Question 1

In a sample of 10 randomly selected women, it was found that their mean height was 63.4 inches. From previous studies, it is assumed that the standard deviation σ is 2.4 and that the population of height
Measurements is normally distributed. Construct the 95% confidence interval for the population mean.
A)(61.9, 64.9)
B)(58.1, 67.3)
C)(59.7, 66.5)
D)(60.8, 65.4)

Accepted Answer

To construct a 95% confidence interval for the population mean when the population standard deviation is known, we use the formula: $\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $Z$ is the Z-value corresponding to the desired confidence level (for 95%, $Z = 1.96$), $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values: $63.4 \pm 1.96 \frac{2.4}{\sqrt{10}}$, we get the interval (61.9, 64.9).

Question 2

The number of wins in a season for 32 randomly selected professional football teams are listed below.
Construct a 90% confidence interval for the true mean number of wins in a season. Assume that $\sigma$ is 2.6.
$$\begin{array} { l l l l l l l l } 
9 & 9 & 9 & 8 & 10 & 9 & 7 & 2 \\
11 & 10 & 6 & 4 & 11 & 9 & 8 & 8 \\
12 & 10 & 7 & 5 & 12 & 6 & 4 & 3 \\
12 & 9 & 9 & 7 & 10 & 7 & 7 & 5
\end{array}$$

Accepted Answer

(7.2, 8.8)

Question 3

A group of 49 randomly selected students has a mean age of 22.4 years. Assume the population standard deviation is 3.8. Construct a 98% confidence interval for the population mean.
A)(21.1, 23.7)
B)(20.3, 24.5)
C)(19.8, 25.1)
D)(18.8, 26.3)

Accepted Answer

We'll use the formula:  
$$(\bar x - z_{\alpha/2} \frac{\sigma}{\sqrt{n}}, \bar x + z_{\alpha/2} \frac{\sigma}{\sqrt{n}})$$

Plugging in the given values, we get:

$$(22.4 - z_{0.01} \frac{3.8}{\sqrt{49}}, 22.4 + z_{0.01} \frac{3.8}{\sqrt{49}})$$

Then, we need to find $z_{0.01}$ using a standard normal table or calculator. We get $z_{0.01} = 2.33$.

Plugging in this value, we get:

$$(22.4 - 2.33 \frac{3.8}{\sqrt{49}}, 22.4 + 2.33 \frac{3.8}{\sqrt{49}})$$

Simplifying, we get:

$$(21.1, 23.7)$$

Therefore, the answer is A). We can use the formula:
Confidence Interval = sample mean ± (critical value) * (standard error)
where the critical value can be found using a t-table with degrees of freedom n-1 (since the population standard deviation is not given).
For a 98% confidence level, the alpha level is 0.02/2 = 0.01 (since it is a two-tailed test). Using a t-table with 48 degrees of freedom (49 - 1 = 48) and alpha = 0.01, we find the critical value to be 2.680.
The standard error can be found using the formula:
standard error = population standard deviation / sqrt(n)
standard error = 3.8 / sqrt(49) = 0.5429
Plugging in the values gives us:
Confidence Interval = 22.4 ± 2.680 * 0.5429
Confidence Interval = (21.1, 23.7)
Therefore, the best choice is A.

Question 4

In a random sample of 60 computers, the mean repair cost was $150. Assume the population standard deviation is $36. Construct a 90% confidence interval for the population mean.
A)($142, $158)
B)($138, $162)
C)($141, $159)
D)($537, $654)

Accepted Answer

The 90% confidence interval for the population mean is calculated using the formula: mean ± Z*(σ/√n), where Z is the Z-score corresponding to the confidence level (for 90%, Z ≈ 1.645), σ is the population standard deviation, and n is the sample size. Plugging in the given values: 150 ± 1.645*(36/√60), we get approximately ($142, $158).

Question 5

A random sample of 40 students has a mean annual earnings of $3120. Assume the population standard deviation is $677. Construct the confidence interval for the population mean, $\mu \text { if } \mathrm { c }$ = 0.95.
A)($2910, $3330)
B)($210, $110)
C)($4812, $5342)
D)($1987, $2346)

Accepted Answer

To construct a 95% confidence interval for the population mean with a sample mean of $3120, a population standard deviation of $677, and a sample size of 40, we use the formula for the confidence interval: $\bar{x} \pm z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the confidence level (for 95% confidence, $z = 1.96$), $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the values, we get the confidence interval as $$3120 \pm 1.96 \times \frac{677}{\sqrt{40}}$, which calculates to approximately $$2910, $3330$.

Question 6

A random sample of 40 students has a test score with $\bar { x }$ = 81.5. Assume the population standard deviation is 10.2. Construct the confidence interval for the population mean, $\mu \text { if } \mathrm { c }$ = 0.90.
A)(78.8, 84.2)
B)(51.8, 92.3)
C)(66.3, 89.1)
D)(71.8, 93.5)

Accepted Answer

To construct a 90% confidence interval for the population mean when the population standard deviation is known, we use the formula for the confidence interval: $\bar{x} \pm z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the desired confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size. For a 90% confidence level, the z-score is approximately 1.645. Plugging in the given values: $81.5 \pm 1.645 \frac{10.2}{\sqrt{40}}$, we calculate the margin of error and find the confidence interval, which matches choice A, (78.8, 84.2).

Question 7

A random sample of 150 students has a grade point average with a mean of 2.86. Assume the population standard deviation is 0.78. Construct the confidence interval for the population mean, $\mu ,$ if c = 0.98.
A)(2.71, 3.01)
B)(2.51, 3.53)
C)(2.43, 3.79)
D)(2.31, 3.88)

Accepted Answer

To construct a 98% confidence interval for the population mean, we use the formula for the confidence interval: $\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $Z$ is the Z-score corresponding to the confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size. Given $\bar{x} = 2.86$, $\sigma = 0.78$, and $n = 150$, and knowing that the Z-score for a 98% confidence level is approximately 2.33 (from Z-tables or standard normal distribution tables), we calculate the margin of error as $2.33 \times \frac{0.78}{\sqrt{150}}$, which gives approximately 0.15. Therefore, the confidence interval is $2.86 \pm 0.15$, which results in the interval (2.71, 3.01).

Question 8

Find the critical value $\mathrm { z } _ { \mathrm { C } }$ that corresponds to a 94% confidence level.
A) $\pm 1.88$
B) $\pm 1.645$
C) $\pm 1.96$
D) $\pm 2.33$

Accepted Answer

The critical value for a 94% confidence level is found using a Z-table or standard normal distribution table. For a 94% confidence level, the tails (both sides) share the remaining 6%, so each tail has 3%. Looking up the Z-value for 0.97 (0.5 + 0.94/2) in the Z-table gives a value close to $\pm 1.88$.

Question 9

The standard IQ test has a mean of 96 and a standard deviation of 14. We want to be 99% certain that we are within 4 IQ points of the true mean. Determine the required sample size.
A)82
B)10
C)178
D)1

Accepted Answer

The answer of The standard IQ test has a mean...

Question 10

In a recent study of 42 eighth graders, the mean number of hours per week that they watched television was 19.6. Assume the population standard deviation is 5.8 hours. Find the 98% confidence interval for the
Population mean.
A)(17.5, 21.7)
B)(14.1, 23.2)
C)(18.3, 20.9)
D)(19.1, 20.4)

Accepted Answer

We can use the formula:

CI = (x̄ - z*(σ/√n), x̄ + z*(σ/√n))

where:
x̄ = sample mean = 19.6
σ = population standard deviation = 5.8
n = sample size = 42
z = z-score for a 98% confidence level = 2.33

Plugging in the values:

CI = (19.6 - 2.33*(5.8/√42), 19.6 + 2.33*(5.8/√42))
CI = (17.5, 21.7)

Therefore, the 98% confidence interval for the population mean is (17.5, 21.7). Choice A is the correct answer. To find the confidence interval, we use the formula:

CI = x̄ ± z* (σ/√n)

where x̄ is the sample mean, z* is the z-value from the standard normal distribution table for the desired confidence level (98% in this case), σ is the population standard deviation, and n is the sample size.

Substituting the given values, we get:

CI = 19.6 ± 2.33 * (5.8/√42)

CI = (17.5, 21.7)

So the best choice is A, because it contains the correct interval based on the given information.

Question 11

In a recent study of 92 eighth graders, the mean number of hours per week that they watched television was
22.6. Assume the population standard deviation is 5.2 hours.
a)Find the 99% confidence interval of the mean.
b)If the standard deviation is doubled to 10.4, what will be the effect on the confidence interval?

Accepted Answer

The answer of In a recent study of 92 eighth...

Question 12

A random sample of 56 fluorescent light bulbs has a mean life of 645 hours. Assume the population standard deviation is 31 hours. Construct a 95% confidence interval for the population mean.
A)(636.9, 653.1)
B)(539.6, 551.2)
C)(112.0, 118.9)
D)(712.0, 768.0)

Accepted Answer

To construct a confidence interval for the population mean, we will use the formula:

CI = X̄ ± Z* (σ/√n)

Where:
X̄ = sample mean = 645
Z* = critical value for a 95% confidence interval = 1.96
σ = population standard deviation = 31
n = sample size = 56

Plugging in the values, we get:

CI = 645 ± 1.96 * (31 / √56)

Simplifying, we get:

CI = 645 ± 7.05

So the 95% confidence interval for the population mean is:

(645 - 7.05, 645 + 7.05) = (636.9, 653.1)

Therefore, the best choice is A) (636.9, 653.1). Since we are dealing with a sample mean and a known population standard deviation, we can use a z-distribution to construct our confidence interval. We use the formula:
CI = X̄ ± z*(σ/√n)
where X̄ is the sample mean, σ is the population standard deviation, n is the sample size, and z is the z-score corresponding to the desired confidence level (95% in this case). 
Plugging in the values, we get:
CI = 645 ± 1.96*(31/√56) 
CI = (636.9, 653.1)
Therefore, the best choice is A.

Question 13

Find the margin of error for the given values of c, $\sigma$, and n. $\mathrm { c } = 0.98 , \sigma = 0.78 , \mathrm { n } = 150$
A)0.15
B)0.08
C)0.11
D)0.12

Accepted Answer

The answer of Find the margin of error for the...

Question 14

In a random sample of 60 computers, the mean repair cost was $150. Assume the population standard
deviation is $36.
a)Construct the 99% confidence interval for the population mean repair cost.
b)If the level of confidence was lowered to 95%, what will be the effect on the confidence interval?

Accepted Answer

The answer of In a random sample of 60 computers,...

Question 15

In a sample of 10 randomly selected women, it was found that their mean height was 63.4 inches. From
previous studies, it is assumed that the standard deviation, $\sigma $ is 2.4 inches and that the population of height
measurements is normally distributed.
a)Construct the 99% confidence interval for the population mean height of women.
b)If the sample size was doubled to 20 women, what will be the effect on the confidence interval?

Accepted Answer

The answer of In a sample of 10 randomly selected...

Question 16

Find the margin of error for the given values of c, $\sigma$ and n. $\mathrm { c } = 0.95 , \sigma = 677 , \mathrm { n } = 40$
A)$210
B)$77
C)$2891
D)$7

Accepted Answer

The margin of error (E) for a given confidence level (c), standard deviation ($\sigma$), and sample size (n) can be calculated using the formula: $E = Z \times \frac{\sigma}{\sqrt{n}}$, where Z is the Z-score corresponding to the confidence level. For a 0.95 confidence level, the Z-score is approximately 1.96. Plugging in the given values: $E = 1.96 \times \frac{677}{\sqrt{40}} \approx 210$.

Question 17

The numbers of advertisements seen or heard in one week for 30 randomly selected people in the United States
are listed below. Construct a 95% confidence interval for the true mean number of advertisements. Assume
that $\sigma$ is 159.5. $$\begin{array} { l l l l l l l l l l } 
598 & 494 & 441 & 595 & 728 & 690 & 684 & 486 & 735 & 808 \\
481 & 298 & 135 & 846 & 764 & 317 & 649 & 732 & 582 & 677 \\
734 & 588 & 590 & 540 & 673 & 727 & 545 & 486 & 702 & 703
\end{array}$$

Accepted Answer

The answer of The numbers of advertisements seen or heard...

Question 18

A group of 40 bowlers showed that their average score was 192. Assume the population standard deviation is 8. Find the 95% confidence interval of the mean score of all bowlers.
A)(189.5, 194.5)
B)(186.5, 197.5)
C)(188.5, 195.6)
D)(187.3, 196.1)

Accepted Answer

We can use the formula for a confidence interval:
CI = X ± (Z α/2)(σ/√n)
where X is the sample mean (192), Z α/2 is the z-score associated with a 95% confidence level (1.96), σ is the population standard deviation (8), and n is the sample size (40).

Plugging in the values, we get:
CI = 192 ± (1.96)(8/√40)
CI = 192 ± 2.95

So the 95% confidence interval for the true mean score of all bowlers is:
CI = (192 - 2.95, 192 + 2.95)
CI = (189.05, 194.95)

Rounded to one decimal place, we get (189.5, 194.5). Therefore, the best choice is A. We can use the formula for confidence interval of the mean:

CI = x̄ ± z*(σ/√n)

where x̄ is the sample mean, σ is the population standard deviation, n is the sample size, and z depends on the level of confidence desired.

For a 95% confidence interval, we can look up the z-value in a standard normal distribution table, which gives us z = 1.96.

Plugging in the given values, we get:

CI = 192 ± 1.96*(8/√40)

Simplifying, we get:

CI = 192 ± 2.81

So the confidence interval is (189.19, 194.81). Rounding to the nearest tenth, we get (189.5, 194.5).

Therefore, the best choice is A.

Question 19

Find the margin of error for the given values of c, $\sigma _ { , }$ and n. $\mathrm { c } = 0.90 , \sigma = 11.5 , \mathrm { n } = 120$
A)1.73
B)0.16
C)1.05
D)0.94

Accepted Answer

The margin of error (E) for a given confidence level (c), standard deviation ($\sigma$), and sample size (n) can be calculated using the formula $E = Z \cdot \frac{\sigma}{\sqrt{n}}$, where Z is the Z-score corresponding to the confidence level. For a 90% confidence level, the Z-score is approximately 1.645. Plugging in the given values: $E = 1.645 \cdot \frac{11.5}{\sqrt{120}} \approx 1.73$.

Question 20

Determine the sampling error if the grade point averages for 10 randomly selected students from a class of 125 students has a mean of $\bar { x }$ = 2.8. Assume the grade point average of the 125 students has a mean of $\mu$ ? = 3.5.
A)0.7
B)3.15
C)-0.7
D)2.45

Accepted Answer

Sampling error is calculated as the difference between the sample mean and the population mean, which is $2.8 - 3.5 = -0.7$. However, the absolute value of this difference is taken to represent the magnitude of the error, making the correct answer $0.7$.

Quiz 6: Confidence Intervals

A group of 49 randomly selected students has a mean age of 22.4 years. Assume the population standard deviation is 3.8. Construct a 98% confidence interval for the population mean.

In a random sample of 60 computers, the mean repair cost was $150. Assume the population standard deviation is $36. Construct a 90% confidence interval for the population mean.

A random sample of 40 students has a mean annual earnings of $3120. Assume the population standard deviation is $677. Construct the confidence interval for the population mean, $\mu \text { if } \mathrm { c }$ = 0.95.

A random sample of 40 students has a test score with $\bar { x }$ = 81.5. Assume the population standard deviation is 10.2. Construct the confidence interval for the population mean, $\mu \text { if } \mathrm { c }$ = 0.90.

A random sample of 150 students has a grade point average with a mean of 2.86. Assume the population standard deviation is 0.78. Construct the confidence interval for the population mean, $\mu ,$ if c = 0.98.

Find the critical value $\mathrm { z } _ { \mathrm { C } }$ that corresponds to a 94% confidence level.

The standard IQ test has a mean of 96 and a standard deviation of 14. We want to be 99% certain that we are within 4 IQ points of the true mean. Determine the required sample size.

In a recent study of 42 eighth graders, the mean number of hours per week that they watched television was 19.6. Assume the population standard deviation is 5.8 hours. Find the 98% confidence interval for the Population mean.

A random sample of 56 fluorescent light bulbs has a mean life of 645 hours. Assume the population standard deviation is 31 hours. Construct a 95% confidence interval for the population mean.

Find the margin of error for the given values of c, $\sigma$ , and n. $\mathrm { c } = 0.98 , \sigma = 0.78 , \mathrm { n } = 150$

In a random sample of 60 computers, the mean repair cost was $150. Assume the population standard deviation is $36. a)Construct the 99% confidence interval for the population mean repair cost. b)If the level of confidence was lowered to 95%, what will be the effect on the confidence interval?

Find the margin of error for the given values of c, $\sigma$ and n. $\mathrm { c } = 0.95 , \sigma = 677 , \mathrm { n } = 40$

A group of 40 bowlers showed that their average score was 192. Assume the population standard deviation is 8. Find the 95% confidence interval of the mean score of all bowlers.

Find the margin of error for the given values of c, $\sigma _ { , }$ and n. $\mathrm { c } = 0.90 , \sigma = 11.5 , \mathrm { n } = 120$

Determine the sampling error if the grade point averages for 10 randomly selected students from a class of 125 students has a mean of $\bar { x }$ = 2.8. Assume the grade point average of the 125 students has a mean of $\mu$ ? = 3.5.

Quiz 6: Confidence Intervals

A group of 49 randomly selected students has a mean age of 22.4 years. Assume the population standard deviation is 3.8. Construct a 98% confidence interval for the population mean.

In a random sample of 60 computers, the mean repair cost was $150. Assume the population standard deviation is $36. Construct a 90% confidence interval for the population mean.

A random sample of 40 students has a mean annual earnings of $3120. Assume the population standard deviation is $677. Construct the confidence interval for the population mean, μ if c\mu \text { if } \mathrm { c }μ if c = 0.95.

A random sample of 40 students has a test score with xˉ\bar { x }xˉ = 81.5. Assume the population standard deviation is 10.2. Construct the confidence interval for the population mean, μ if c\mu \text { if } \mathrm { c }μ if c = 0.90.

A random sample of 150 students has a grade point average with a mean of 2.86. Assume the population standard deviation is 0.78. Construct the confidence interval for the population mean, μ,\mu ,μ, if c = 0.98.

Find the critical value zC\mathrm { z } _ { \mathrm { C } }zC​ that corresponds to a 94% confidence level.

The standard IQ test has a mean of 96 and a standard deviation of 14. We want to be 99% certain that we are within 4 IQ points of the true mean. Determine the required sample size.

In a recent study of 42 eighth graders, the mean number of hours per week that they watched television was 19.6. Assume the population standard deviation is 5.8 hours. Find the 98% confidence interval for the Population mean.

A random sample of 56 fluorescent light bulbs has a mean life of 645 hours. Assume the population standard deviation is 31 hours. Construct a 95% confidence interval for the population mean.

Find the margin of error for the given values of c, σ\sigmaσ , and n. c=0.98,σ=0.78,n=150\mathrm { c } = 0.98 , \sigma = 0.78 , \mathrm { n } = 150c=0.98,σ=0.78,n=150

In a random sample of 60 computers, the mean repair cost was $150. Assume the population standard deviation is $36. a)Construct the 99% confidence interval for the population mean repair cost. b)If the level of confidence was lowered to 95%, what will be the effect on the confidence interval?

Find the margin of error for the given values of c, σ\sigmaσ and n. c=0.95,σ=677,n=40\mathrm { c } = 0.95 , \sigma = 677 , \mathrm { n } = 40c=0.95,σ=677,n=40

A group of 40 bowlers showed that their average score was 192. Assume the population standard deviation is 8. Find the 95% confidence interval of the mean score of all bowlers.

Find the margin of error for the given values of c, σ,\sigma _ { , }σ,​ and n. c=0.90,σ=11.5,n=120\mathrm { c } = 0.90 , \sigma = 11.5 , \mathrm { n } = 120c=0.90,σ=11.5,n=120

Determine the sampling error if the grade point averages for 10 randomly selected students from a class of 125 students has a mean of xˉ\bar { x }xˉ = 2.8. Assume the grade point average of the 125 students has a mean of μ\muμ ? = 3.5.

A random sample of 40 students has a mean annual earnings of $3120. Assume the population standard deviation is $677. Construct the confidence interval for the population mean, $\mu \text { if } \mathrm { c }$ = 0.95.

A random sample of 40 students has a test score with $\bar { x }$ = 81.5. Assume the population standard deviation is 10.2. Construct the confidence interval for the population mean, $\mu \text { if } \mathrm { c }$ = 0.90.

A random sample of 150 students has a grade point average with a mean of 2.86. Assume the population standard deviation is 0.78. Construct the confidence interval for the population mean, $\mu ,$ if c = 0.98.

Find the critical value $\mathrm { z } _ { \mathrm { C } }$ that corresponds to a 94% confidence level.

Find the margin of error for the given values of c, $\sigma$ , and n. $\mathrm { c } = 0.98 , \sigma = 0.78 , \mathrm { n } = 150$

Find the margin of error for the given values of c, $\sigma$ and n. $\mathrm { c } = 0.95 , \sigma = 677 , \mathrm { n } = 40$

Find the margin of error for the given values of c, $\sigma _ { , }$ and n. $\mathrm { c } = 0.90 , \sigma = 11.5 , \mathrm { n } = 120$

Determine the sampling error if the grade point averages for 10 randomly selected students from a class of 125 students has a mean of $\bar { x }$ = 2.8. Assume the grade point average of the 125 students has a mean of $\mu$ ? = 3.5.