Briefly explain the meaning of an estimator and an estimate
An estimator is a function or a rule that is used to estimate an unknown parameter or feature of a
population using the sample data. In other words, an estimator is a mathematical formula or
algorithm that is used to calculate the value of an unknown parameter based on a set of observed
data.
An estimate, on the other hand, is the result of applying an estimator to a particular set of data. It is
the numerical value that is obtained by using an estimator to predict or approximate an unknown
parameter or feature of the population. An estimate is a single value that represents the best guess
or approximation of the true value of the parameter based on the available data. It is important to
note that an estimate is not the same as the true value of the parameter, but it is the best
approximation of that value based on the information available.
Explain the meaning of a point estimate and an interval estimate
A point estimate is a single numerical value that is used to estimate an unknown parameter of a
population based on a sample data. The point estimate is usually obtained by applying a statistical
estimator to the sample data, and it represents the best guess or approximation of the true value of
the parameter. For example, if we want to estimate the mean income of a population based on a
sample of data, we can use the sample mean as a point estimate of the population mean.
An interval estimate, on the other hand, is a range of values that is used to estimate an unknown
parameter of a population based on a sample data. An interval estimate provides a range of possible
values for the parameter, along with a measure of confidence or probability that the true value of the
parameter lies within the range. For example, if we want to estimate the mean income of a
population with a certain level of confidence, we can use a confidence interval as an interval
estimate. A confidence interval provides a range of values that is likely to contain the true value of
the parameter, along with a measure of confidence that the true value is within that range.
In summary, a point estimate provides a single numerical value as an estimate of an unknown
parameter, while an interval estimate provides a range of values along with a measure of confidence
or probability that the true value of the parameter lies within that range.
What is the point estimator of the population mean, ? How would you calculate the margin of error
for an estimate of ?
The point estimator of the population mean, μ, is the sample mean, x̄. The sample mean is calculated
by summing up all the values in the sample and dividing by the sample size, n.
That is, x̄ = (x1 + x2 + ... + xn) / n, where xi is the i-th value in the sample and n is the sample size.
To calculate the margin of error for an estimate of μ, we need to use a confidence interval. The
margin of error is the amount by which the point estimate may differ from the true value of the
population mean, given a certain level of confidence.
The formula for calculating the margin of error is:
Margin of error = z* * (s / √n)
Where:



z* is the critical value from the standard normal distribution corresponding to the desired level of
confidence
s is the sample standard deviation
n is the sample size
For example, if we want to estimate the population mean with a 95% confidence level, the
corresponding z*-value is 1.96. If the sample size is 100 and the sample standard deviation is 10, the
margin of error would be:
Margin of error = 1.96 * (10 / √100) = 1.96 * 1 = 1.96
Therefore, the 95% confidence interval for the population mean would be x̄ ± 1.96, or [x̄ - 1.96, x̄ +
1.96].
Explain the various alternatives for decreasing the width of a confidence interval. Which is the best
alternative
There are several alternatives for decreasing the width of a confidence interval, including:
1. Increasing the sample size: A larger sample size can decrease the standard error and therefore
decrease the width of the confidence interval.
2. Decreasing the confidence level: A lower confidence level will result in a narrower interval, as it
requires less certainty that the true parameter lies within the interval.
3. Decreasing the variability of the data: A smaller standard deviation will result in a narrower interval.
4. Using a more precise estimator: Some estimators may provide more precise estimates of the
parameter, resulting in a narrower interval.
The best alternative for decreasing the width of a confidence interval depends on the specific
situation and the resources available. In general, increasing the sample size is often the most
effective way to decrease the width of a confidence interval, as it directly decreases the standard
error. However, increasing the sample size may not always be practical or feasible.
If increasing the sample size is not possible, decreasing the confidence level or reducing the
variability of the data can be effective alternatives. However, decreasing the confidence level should
only be done if a lower level of confidence is acceptable for the situation. Using a more precise
estimator can also be effective but may require additional data or resources.
It is important to note that decreasing the width of a confidence interval should not be the sole focus
of the analysis. The primary goal should be to obtain an accurate and reliable estimate of the
parameter, even if the resulting confidence interval is wider.
Briefly explain how the width of a confidence interval decreases with an increase in the sample size.
Give an example
The width of a confidence interval decreases with an increase in the sample size because a larger
sample size results in a smaller standard error of the estimate. The standard error measures the
variability of the sample mean and reflects the degree of uncertainty in the estimate of the
population parameter. As the sample size increases, the standard error decreases, which in turn
decreases the width of the confidence interval.
For example, suppose we want to estimate the mean weight of all apples produced by a particular
farm. We take a random sample of 25 apples and find that the sample mean weight is 150 grams,
with a standard deviation of 10 grams. If we construct a 95% confidence interval for the population
mean weight, we obtain:
Margin of error = 1.96 * (10 / √25) = 3.92 grams
95% confidence interval = 150 ± 3.92, or [146.08, 153.92]
Now suppose we increase the sample size to 100 apples and find that the sample mean weight is
152 grams, with the same standard deviation of 10 grams. If we construct a 95% confidence interval
for the population mean weight, we obtain:
Margin of error = 1.96 * (10 / √100) = 1.96 grams
95% confidence interval = 152 ± 1.96, or [150.04, 153.96]
As we can see, the width of the confidence interval is smaller with the larger sample size, indicating a
more precise estimate of the population mean.
Briefly explain how the width of a confidence interval decreases with a decrease in the confidence
level. Give an example
The width of a confidence interval decreases with a decrease in the confidence level because a lower
confidence level requires less certainty that the true population parameter lies within the interval. As
a result, the interval can be narrower and more precise.
For example, suppose we want to estimate the mean height of all adults in a certain city. We take a
random sample of 100 adults and find that the sample mean height is 170 cm, with a standard
deviation of 5 cm. If we construct a 95% confidence interval for the population mean height, we
obtain:
Margin of error = 1.96 * (5 / √100) = 0.98 cm
95% confidence interval = 170 ± 0.98, or [169.02, 170.98]
Now suppose we construct a 90% confidence interval instead of a 95% confidence interval. The
critical value from the standard normal distribution for a 90% confidence level is 1.645 instead of
1.96. If we construct a 90% confidence interval for the population mean height, we obtain:
Margin of error = 1.645 * (5 / √100) = 0.82 cm
90% confidence interval = 170 ± 0.82, or [169.18, 170.82]
As we can see, the width of the confidence interval is narrower with the lower confidence level,
indicating a more precise estimate of the population mean. However, it is important to note that
decreasing the confidence level may also increase the risk of making a Type I error, which is the error
of rejecting a true null hypothesis. Therefore, the choice of confidence level should be based on the
specific situation and the acceptable level of risk.
Briefly explain the difference between a confidence level and a confidence interval
A confidence level and a confidence interval are two related concepts used in statistical inference.
A confidence level is the probability that a confidence interval will contain the true population
parameter. It is expressed as a percentage or a decimal between 0 and 1. For example, a 95%
confidence level means that if we repeated the sampling process many times and constructed a 95%
confidence interval for each sample, we would expect 95% of those intervals to contain the true
population parameter.
A confidence interval, on the other hand, is a range of values that is likely to contain the true
population parameter with a certain level of confidence. It is calculated from a sample of data and is
used to estimate the unknown population parameter. For example, a 95% confidence interval means
that we are 95% confident that the true population parameter lies within the interval.
In summary, a confidence level is a probability that quantifies the level of confidence we have in the
method of constructing the confidence interval. A confidence interval is a range of values that
provides an estimate of the population parameter with a certain level of confidence.
What is the margin of error of estimate for when
is known? How is it calculated?
When the population standard deviation (σ) is known, the margin of error (ME) for the estimate of
the population mean (μ) can be calculated using the following formula:
ME = z* (σ / √n)
where z* is the critical value from the standard normal distribution corresponding to the desired level
of confidence, σ is the known population standard deviation, and n is the sample size.
For example, suppose we want to estimate the mean height of all students in a particular university,
and we know that the population standard deviation is 4 cm. We take a random sample of 100
students and find that the sample mean height is 170 cm. If we construct a 95% confidence interval
for the population mean height, the critical value from the standard normal distribution
corresponding to a 95% confidence level is 1.96. Using the formula above, we can calculate the
margin of error as:
ME = 1.96 * (4 / √100) = 0.784
Therefore, the 95% confidence interval for the population mean height is:
170 ± 0.784, or [169.216, 170.784]
This means that we are 95% confident that the true population mean height lies within the interval
[169.216, 170.784].
How will you interpret a 99% confidence interval for ? Explain.
A 99% confidence interval for the population mean (μ) is a range of values that provides an estimate
of the population mean with 99% confidence. This means that if we were to repeat the sampling
process many times and construct a 99% confidence interval for each sample, we would expect 99%
of those intervals to contain the true population mean.
For example, suppose we want to estimate the mean weight of all people in a certain city. We take a
random sample of 200 people and find that the sample mean weight is 75 kg, with a standard
deviation of 10 kg. If we construct a 99% confidence interval for the population mean weight, we
might obtain an interval of [72.5, 77.5]. This means that we are 99% confident that the true
population mean weight lies within this interval.
To interpret this interval, we can say that if we were to take many samples of 200 people from the
same city and construct a 99% confidence interval for each sample, we would expect 99% of those
intervals to contain the true population mean weight. In other words, we can be very confident that
the true population mean weight is somewhere between 72.5 kg and 77.5 kg.
It is important to note that the 99% confidence level implies a wider interval than a lower confidence
level, such as 95% or 90%. This means that we have a higher degree of certainty that the true
population mean lies within the interval, but the interval itself is wider and less precise. The choice of
confidence level depends on the specific situation and the acceptable level of risk.
Find z for each of the following confidence levels. a. 90% b. 95% c. 96% d. 97% e. 98% f. 99%
We can use the standard normal distribution (also known as the Z-distribution) to find the value of z
corresponding to a given confidence level. The values of z for some common confidence levels are:
a. 90% confidence level: z = 1.645 b. 95% confidence level: z = 1.96 c. 96% confidence level: z = 2.05
d. 97% confidence level: z = 2.17 e. 98% confidence level: z = 2.33 f. 99% confidence level: z = 2.58
These values represent the number of standard deviations from the mean for a normal distribution
that corresponds to the given confidence level. They are commonly used in constructing confidence
intervals and hypothesis testing in statistics.
For a data set obtained from a sample, n 20 and . It is known that
3.1. The population is normally distributed. a. What is the point estimate of ? b. Make a 99%
confidence interval for . c. What is the margin of error of estimate for part b?
a. The point estimate of the population mean (μ) is equal to the sample mean (x̄), which is given as
78.5.
b. To make a 99% confidence interval for the population mean, we can 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
critical value from the standard normal distribution corresponding to the desired level of confidence.
For a 99% confidence level, the critical value is 2.576.
Substituting the given values, we get:
CI = 78.5 ± 2.576*(3.1 / √20) = 78.5 ± 1.390
Therefore, the 99% confidence interval for the population mean is [77.11, 79.89].
c. The margin of error (ME) for the estimate of the population mean is given by:
ME = z* (σ / √n)
where z* is the critical value from the standard normal distribution corresponding to the desired level
of confidence, σ is the known population standard deviation, and n is the sample size. Substituting
the given values, we get:
ME = 2.576*(3.1 / √20) = 1.390
Therefore, the margin of error of estimate for the 99% confidence interval in part (b) is 1.390.
For a data set obtained from a sample, n 81 and . It is known that
4.8. a. What is the point estimate of ? b. Make a 95% confidence interval for . c. What is the margin of
error of estimate for part b?
a. The point estimate of the population mean (μ) is equal to the sample mean (x̄), which is given as
73.
b. To make a 95% confidence interval for the population mean, we can 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
critical value from the standard normal distribution corresponding to the desired level of confidence.
For a 95% confidence level, the critical value is 1.96.
Substituting the given values, we get:
CI = 73 ± 1.96*(4.8 / √81) = 73 ± 1.058
Therefore, the 95% confidence interval for the population mean is [71.94, 74.06].
c. The margin of error (ME) for the estimate of the population mean is given by:
ME = z* (σ / √n)
where z* is the critical value from the standard normal distribution corresponding to the desired level
of confidence, σ is the known population standard deviation, and n is the sample size. Substituting
the given values, we get:
ME = 1.96*(4.8 / √81) = 1.058
Therefore, the margin of error of estimate for the 95% confidence interval in part (b) is 1.058.
3 The standard deviation for a population is
15.3. A sample of 36 observations selected from this population gave a mean equal to 74.8. a. Make
a 90% confidence interval for . b. Construct a 95% confidence interval for . c. Determine a 99%
confidence interval for . d. Does the width of the confidence intervals constructed in parts a through
c increase as the confidence level increases? Explain your answer
a. To make a 90% confidence interval for the population mean, we can 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
critical value from the standard normal distribution corresponding to the desired level of confidence.
For a 90% confidence level, the critical value is 1.645.
Substituting the given values, we get:
CI = 74.8 ± 1.645*(15.3 / √36) = 74.8 ± 5.02
Therefore, the 90% confidence interval for the population mean is [69.78, 79.82].
b. To construct a 95% confidence interval, we can use the same formula, but with a different critical
value. For a 95% confidence level, the critical value is 1.96.
Substituting the given values, we get:
CI = 74.8 ± 1.96*(15.3 / √36) = 74.8 ± 5.82
Therefore, the 95% confidence interval for the population mean is [68.98, 80.62].
c. To determine a 99% confidence interval, we can again use the same formula, but with a different
critical value. For a 99% confidence level, the critical value is 2.576.
Substituting the given values, we get:
CI = 74.8 ± 2.576*(15.3 / √36) = 74.8 ± 7.22
Therefore, the 99% confidence interval for the population mean is [67.58, 82.02].
d. Yes, the width of the confidence intervals increases as the confidence level increases. This is
because the critical value z* is larger for higher confidence levels, and therefore the margin of error
(which is proportional to z* / √n) increases, leading to wider confidence intervals.
The standard deviation for a population is
14.8. A sample of 25 observations selected from this population gave a mean equal to 143.72. The
population is known to have a normal distribution. a. Make a 99% confidence interval for . b.
Construct a 95% confidence interval for . c. Determine a 90% confidence interval for . d. Does the
width of the confidence intervals constructed in parts a through c decrease as the confidence level
decreases? Explain your answer
a. To make a 99% confidence interval for the population mean, we can 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
critical value from the standard normal distribution corresponding to the desired level of confidence.
For a 99% confidence level, the critical value is 2.576.
Substituting the given values, we get:
CI = 143.72 ± 2.576*(14.8 / √25) = 143.72 ± 10.16
Therefore, the 99% confidence interval for the population mean is [133.56, 153.88].
b. To construct a 95% confidence interval, we can use the same formula, but with a different critical
value. For a 95% confidence level, the critical value is 1.96.
Substituting the given values, we get:
CI = 143.72 ± 1.96*(14.8 / √25) = 143.72 ± 8.16
Therefore, the 95% confidence interval for the population mean is [135.56, 151.88].
c. To determine a 90% confidence interval, we can again use the same formula, but with a different
critical value. For a 90% confidence level, the critical value is 1.645.
Substituting the given values, we get:
CI = 143.72 ± 1.645*(14.8 / √25) = 143.72 ± 7.19
Therefore, the 90% confidence interval for the population mean is [136.53, 150.91].
d. Yes, the width of the confidence intervals decreases as the confidence level decreases. This is
because the critical value z* is smaller for lower confidence levels, and therefore the margin of error
(which is proportional to z* / √n) decreases, leading to narrower confidence intervals.
The standard deviation for a population is
6.30. A random sample selected from this population gave a mean equal to 81.90. The population is
known to be normally distributed. a. Make a 99% confidence interval for assuming n 16. b. Construct
a 99% confidence interval for assuming n 20. c. Determine a 99% confidence interval for assuming n
25. d. Does the width of the confidence intervals constructed in parts a through c decrease as the
sample size increases? Explain
a. For a 99% confidence interval assuming n = 16, the critical value for the t-distribution with 15
degrees of freedom is 2.602. Therefore, the margin of error is:
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Margin of error = 2.602 * ( 6.30 / sqrt ( 16 )) = 3.239
The confidence interval is:
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81.90 ± 3.239 , or ( 78.661 , 85.139 )
b. For a 99% confidence interval assuming n = 20, the critical value for the t-distribution with 19
degrees of freedom is 2.539. Therefore, the margin of error is:
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Margin of error = 2.539 * ( 6.30 / sqrt ( 20 )) = 2.834
The confidence interval is:
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81.90 ± 2.834 , or ( 79.066 , 84.734 )
c. For a 99% confidence interval assuming n = 25, the critical value for the t-distribution with 24
degrees of freedom is 2.492. Therefore, the margin of error is:
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Margin of error = 2.492 * ( 6.30 / sqrt ( 25 )) = 3.146
The confidence interval is:
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81.90 ± 3.146 , or ( 78.754 , 85.046 )
d. As the sample size increases, the width of the confidence interval generally decreases, assuming
that the population standard deviation is known. This is because a larger sample size decreases the
standard error of the mean, which results in a narrower interval.
The standard deviation for a population is
7.14. A random sample selected from this population gave a mean equal to 48.52. a. Make a 95%
confidence interval for assuming n 196. b. Construct a 95% confidence interval for assuming n 100. c.
Determine a 95% confidence interval for assuming n 49. d. Does the width of the confidence intervals
constructed in parts a through c increase as the sample size decreases? Explain.
Yes, the width of the confidence intervals constructed in parts a through c increases as the sample
size decreases. This is because as the sample size decreases, the standard error of the mean
increases, and therefore the margin of error of the estimate also increases.
a. For n = 196, we have:
Standard error of the mean = 7.14 / sqrt(196) = 0.51
Margin of error = z* * (standard error of the mean) = 1.96 * 0.51 = 1.00
99% confidence interval = (48.52 - 1.00, 48.52 + 1.00) = (47.52, 49.52)
b. For n = 100, we have:
Standard error of the mean = 7.14 / sqrt(100) = 0.71
Margin of error = z* * (standard error of the mean) = 1.96 * 0.71 = 1.39
99% confidence interval = (48.52 - 1.39, 48.52 + 1.39) = (47.13, 49.91)
c. For n = 49, we have:
Standard error of the mean = 7.14 / sqrt(49) = 1.02
Margin of error = z* * (standard error of the mean) = 1.96 * 1.02 = 2.00
99% confidence interval = (48.52 - 2.00, 48.52 + 2.00) = (46.52, 50.52)
For a population, the value of the standard deviation is 2.65. A sample of 35 observations taken from
this population produced the following data. 42 51 42 31 28 36 49 29 46 37 32 27 33 41 47 41 28 46
34 39 48 26 35 37 38 46 48 39 29 31 44 41 37 38 46 a. What is the point estimate of ? b. Make a 98%
confidence interval for . c. What is the margin of error of estimate for part b?
a. The point estimate of the population mean can be calculated as the sample mean:
x̄ =
(42+51+42+31+28+36+49+29+46+37+32+27+33+41+47+41+28+46+34+39+48+26+35+37+38+
46+48+39+29+31+44+41+37+38+46)/35 = 38.77
Thus, the point estimate of the population mean is 38.77.
b. To calculate the 98% confidence interval for the population mean, we first need to determine the
critical value for a t-distribution with 34 degrees of freedom (since n = 35):
t = t(0.99, 34) ≈ 2.429
where t(0.99, 34) is the critical value of the t-distribution with 34 degrees of freedom for a 98%
confidence level.
The margin of error for the estimate can be calculated as:
E = t* * (s/√n)
where s is the sample standard deviation and n is the sample size. Here:
s = √[Σ(xi - x̄)²/(n-1)] = √[1220.5143/(35-1)] ≈ 3.47
Therefore, the 98% confidence interval is:
x̄ ± E = 38.77 ± 2.429*(3.47/√35) ≈ 38.77 ± 1.72
So, the 98% confidence interval is (37.05, 40.49).
c. The margin of error of estimate for part b is 1.72.
For a population, the value of the standard deviation is 4.96. A sample of 32 observations taken from
this population produced the following data. 74 85 72 73 86 81 77 60 83 78 79 88 76 73 84 78 81 72
82 81 79 83 88 86 78 83 87 82 80 84 76 74 a. What is the point estimate of ? b. Make a 99%
confidence interval for . c. What is the margin of error of estimate for part b?
a. The point estimate of the population mean is the sample mean:
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x̄ =
( 74 + 85 + 72 + 73 + 86 + 81 + 77 + 60 + 83 + 78 + 79 + 88 + 76 + 73 + 84 + 78 + 81 + 72 + 82 + 81 + 79 + 83 + 88 + 86 + 78
+ 83 + 87 + 82 + 80 + 84 + 76 + 74 )/ 32 = 79.25 Therefore, the point estimate of μ is 79.25 .
b. To make a 99% confidence interval for μ, we can use the formula:
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x̄ ± z(α/ 2 ) * σ/√n where x̄ is the sample mean, σ is the population standard deviation, n is the sample size,
z(α/ 2 ) is the z-score corresponding to the desired confidence level, and α is the significance level. For a 99 %
confidence level, α = 0.01 and z(α/2) = 2.576 ( from a standard normal distribution table). Substituting the given
values, we get : 79.25 ± 2.576 * 4.96 /√ 32 Simplifying the above expression, we get the 99 % confidence
interval for μ as : ( 74.47 , 84.03 ) Therefore, we are 99 % confident that the true population mean μ lies between
74.47 and 84.03 .
c. The margin of error of estimate is the distance between the point estimate and the upper or lower
bound of the confidence interval, divided by 2. In this case, the margin of error is:
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( 84.03 - 79.25 )/ 2 = 2.39 Therefore, the margin of error of estimate for the 99% confidence interval is 2.39 .
For a population data set, a. How large a sample should be selected so that the margin of error of
estimate for a 99% confidence interval for is 2.50? b. How large a sample should be selected so that
the margin of error of estimate for a 96% confidence interval for is 3.20?
To answer this question, we need to use the formula for the margin of error of estimate:
Margin of error = z *
(
/sqrt(n))
where z is the z-score corresponding to the desired confidence level,
is the standard deviation of the population, and n is the sample size.
a. For a 99% confidence interval and a margin of error of 2.50, we have:
2.50 = z *
(
/sqrt(n))
We need to find n. To do so, we need to know the value of z corresponding to a 99% confidence
level. From a standard normal distribution table, we find that z = 2.576.
Substituting these values, we get:
2.50 = 2.576 *
(
/sqrt(n))
Solving for n, we get:
n = ((2.576 *
) / (2.50)) ^ 2 n = 107.47
Therefore, we need to select a sample size of at least 108.
b. For a 96% confidence interval and a margin of error of 3.20, we have:
3.20 = z *
(
/sqrt(n))
Again, we need to find n. From a standard normal distribution table, we find that z = 1.751.
Substituting these values, we get:
3.20 = 1.751 *
(
/sqrt(n))
Solving for n, we get:
n = ((1.751 *
) / (3.20)) ^ 2 n = 37.88
Therefore, we need to select a sample size of at least 38.
For a population data set, a. What should the sample size be for a 98% confidence interval for to
have a margin of error of estimate equal to 5.50? b. What should the sample size be for a 95%
confidence interval for to have a margin of error of estimate equal to 4.25
To find the sample size for a given margin of error and confidence level, we can use the formula:
n = (z^2 * s^2) / E^2
where n is the sample size, z is the z-score associated with the desired confidence level, s is the
standard deviation of the population, and E is the desired margin of error.
a. For a 98% confidence interval and a margin of error of 5.50, we have:
z = 2.33 (from standard normal distribution table) s = ? (not given) E = 5.50
Since we do not have the standard deviation of the population, we cannot calculate the sample size.
b. For a 95% confidence interval and a margin of error of 4.25, we have:
z = 1.96 (from standard normal distribution table) s = ? (not given) E = 4.25
Again, since we do not have the standard deviation of the population, we cannot calculate the
sample size.