The EAI data has information on the annual incomes of managers and whether they have attended the training program or not. This
data comprise all the 2500 managers that work for this organization. Using this information, address the following questions: Select a simple random sample of 150 managers and another of 250 managers and calculate the point estimates for the population mean, standard deviation, and proportion. How do the results you obtained for n = 150 and n = 250 compare to the population information? Provide a descriptive assessment of the incomes of those managers with training and those without training. Is your assessment in line with the population information? Referring to c), what do you need to do to be able to make a definitive conclusion of the relationship between training and the mean salary of the managers?
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Answer:
Answer for the question:
The EAI data has information on the annual incomes of managers and whether they have attended the training program or not. This data comprise all the 2500 managers that work for this organization. Using this information, address the following questions: Select a simple random sample of 150 managers and another of 250 managers and calculate the point estimates for the population mean, standard deviation, and proportion. How do the results you obtained for n = 150 and n = 250 compare to the population information? Provide a descriptive assessment of the incomes of those managers with training and those without training. Is your assessment in line with the population information? Referring to c), what do you need to do to be able to make a definitive conclusion of the relationship between training and the mean salary of the managers?
is given in the attachment.
Step-by-step explanation:
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Answer:
a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.
b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?
This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.
Due to the Central Limit Theorem, Z is:
X = 205
has a pvalue of 0.8413.
X = 195
has a pvalue of 0.1587.
0.8413 - 0.1587 = 0.6426
0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.
b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?
This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.
X = 210
has a pvalue of 0.9772.
X = 195
has a pvalue of 0.0228.
0.9772 - 0.0228 = 0.9544
0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.