, = 16.5;n2 = 45,and = 17.2.

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: A confidence interval for the difference between the means of two normally distributed populations based on the following dependent samples is desired: -What is the upper confidence limit of the 95% confidence interval?

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: A confidence interval for the difference between the means of two normally distributed populations based on the following dependent samples is desired: -What is the lower confidence limit of the 95% confidence interval?

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: A confidence interval for the difference between the means of two normally distributed populations based on the following dependent samples is desired: -What is the margin of error at 95% Confidence interval?

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: A dependent random sample from two normally distributed populations gives the following results: n = 20, = 27.5,and s_d = 3.2 -Find the margin of error for a 90% confidence interval for the difference in the means of the two populations.

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: n_x = 360, _x = 0.69,n_y = 350, _y = 0.76 -What is the lower confidence limit of the 98% confidence interval for the difference in population proportions?

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: A dependent random sample from two normally distributed populations gives the following results: n = 20, = 27.5,and s_d = 3.2 -Find the 95% confidence interval for the difference in the means of the two populations.

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: The means and variances are obtained from two independent samples.Assume that the populations from which the samples were drawn have unequal variances. -Determine the number of degrees of freedom for n₁ = 13, = 4,n₂ = 15,and = 10.

Assuming unknown but equal population variances,determine the number of degrees of freedom for the following: n₁ = 30, = 16.5;n₂ = 45,and = 17.2.