Annual starting salaries for college graduates with degrees in business administration aregenerally expected to be between $30,000 and $45,000. Assume that a 95% confidenceinterval estimate of the population mean annual starting salary is desired. What is the planning value for the population standard deviation? How large a sample should be taken ifthe desired margin of error isa. $500?b. $200?c. $100?d. Would you recommend trying to obtain the $100 margin of error? Explain
Contingency Table
A contingency table can be defined as the visual representation of the relationship between two or more categorical variables that can be evaluated and registered. It is a categorical version of the scatterplot, which is used to investigate the linear relationship between two variables. A contingency table is indeed a type of frequency distribution table that displays two variables at the same time.
Binomial Distribution
Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.
Annual starting salaries for college graduates with degrees in business administration are
generally expected to be between $30,000 and $45,000. Assume that a 95% confidence
interval estimate of the population mean annual starting salary is desired. What is the planning value for the population standard deviation? How large a sample should be taken if
the desired margin of error is
a. $500?
b. $200?
c. $100?
d. Would you recommend trying to obtain the $100 margin of error? Explain
Note:
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a.
The starting salaries are expected to be between $30,000, and $45,000.
Thus, the range of starting salaries is:
Range = $45,000 – $30,000 = $15,000.
Now, by the rule of thumb, an estimate of the standard deviation is one-fourth of the range.
Thus, the planning value of the population standard deviation is, σ = $3,750 [= $15,000 / 4].
The 100 (1 – α) % confidence interval for the population mean, μ, when the population standard deviation, σ is known, is the confidence interval assuming a normal distribution, as follows:
Here, n is the sample size, x̄ is the sample mean, and zα/2 is the critical value of the standard normal distribution, above which, 100 (α/2) % or α/2 proportion of the observations lie, and below which, 100 (1 – α + α/2) % = 100 (1 – α/2) % or (1 – α/2) proportion of the observations lie.
Denote X as the random variable of interest.
The desired confidence level is 95%. Thus,
For margin of error $500:
Since the sample size, n can only be an integer, the value of n would actually be the next higher integer, which is, 217.
Hence, the required sample size for $500 margin of error is 217.
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