/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 If random samples of the given s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If random samples of the given size are drawn from a population with the given mean and standard deviation, find the standard error of the distribution of sample means. Samples of size 1000 from a population with mean 28 and standard deviation 5

Short Answer

Expert verified
The standard error of the mean (SEM) for the given population and sample size is approximately 0.158.

Step by step solution

01

Identify the population mean, population standard deviation, and sample size

From the exercise, the population mean (μ) is given as 28, the population standard deviation (σ) is given as 5, and the sample size (n) being drawn from the population is 1000.
02

Substitute the values into the standard error formula

Next, substitute these values into the formula for the standard error of the mean (σ / √n). Substitute 5 for σ (population standard deviation), and 1000 for n (sample size). This gives us: SEM = 5 / √1000
03

Perform the calculations

Next, perform the calculation. First, take the square root of 1000 to give √1000 = 31.62, correct to two decimal places. Then, divide the population standard deviation by this square root to give the standard error of the mean: SEM = 5 / 31.62 = 0.158, correct to three decimal places.

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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, also known by the symbol \( \mu \), is a measure that tells us the average value of a population's characteristic. It is vital to understand that the population mean includes every single individual in the group. For example, if we want to establish the average age of all the people in a specific town, every person's age needs to be summed up and then divided by the total number of individuals living there.

To give a concrete example from our exercise, the population mean given is 28. This number represents the central point of a set of data, from which we can measure variations or deviations.
Population Standard Deviation
Population standard deviation, denoted as \( \sigma \) is a statistical term that measures the amount of variety or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean of the set, while a high standard deviation indicates that the values are spread out over a wider range.

Coming back to our task, a standard deviation of 5 indicates a relatively moderate spread of values around the mean. It is critical when calculating the standard error because it reflects the extent to which individuals within a population vary from the average, or the mean.
Sample Size
The sample size, represented as \( n \), is simply the number of observations or replicates to include in a statistical sample. The greater the sample size, the more reliable the inferences and conclusions drawn from that sample are likely to be since it reduces the margin of error and the influence of outliers on your data.

In the context of our exercise, a sample size of 1000 is quite large, which allows for a more accurate estimation of the population mean. This large sample size would lead to a smaller standard error, confirming the notion that larger samples give us a clearer picture of the population's characteristics.
Standard Error Formula
The standard error (SE) is paramount in assessing the precision of an estimated population mean derived from a sample. The formula given for calculating the standard error is \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation and \( n \) is the sample size. In less technical terms, it indicates how far off we expect our sample mean to be from the actual population mean.

Our exercise showcases the practical application of this formula. We take the population standard deviation of 5 and divide by the square root of the sample size, which is 1000. This calculation provides us with a measure of how much variability we can expect by chance alone in our sample estimates. The smaller the standard error, the more confidence we can have in our sample mean as a representative of the population mean.

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Most popular questions from this chapter

Who Exercises More: Males or Females? The dataset StudentSurvey has information from males and females on the number of hours spent exercising in a typical week. Computer output of descriptive statistics for the number of hours spent exercising, broken down by gender, is given: \(\begin{array}{l}\text { Descriptive Statistics: Exercise } \\ \text { Variable } & \text { Gender } & \mathrm{N} & \text { Mean } & \text { StDev } \\\ \text { Exercise } & \mathrm{F} & 168 & 8.110 & 5.199 \\ & \mathrm{M} & 193 & 9.876 & 6.069\end{array}\) \(\begin{array}{rrrrr}\text { Minimum } & \text { Q1 } & \text { Median } & \text { Q3 } & \text { Maximum } \\ 0.000 & 4.000 & 7.000 & 12.000 & 27.000 \\\ 0.000 & 5.000 & 10.000 & 14.000 & 40.000\end{array}\) (a) How many females are in the dataset? How many males? (b) In the sample, which group exercises more, on average? By how much? (c) Use the summary statistics to compute a \(95 \%\) confidence interval for the difference in mean number of hours spent exercising. Be sure to define any parameters you are estimating. (d) Compare the answer from part (c) to the confidence interval given in the following computer output for the same data: Two-sample \(\mathrm{T}\) for Exercise Gender N Mean StDev SE Mean \(\begin{array}{lllll}\mathrm{F} & 168 & 8.11 & 5.20 & 0.40 \\ \mathrm{M} & 193 & 9.88 & 6.07 & 0.44\end{array}\) Difference \(=\operatorname{mu}(F)-\operatorname{mu}(M)\) Estimate for difference: -1.766 \(95 \%\) Cl for difference: (-2.932,-0.599)

Use a t-distribution and the given matched pair sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distribution of the differences is relatively normal. Assume that differences are computed using \(d=x_{1}-x_{2}\). Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}>\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lllllllllll} \hline \text { Situation } & 1 & 125 & 156 & 132 & 175 & 153 & 148 & 180 & 135 & 168 & 157 \\ \text { Situation } & 2 & 120 & 145 & 142 & 150 & 160 & 148 & 160 & 142 & 162 & 150 \\ \hline \end{array} $$

Restaurant Bill by Gender In the study described in Exercise 6.224 the diners were also chosen so that half the people at each table were female and half were male. Thus we can also test for a difference in mean meal cost between females \(\left(n_{f}=24, \bar{x}_{f}=44.46, s_{f}=15.48\right)\) and males \(\left(n_{m}=24, \bar{x}_{m}=43.75, s_{m}=14.81\right) .\) Show all details for doing this test.

Gender Bias In a study \(^{52}\) examining gender bias, a nationwide sample of 127 science professors evaluated the application materials of an undergraduate student who had ostensibly applied for a laboratory manager position. All participants received the same materials, which were randomly assigned either the name of a male \(\left(n_{m}=63\right)\) or the name of a female \(\left(n_{f}=64\right) .\) Participants believed that they were giving feedback to the applicant, including what salary could be expected. The average salary recommended for the male applicant was \(\$ 30,238\) with a standard deviation of \(\$ 5152\) while the average salary recommended for the (identical) female applicant was \(\$ 26,508\) with a standard deviation of \(\$ 7348\). Does this provide evidence of a gender bias, in which applicants with male names are given higher recommended salaries than applicants with female names? Show all details of the test.

In Exercise \(6.107,\) we see that plastic microparticles are contaminating the world's shorelines and that much of the pollution appears to come from fibers from washing polyester clothes. The same study referenced in Exercise 6.107 also took samples from ocean beaches. Five samples were taken from each of 18 different shorelines worldwide, for a total of 90 samples of size \(250 \mathrm{~mL}\). The mean number of plastic microparticles found per \(250 \mathrm{~mL}\) of sediment was 18.3 with a standard deviation of 8.2 . (a) Find and interpret a \(99 \%\) confidence interval for the mean number of polyester microfibers per \(250 \mathrm{~mL}\) of beach sediment. (b) What is the margin of error? (c) If we want a margin of error of only ±1 with \(99 \%\) confidence, what sample size is needed?

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