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A recent survey of 50 executives who were laid off from their previous position revealed it took a mean of 26 weeks for them to find another position. The standard deviation of the sample was 6.2 weeks. Construct a 95 percent confidence interval for the population mean. Is it reasonable that the population mean is 28 weeks? Justify your answer.

Short Answer

Expert verified
The population mean is likely not 28 weeks, as it falls outside the 95% confidence interval (24.28, 27.72).

Step by step solution

01

Determine the Confidence Level and Z-score

First, we need to determine the confidence level, which is given as 95%. Using a standard normal distribution table, the Z-score corresponding to a 95% confidence level (where the significance level \( \alpha = 0.05 \)) is approximately 1.96. This value will be used in our confidence interval calculation.
02

Calculate the Standard Error

The standard error (SE) of the mean is calculated using the formula:\[ SE = \frac{s}{\sqrt{n}} \]where \( s = 6.2 \) is the standard deviation and \( n = 50 \) is the sample size. Therefore:\[ SE = \frac{6.2}{\sqrt{50}} \approx 0.876 \]
03

Calculate the Margin of Error

The margin of error (ME) is calculated using the Z-score and the standard error:\[ ME = Z \times SE = 1.96 \times 0.876 \approx 1.71776 \]
04

Calculate the Confidence Interval

The confidence interval for the population mean is given by:\[ \text{Mean} \pm \text{Margin of Error} \]Substituting the given mean of 26 weeks and the calculated margin of error:\[ 26 \pm 1.71776 \]Thus, the confidence interval is approximately (24.28, 27.72) weeks.
05

Interpret the Confidence Interval

The calculated 95% confidence interval is (24.28, 27.72) weeks. This means we are 95% confident that the true population mean of the time taken to find a new job lies within this interval.
06

Evaluate the Reasonableness of the Population Mean Being 28 Weeks

Since 28 weeks is not within the calculated confidence interval of (24.28, 27.72), it is not reasonable to assume that the population mean is 28 weeks based on the given data and a 95% confidence level.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Population Mean
The population mean refers to the average of a set of values, which in this case is the average time it takes for all executives, not just those surveyed, to find a new position. It is a theoretical value and can be estimated using the sample mean if the sample is representative. In the context of our problem, we calculated the sample mean as 26 weeks based on a survey of executives. This sample mean serves as an estimate to infer the true population mean. When dealing with samples, it's important to remember that the sample mean is just one possible estimate of the population mean, and it can vary from one sample to another.
Standard Deviation
Standard deviation measures how spread out the values in a data set are around the mean. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation shows that the values are spread out over a wider range. In our exercise, the standard deviation of the sample is 6.2 weeks. This tells us that the time it took for the surveyed executives to find a new job varied by about 6.2 weeks on average from the sample mean of 26 weeks. Standard deviation is essential in confidence interval calculations as it is used to calculate the standard error, which in turn helps determine the margin of error.
Sample Size
Sample size is the number of observations in a sample used for statistical analysis. In the confidence interval example, the sample size is 50, meaning 50 executives were surveyed. A larger sample size generally leads to more reliable and accurate estimates of the population parameters. This is because a larger sample provides a better approximation of the population's characteristics. The sample size impacts the calculation of the standard error, which inversely affects the margin of error—this means a larger sample size typically results in a smaller margin of error, making the confidence interval more precise.
Margin of Error
The margin of error represents the range above and below the sample statistic in which we expect the true population parameter to lie. It is calculated as the product of the Z-score and the standard error. For instance, in our problem, the margin of error was found to be approximately 1.71776 weeks. When we add and subtract this margin of error from the sample mean of 26 weeks, we obtain the confidence interval (24.28, 27.72 weeks). This interval suggests that if we could sample the whole population of executives, the true mean time it takes to find a new job would fall within this range 95% of the time. The margin of error thus provides a buffer zone that accounts for the natural variability in the data and sample discrepancies.

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Most popular questions from this chapter

Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It found that the mean annual health insurance premium for a family with coverage through an employer was \(\$ 10,979 .\) The standard deviation of the sample was \(\$ 1,000 .\) a. Based on this sample information, develop a 90 percent confidence interval for the population mean yearly premium. b. How large a sample is needed to find the population mean within \(\$ 250\) at 99 percent confidence?

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