Question 1

As the sample size increases,the variability of the sampling distribution of the mean ______________.

Accepted Answer

As the sample size increases, the variability of the sampling distribution of the mean decreases. This is due to the central limit theorem, which states that as the sample size increases, the distribution of sample means approaches a normal distribution with a smaller standard deviation. Therefore, with a larger sample size, the sample mean is more likely to be closer to the true population mean, and there is less variability in the estimates of the sample mean.

Question 2

The sampling distribution of the sample mean is a normal distribution for ________ sample sizes,regardless of the shape of the corresponding population distribution.

Accepted Answer

The Central Limit Theorem states that the sampling distribution of the sample mean will be approximately normal if the sample size is large enough, regardless of the population's distribution shape.

Question 3

A random sample of size 36 is taken from a population with a mean of 50 and a standard deviation of 5.The sampling distribution of $\bar { X }$
____________________.

Accepted Answer

According to the Central Limit Theorem, the sampling distribution of sample means will be approximately normal with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size. In this case, the mean is 50 and the standard deviation is 5/sqrt(36) = 5/6, which is not too different from the standard deviation we would expect if the sample size was much larger. Therefore, we can reasonably assume that the sampling distribution of sample means is approximately normal.

Question 4

The population of all _________________ proportions is described by the sampling distribution of $\hat { p }$.

Accepted Answer

The sampling distribution of $\hat{p}$ describes the distribution of sample proportions, not population proportions.

Question 5

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.3 inches.What is the probability that the average length of a steel sheet from a sample of 9 units is more than 29.95 inches long?

Accepted Answer

The sample mean has a normal distribution with the same mean (30.05 inches) and a standard deviation of 0.3/sqrt(9) = 0.1 inches. The Z-score for 29.95 is (29.95 - 30.05) / 0.1 = -1. The probability of Z being greater than -1 is 0.8413.

Question 6

The notation for the standard deviation of the sampling distribution of the sample mean is __________.

Accepted Answer

The notation for the standard deviation of the sampling distribution of the sample mean is either $\sigma _ { \bar { X } }$ or $s _ { \bar { X } }$, depending on whether the population standard deviation or the sample standard deviation is used in the calculation. Choice B ($\sigma _ { x }$) is incorrect since that refers to the population standard deviation of the variable x, not the standard deviation of the sampling distribution of the sample mean. Choice C ($\sigma _ { \bar { X } / \mathbf { n } }$) is also incorrect since that notation represents the standard error of the mean, not the standard deviation of the sampling distribution of the sample mean. Choice D ($\mu$) is the symbol used for the population mean, not the standard deviation of the sampling distribution of the sample mean.

Question 7

Find $\mathrm { P } ( \bar { X } < 35 )$ If ? = 40,?_x = 16,n = 16.

Accepted Answer

The answer of Find \(\mathrm { P } ( \bar...

Question 8

A PGA (Professional Golf Association)tournament organizer is attempting to determine whether hole (pin)placement has a significant impact on the average number of strokes for the 13^th hole on a given golf course.Historically,the pin has been placed in the front right corner of the green,and the historical mean number of strokes for the hole has been 4.25,with a standard deviation of 1.6 strokes.On a particular day during the most recent golf tournament,the organizer placed the hole (pin)in the back left corner of the green.64 golfers played the hole with the new placement on that day.Determine the probability of the sample average number of strokes exceeding 4.75.

Accepted Answer

We are given:
Historical mean number of strokes = μ1 = 4.25
Standard deviation = σ = 1.6 
Sample size = n = 64
New mean number of strokes = μ2 = ?
We can use the central limit theorem to approximate the sampling distribution of the sample mean. Since the sample size is large enough, we can assume that the sampling distribution follows a normal distribution with mean μ2 and standard error (σ/√n).
To find the new mean, we can use the fact that the sample average is an unbiased estimator of the population mean. Therefore, we can assume that the sample average is equal to μ2, and we can use it to estimate the probability of the sample average exceeding 4.75 strokes.
z = (X̄ - μ) / (σ / √n)
z = (4.75 - 4.25) / (1.6 / √64)
z = 2.5
Using a standard normal distribution table, we can find the probability of z being greater than 2.5. The probability is:
P(z > 2.5) = 0.0062
Therefore, the answer is C) 0.0062.

Question 9

The ______________ of a sample statistic is the probability distribution of the population of all possible values of the sample statistic.

Accepted Answer

The sampling distribution is a probability distribution that represents the possible values of a sample statistic, such as the sample mean or sample proportion. It is important to understand the sampling distribution in order to make statistical inferences about the population based on a sample.

Question 10

As the sample size ___________,the standard deviation of the population of all sample proportions increases.

Accepted Answer

As the sample size increases, the standard deviation of the population of all sample proportions decreases due to the inverse relationship between sample size and standard error.

Question 11

A theorem that allows us to use the normal probability distribution to approximate the sampling distribution of sample means and sample proportions whenever the sample size is large is known as _______________.

Accepted Answer

The Central Limit Theorem states that the sampling distribution of sample means and sample proportions will be approximately normally distributed when the sample size is large enough. This allows us to use the normal probability distribution to make inferences about the population based on the sample data.

Question 12

Find $\mathrm { P } ( \bar { X } > 172 )$ If ? = 175 and $\sigma _ { \bar { X } } ^ { 2 } = 9$

Accepted Answer

The standard deviation of the sample mean is the square root of the variance, which is 3. The z-score is (172-175)/3 = -1. The probability of a z-score greater than -1 is 0.8413.

Question 13

Find $\mathrm { P } ( \bar { X } < 25 )$ If ? = 16 and $\sigma _ { \bar { X } } = 4$

Accepted Answer

To begin, we need to standardize the variable of interest, which is the sample mean. Since we know that the population mean is 16 and the standard deviation of the sample mean is 4, we can use the formula:

$$Z = \frac{\bar{X} - \mu}{\sigma_{\bar{X}}} = \frac{25-16}{4} = 2.25$$

where Z is the standard normal variable.

Next, we want to find the probability that the sample mean is less than 25, which is equivalent to finding the probability that Z is less than:

$$Z = \frac{25-16}{4} = 2.25$$

Using a standard normal table or calculator, we can find that the area to the left of 2.25 is approximately 0.9878.

Finally, we want the probability that the sample mean is greater than 25, which is the complement of the probability that the sample mean is less than or equal to 25. Therefore:

$$\mathrm{P}(\bar{X} < 25) = 0.9878$$

So the answer is C.

Question 14

An unbiased estimate of ?² is _____.

Accepted Answer

The unbiased estimate of the population variance ?² is the sample variance s².

Question 15

Find $\mathrm { P } ( \bar { X } > 2,510 )$
If ? = 2,500 and $\sigma _ { \bar { x } } = 7$

Accepted Answer

The probability $\mathrm { P } ( \bar { X } > 2,510 )$ can be found using the standard normal distribution. First, calculate the z-score: $ z = \frac{2,510 - 2,500}{7} = \frac{10}{7} \approx 1.43 $. Using the standard normal distribution table, the probability of $ Z > 1.43 $ is approximately 0.0764.

Question 16

A random sample of size 1,000 is taken from a population where p = .20.Describe the sampling distribution of $\hat { p }$.

Accepted Answer

According to the central limit theorem, when the sample size is large enough (n > 30) and the population is not extremely skewed, the sampling distribution of the sample proportion will be approximately normal. In this case, n = 1,000 which is sufficiently large and p = 0.20 is not extremely skewed. Therefore, the sampling distribution of the sample proportion will be approximately normal.

Question 17

The ___________ is a minimum-variance unbiased point estimate of the mean of a normally distributed population.

Accepted Answer

The sample mean is an unbiased estimator of the population mean, meaning that on average, it will give the correct estimate. It is also the minimum-variance estimator, meaning that there is no other unbiased estimator that has a lower variance than the sample mean.

Question 18

Packages of sugar bags for Sweeter Sugar Inc.have an average weight of 16 ounces and a standard deviation of 0.2 ounces.The weights of the sugar packages are normally distributed.What is the probability that 16 randomly selected packages will have a weight in excess of 16.075 ounces?

Accepted Answer

To solve this problem, we need to use the formula for the z-score:

z = (x - μ) / σ

where x is the value we are interested in (16.075 ounces), μ is the mean weight of the sugar packages (16 ounces), and σ is the standard deviation (0.2 ounces).

Using the formula, we get:

z = (16.075 - 16) / 0.2 = 0.375

This means that the weight of 16.075 ounces is 0.375 standard deviations above the mean.

We want to find the probability that 16 randomly selected packages will have a weight in excess of 16.075 ounces. This can be translated to finding the probability of getting a z-score greater than 0.375, which can be looked up in a standard normal distribution table or calculated using a calculator.

Using either method, we find that the probability is approximately 0.0668, which corresponds to answer choice D. Therefore, D is the best choice.

Question 19

Find ?_x,if ? = 400, $\mathrm { P } ( \bar { X } < 396 ) = .0228$ And n = 100.

Accepted Answer

We know that
μ = 400 ,
$\mathrm { P } ( \bar { X } < 396 ) = .0228$ ,
n = 100,
σ = ?
Now, we can use the formula for the standardized variable (z-score):
$$
z = \frac{\bar{X}-\mu}{\sigma/\sqrt{n}}
$$
We can rearrange this to solve for sigma:
$$
\sigma = \frac{(\bar{X}-\mu)}{z_{\alpha/2} /\sqrt{n}}
$$
where $z_{\alpha/2}$ is the z-score corresponding to the given confidence level and is calculated as:
$$
z_{\alpha/2} = -z_\alpha = -z_{0.0228/2} \approx -2.19
$$
Now, substituting the values:
$$
\sigma = \frac{(396-400)}{-2.19/\sqrt{100}}\approx 20
$$
Therefore, the answer is C) 20.

Question 20

Find $\mathrm { P } ( \bar { X } < 402 )$ If ? = 400,?_x = 200,and n = 100.

Accepted Answer

First we need to find the z-score for 402:

\begin{align*}
z &= \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}} \
&= \frac{402 - 400}{\frac{200}{\sqrt{100}}} \
&= \frac{2}{20} \
&= 0.1
\end{align*}

Then we need to find the area to the left of this z-score using a standard normal distribution table, or a calculator:

\begin{align*}
P(\bar{X} < 402) &= P(z < 0.1) \
&\approx 0.5398
\end{align*}

Therefore, the best choice is C, with a probability of 0.5398.

Quiz 7: Sampling and Sampling Distributions

As the sample size increases,the variability of the sampling distribution of the mean ______________.

The sampling distribution of the sample mean is a normal distribution for ________ sample sizes,regardless of the shape of the corresponding population distribution.

A random sample of size 36 is taken from a population with a mean of 50 and a standard deviation of 5.The sampling distribution of $\bar { X }$ ____________________.

The population of all _________________ proportions is described by the sampling distribution of $\hat { p }$ .

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.3 inches.What is the probability that the average length of a steel sheet from a sample of 9 units is more than 29.95 inches long?

The notation for the standard deviation of the sampling distribution of the sample mean is __________.

Find $\mathrm { P } ( \bar { X } < 35 )$ If ? = 40,?_x = 16,n = 16.

The ______________ of a sample statistic is the probability distribution of the population of all possible values of the sample statistic.

As the sample size ___________,the standard deviation of the population of all sample proportions increases.

A theorem that allows us to use the normal probability distribution to approximate the sampling distribution of sample means and sample proportions whenever the sample size is large is known as _______________.

Find $\mathrm { P } ( \bar { X } > 172 )$ If ? = 175 and $\sigma _ { \bar { X } } ^ { 2 } = 9$

Find $\mathrm { P } ( \bar { X } < 25 )$ If ? = 16 and $\sigma _ { \bar { X } } = 4$

An unbiased estimate of ?² is _____.

Find $\mathrm { P } ( \bar { X } > 2,510 )$ If ? = 2,500 and $\sigma _ { \bar { x } } = 7$

A random sample of size 1,000 is taken from a population where p = .20.Describe the sampling distribution of $\hat { p }$ .

The ___________ is a minimum-variance unbiased point estimate of the mean of a normally distributed population.

Packages of sugar bags for Sweeter Sugar Inc.have an average weight of 16 ounces and a standard deviation of 0.2 ounces.The weights of the sugar packages are normally distributed.What is the probability that 16 randomly selected packages will have a weight in excess of 16.075 ounces?

Find ?_x,if ? = 400, $\mathrm { P } ( \bar { X } < 396 ) = .0228$ And n = 100.

Find $\mathrm { P } ( \bar { X } < 402 )$ If ? = 400,?_x = 200,and n = 100.

Quiz 7: Sampling and Sampling Distributions

As the sample size increases,the variability of the sampling distribution of the mean ______________.

The sampling distribution of the sample mean is a normal distribution for ________ sample sizes,regardless of the shape of the corresponding population distribution.

A random sample of size 36 is taken from a population with a mean of 50 and a standard deviation of 5.The sampling distribution of Xˉ\bar { X }Xˉ ____________________.

The population of all _________________ proportions is described by the sampling distribution of p^\hat { p }p^​ .

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.3 inches.What is the probability that the average length of a steel sheet from a sample of 9 units is more than 29.95 inches long?

The notation for the standard deviation of the sampling distribution of the sample mean is __________.

Find P(Xˉ<35) \mathrm { P } ( \bar { X } < 35 ) P(Xˉ<35) If ? = 40,?x = 16,n = 16.

The ______________ of a sample statistic is the probability distribution of the population of all possible values of the sample statistic.

As the sample size ___________,the standard deviation of the population of all sample proportions increases.

A theorem that allows us to use the normal probability distribution to approximate the sampling distribution of sample means and sample proportions whenever the sample size is large is known as _______________.

Find P(Xˉ>172) \mathrm { P } ( \bar { X } > 172 ) P(Xˉ>172) If ? = 175 and σXˉ2=9\sigma _ { \bar { X } } ^ { 2 } = 9σXˉ2​=9

Find P(Xˉ<25) \mathrm { P } ( \bar { X } < 25 ) P(Xˉ<25) If ? = 16 and σXˉ=4\sigma _ { \bar { X } } = 4σXˉ​=4

An unbiased estimate of ?2 is _____.

Find P(Xˉ>2,510) \mathrm { P } ( \bar { X } > 2,510 ) P(Xˉ>2,510) If ? = 2,500 and σxˉ=7\sigma _ { \bar { x } } = 7σxˉ​=7

A random sample of size 1,000 is taken from a population where p = .20.Describe the sampling distribution of p^\hat { p }p^​ .

The ___________ is a minimum-variance unbiased point estimate of the mean of a normally distributed population.

Packages of sugar bags for Sweeter Sugar Inc.have an average weight of 16 ounces and a standard deviation of 0.2 ounces.The weights of the sugar packages are normally distributed.What is the probability that 16 randomly selected packages will have a weight in excess of 16.075 ounces?

Find ?x,if ? = 400, P(Xˉ<396) =.0228\mathrm { P } ( \bar { X } < 396 ) = .0228P(Xˉ<396) =.0228 And n = 100.

Find P(Xˉ<402) \mathrm { P } ( \bar { X } < 402 ) P(Xˉ<402) If ? = 400,?x = 200,and n = 100.

A random sample of size 36 is taken from a population with a mean of 50 and a standard deviation of 5.The sampling distribution of $\bar { X }$ ____________________.

The population of all _________________ proportions is described by the sampling distribution of $\hat { p }$ .

Find $\mathrm { P } ( \bar { X } < 35 )$ If ? = 40,?_x = 16,n = 16.

Find $\mathrm { P } ( \bar { X } > 172 )$ If ? = 175 and $\sigma _ { \bar { X } } ^ { 2 } = 9$

Find $\mathrm { P } ( \bar { X } < 25 )$ If ? = 16 and $\sigma _ { \bar { X } } = 4$

An unbiased estimate of ?² is _____.

Find $\mathrm { P } ( \bar { X } > 2,510 )$ If ? = 2,500 and $\sigma _ { \bar { x } } = 7$

A random sample of size 1,000 is taken from a population where p = .20.Describe the sampling distribution of $\hat { p }$ .

Find ?_x,if ? = 400, $\mathrm { P } ( \bar { X } < 396 ) = .0228$ And n = 100.

Find $\mathrm { P } ( \bar { X } < 402 )$ If ? = 400,?_x = 200,and n = 100.