A simple random sample of 40 items resulted in a sample mean of \(25 .\) The population standard deviation is \(\sigma=5\) a. What is the standard error of the mean, \(\sigma_{\bar{x}} ?\) b. At \(95 \%\) confidence, what is the margin of error?

Short Answer

Expert verified
The standard error of the mean is approximately \(0.79\), and the margin of error at 95% confidence is approximately \(1.55\).

Step by step solution

01

Calculate the Standard Error of the Mean

Using the formula for the standard error of the mean, we get the following: \[\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\] Plug in the given values, \(\sigma = 5\) and \(n = 40\): \[\sigma_{\bar{x}} = \frac{5}{\sqrt{40}}\] Now, calculate the value: \[\sigma_{\bar{x}} \approx 0.79\] The standard error of the mean is approximately 0.79.
02

Find the z-score for the 95% Confidence Interval

A 95% confidence interval corresponds to the area within 1.96 standard errors from the mean in a standard normal distribution. This is because 95% of the area under a standard normal curve lies between z-scores of -1.96 and +1.96. Therefore, the z-score associated with a 95% confidence interval is: \[z = 1.96\]
03

Calculate the Margin of Error

Now, let's use the margin of error formula: \[E = z * \sigma_{\bar{x}}\] Plug in the values we found earlier, \(z = 1.96\) and \(\sigma_{\bar{x}} \approx 0.79\): \[E = 1.96 * 0.79\] Now, calculate the value: \[E \approx 1.55\] The margin of error for the 95% confidence interval is approximately 1.55. In conclusion, the standard error of the mean is approximately 0.79, and the margin of error at 95% confidence is approximately 1.55.

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Key Concepts

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

Population Standard Deviation
Understanding the population standard deviation is crucial when dealing with statistical data. It is the measure that tells us how much the values in a population deviate from the mean, or average, of the population. The formula to calculate the standard deviation of a population, denoted as \(\sigma\), is given by taking the square root of the variance. The variance is the average of the squared differences between each data point and the mean of the population.

The population standard deviation is useful as it gives an overall picture of the spread of data points. In the context of our exercise, the population standard deviation is given as \(\sigma = 5\), which means that, on average, the data points deviate from the population mean by 5 units.
Sample Mean
The sample mean, denoted as \(\bar{x}\), is simply the average of all the data points in a sample. It is calculated by adding up all of the observed values and dividing by the number of observations. The sample mean is a very important statistic because it represents the center point of a data set.

In our exercise, the sample mean is reported as 25. This figure is critical when applying the central limit theorem, which tells us that the distribution of sample means will be normal or nearly normal if the sample size is large enough. The sample mean is often used along with the standard error of the mean to estimate the mean of the population from which the sample is drawn.
Confidence Interval
A confidence interval is a range of values that is used to estimate the true value of a population parameter. It's based on the sample data and gives an interval, which is expected, with a certain degree of confidence, to contain the actual unknown parameter of the population. The more common levels of confidence are 90%, 95%, and 99%.

In constructing a confidence interval, we need a central point (often the sample mean) and a margin of error. The actual interval is then the range from the central point minus the margin of error to the central point plus the margin of error. In our exercise, we use the sample mean and the margin of error (which we will discuss next) to construct a 95% confidence interval.
Margin of Error
The margin of error represents how much we expect our estimate to vary from the true population parameter. It quantifies the uncertainty in the estimate. The margin of error depends on the standard error of the mean and the level of confidence desired.

Using the z-score, which corresponds to the desired confidence level, we multiply it by the standard error of the mean to find the margin of error. For the exercise in question, a z-score of 1.96 was used for 95% confidence, and when multiplied by the standard error of 0.79, the margin of error is approximately 1.55. This tells us that the true population mean is likely within 1.55 units of our sample mean, 25, with 95% confidence.
Z-score
A z-score, or standard score, is a numerical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. If the z-score is 0, it is the mean value. A positive z-score indicates that the value is above the mean, and a negative z-score indicates it is below the mean.

In the context of confidence intervals, the z-score helps us determine how far we need to go from the mean to encompass the central 95% (for a 95% confidence interval) of the standard normal distribution. The z-score of 1.96, used in this exercise for a 95% confidence interval, means we include values within approximately 1.96 standard deviations from the sample mean in both directions.

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Most popular questions from this chapter

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 65 weekly reports showed a sample mean of 19.5 customer contacts per week. The sample standard deviation was \(5.2 .\) Provide \(90 \%\) and \(95 \%\) confidence intervals for the population mean number of weekly customer contacts for the sales personnel.

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In an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant, data were collected for a sample of 49 customers. Assume a population standard deviation of \$5. a. At \(95 \%\) confidence, what is the margin of error? b. If the sample mean is \(\$ 24.80\), what is the \(95 \%\) confidence interval for the population mean?

An online survey by ShareBuilder, a retirement plan provider, and Harris Interactive reported that \(60 \%\) of female business owners are not confident they are saving enough for retirement (SmallBiz, Winter 2006). Suppose we would like to do a follow-up study to determine how much female business owners are saving each year toward retirement and want to use \(\$ 100\) as the desired margin of error for an interval estimate of the population mean. Use \(\$ 1100\) as a planning value for the standard deviation and recommend a sample size for each of the following situations. a A \(90 \%\) confidence interval is desired for the mean amount saved. b. \(A 95 \%\) confidence interval is desired for the mean amount saved. c. \(\quad\) A \(99 \%\) confidence interval is desired for the mean amount saved. d. When the desired margin of error is set, what happens to the sample size as the confidence level is increased? Would you recommend using a \(99 \%\) confidence interval in this case? Discuss.

Disney's Hannah Montana: The Movie opened on Easter weekend in April 2009. Over the three-day weekend, the movie became the number-one box office attraction (The Wall Street Journal, April 13,2009 ). The ticket sales revenue in dollars for a sample of 25 theaters is as follows. $$\begin{array}{rrrrr} 20,200 & 10,150 & 13,000 & 11,320 & 9,700 \\ 8,350 & 7,300 & 14,000 & 9,940 & 11,200 \\ 10,750 & 6,240 & 12,700 & 7,430 & 13,500 \\ 13,900 & 4,200 & 6,750 & 6,700 & 9,330 \\ 13,185 & 9,200 & 21,400 & 11,380 & 10,800 \end{array}$$ a. What is the \(95 \%\) confidence interval estimate for the mean ticket sales revenue per theater? Interpret this result. b. Using the movie ticket price of \(\$ 7.16\) per ticket, what is the estimate of the mean number of customers per theater? c. The movie was shown in 3118 theaters. Estimate the total number of customers who saw Hannah Montana: The Movie and the total box office ticket sales for the threeday weekend.

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