The distribution of weekly salaries at a large company is reverse J-shaped with a mean of $1000 and a standard deviation of $370. What is the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 80 employees will be at most $75?

Problem 5

(a) A researcher obtained estimates for the mean age, μ, of all U.S. millionaires. He randomly selected 36 U.S. millionaires and found that the mean age is 58.53 years. Assume that the standard deviation of ages of all U.S. millionaires is 13.0 years.
(i) Determine the margin of error, E, for a 98% confidence interval. (2%)
(ii) Determine the sample size required to have a margin of error of 4.5 years with a 95.44% confidence level. (2%)

(b) A U.S. researcher compared the rental rates of commercial real estates in Asia and Europe in 2018. Annual lease rates for a sample of 30 commercial properties in Hong Kong showed a mean of $1458 per square meter with a standard deviation of $269. Annual lease rates for a sample of 40 commercial properties in Paris showed a mean lease rate of $1279 per square meter with a standard deviation of $215. Obtain a 98% confidence interval for the difference between the mean annual lease rates in Hong Kong and Paris. Preliminary data analyses indicate that you can reasonably use the required procedure.

To find the probability that the sampling error will be at most $75, we need to use the Central Limit Theorem.

According to the Central Limit Theorem, the distribution of sample means will be approximately normal if the sample size is large enough (n > 30).

Given that the sample size is 80, we can use the formula for the standard deviation of the sampling distribution of sample means:

Standard Deviation (σx̄) = Standard Deviation (σ) / √n

= 370 / √80

= 370 / 8.94

= 41.38

Now, we need to find the z-score for a sampling error of $75:

Z = (x - μ) / σx̄

= 75 / 41.38

= 1.81

Using a z-table, we can find the probability associated with a z-score of 1.81. However, since the distribution is reversed J-shaped, we need to calculate the probability from both tails of the distribution.

P(Z ≤ -1.81 or Z ≥ 1.81) = P(Z ≤ -1.81) + P(Z ≥ 1.81)

Looking up these values in the z-table, we find that:

P(Z ≤ -1.81) = 0.0359
P(Z ≥ 1.81) = 0.0359

Adding these probabilities together:

0.0359 + 0.0359 = 0.0718

The probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 80 employees will be at most $75 is approximately 0.0718, or 7.18%.

To solve this problem, we need to use the concept of sampling distribution. The sampling distribution of the mean is the distribution of all possible sample means that could be drawn from a population.

In this case, we are given that the distribution of weekly salaries is reverse J-shaped, with a mean of $1000 and a standard deviation of $370. We want to find the probability that the sampling error made in estimating the mean weekly salary for all employees by the mean of a random sample of 80 employees will be at most $75.

Sampling error refers to the difference between the sample mean and the population mean. The formula for the standard deviation of the sampling distribution (also known as the standard error) is given by:

Standard error = population standard deviation / square root of sample size

Let's calculate the standard error:

Standard error = $370 / sqrt(80)
Standard error ≈ $41.38

Now, we need to find the probability that the sampling error is at most $75. In other words, we want to find the area under the sampling distribution curve where the sampling error is less than or equal to $75.

To find this probability, we can use a normal distribution table or a statistical software. We'll use a normal distribution table for this explanation.

We need to convert the sampling error of $75 into a standardized z-score. The z-score formula is:

z = (x - μ) / σ

Where:
x is the value we want to standardize (in this case, $75),
μ is the mean of the sampling distribution (which is equal to the population mean),
and σ is the standard error.

Let's calculate the z-score:

z = ($75 - $1000) / $41.38
z ≈ -22.56

Now, let's find the probability corresponding to this z-score from the standard normal distribution table. The standard normal distribution table gives us the area under the curve to the left of a given z-score.

In this case, we want to find the probability to the left of z = -22.56. However, the table only provides values for positive z-scores. To find the probability for negative z-scores, we can use the symmetry property of the standard normal distribution.

The symmetry property states that the area to the left of z is equal to 1 minus the area to the right of z.

Therefore, the probability to the left of z = -22.56 is equal to 1 minus the probability to the right of z = 22.56.

Using the standard normal distribution table, the probability to the right of z = 22.56 is practically 0 (since z-scores at the tails are very rare).

Hence, the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of 80 employees will be at most $75 is approximately 1.

The distribution of weekly salaries at a large company is reverse J-shaped with a mean of $1000 and a standard deviation of $370. What is the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 80 employees will be at most $75?

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n = [z*s/E]^2
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80 = [z*370/75]^2
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sqrt(80) = 4.93z
z = 1.81
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P(z > 1.81) = 0.035
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Ans:: P(condition described) = 2*0.035) = 7%
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