/*! 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 27 Wireless routers. John is shoppi... [FREE SOLUTION] | 91Ó°ÊÓ

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Wireless routers. John is shopping for wireless routers and is overwhelmed by the number of available pptions. In order to get a feel for the average price, he takes a random sample of 75 routers and finds that the average price for this sample is \(\$ 75\) and the standard deviation is \(\$ 25\). (a) Based on this information, how much variability should he expect to see in the mean prices of repeated samples, each containing 75 randomly selected wireless routers? b) A consumer website claims that the average price of routers is \(\$ 80 .\) Is a true average of \(\$ 80\) consistent with John's sample?

Short Answer

Expert verified
The variability expected is approximately $2.89. A true average of $80 is consistent with John's sample.

Step by step solution

01

Understand the Concepts

We are asked to find the variability expected in the mean prices of 75 routers in repeated samples and assess if a claimed mean price of $80 is consistent with the sample data provided. We will use standard deviation of the sample mean (also called the standard error) to quantify this variability.
02

Compute the Standard Error

The standard error (SE) of the mean is given by the formula \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the standard deviation of the sample, and \( n \) is the sample size. Here, \( \sigma = 25 \) and \( n = 75 \). So, \( SE = \frac{25}{\sqrt{75}} \approx 2.8867 \). This value represents the expected variability in the mean prices of repeated samples of 75 routers.
03

Formulate the Hypothesis for Consistency Check

We want to test if a true average price of \$80 is consistent with John's sample. We do this using hypothesis testing, framing our null hypothesis as \( H_0: \mu = 80 \) and the alternative hypothesis as \( H_1: \mu eq 80 \), where \( \mu \) represents the true mean price.
04

Calculate the Z-score

The Z-score helps measure how far the sample mean is from the claimed population mean in units of the standard error. The formula for the Z-score is \( Z = \frac{\bar{x} - \mu}{SE} \), where \( \bar{x} = 75 \) and \( \mu = 80 \). Substituting the values, \( Z = \frac{75 - 80}{2.8867} \approx -1.73 \).
05

Determine Statistical Significance

We compare the computed Z-score to a standard Z-table to determine if the result is statistically significant. Typically, we use a significance level of 0.05. A Z-score of \(-1.73\) does not exceed the critical Z-value of approximately ±1.96 for a two-tailed test at this significance level, suggesting that the result is not significant enough to reject the null hypothesis.

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Key Concepts

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

Sampling Variability
When we talk about sampling variability, we are referring to the natural variation that occurs in statistics calculated from different random samples. In John's case, because he selected 75 routers, each sample he takes could yield a different average price. This is because each sample will likely include a slightly different mix of individual router prices.
Sampling variability is an important concept in statistical inference because it helps us understand why variations occur in data. It also affects how confident we can be in our estimates. Even though John finds an average of $75 for his sample, another sample might have a slightly higher or lower average.
  • Sampling variability results from innate differences in individual data points.
  • Larger samples tend to have less variability, making estimates more reliable.
  • Understanding this concept helps in assessing the reliability of statistical results.
Standard Error
The standard error (SE) of a sample mean is the standard deviation of the sample means. It's an estimation of the divergence of sample means from the true population mean.
In our exercise, the standard error can be calculated using the formula:\[ SE = \frac{\sigma}{\sqrt{n}} \]where \( \sigma = 25 \) (the sample standard deviation) and \( n = 75 \) (the sample size). By calculating the SE to be approximately 2.8867, we can approximate how much sample means will differ from the true population mean.
  • Smaller standard error implies the sample mean is closer to the population mean.
  • It's crucial for understanding how variable our sample mean is likely to be.
This means that any given sample mean is expected to fall within about 2.8867 units of the true mean due to random sampling errors alone.
Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about population parameters based on sample data. In the exercise, we want to determine if the sample mean of \\(75 provides enough evidence to dispute the consumer website's claim that the average price is \\)80.
We start with forming two hypotheses:
  • Null hypothesis \( (H_0) \): The true mean price is \\(80.
  • Alternative hypothesis \( (H_1) \): The true mean price is not \\)80.
Hypothesis testing involves comparing our sample data against the null hypothesis. If the evidence is strong enough, we may reject the null hypothesis; if not, we fail to reject it. This process helps determine if our sample results are typical of the population or if they're so unusual, it casts doubt on the initial claim.
Z-score
In statistics, the Z-score is a measure that describes how many standard deviations a data point is from the mean. In this exercise, it helps us determine how far John's sample mean of \\(75 is from the hypothesized population mean of \\)80.
The Z-score formula used here is:\[ Z = \frac{\bar{x} - \mu}{SE} \]where \( \bar{x} = 75 \) is the sample mean, \( \mu = 80 \) is the population mean, and the SE is the standard error previously calculated as 2.8867. John's Z-score of approximately -1.73 suggests that the sample mean is 1.73 standard errors below the claimed population mean.
  • Z-scores help us understand the extremeness of our sample mean.
  • It allows us to compare the mean using standard units.
In this context, the Z-score indicates that John's sample mean is not significantly different enough from the website's claim to refute it. This is because the Z-score does not exceed our critical value for significance in a two-tailed test.

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Most popular questions from this chapter

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