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It is reported \(^{l}\) that the average or mean number of Facebook friends is \(155 .\) Suppose that when 50 randomly chosen Facebook users are polled regarding the number of their friends, the average number of their friends was reported to be \(\bar{x}=149\) with a standard deviation of \(s=29.7\). Use this information to answer the questions in Exercises \(13-15 .\) Calculate the value of the \(z\) -statistic based on the sample mean, \(\bar{x}\). Is this an unusual value of \(z\) ?

Short Answer

Expert verified
Answer: No, the sample mean of 149 is not unusual with respect to the population mean of 155, as the calculated z-statistic is approximately -1.43, which is less than the absolute value of 2.

Step by step solution

01

Identify the given values and formula

We are given the sample mean \(\bar{x} = 149\), population mean \(\mu = 155\), sample standard deviation \(s = 29.7\), and the sample size \(n = 50\). The z-statistic formula is given by: $$z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$$
02

Plug in the values into the z-statistic formula

Now, plug in the given values into the formula: $$z = \frac{149 - 155}{\frac{29.7}{\sqrt{50}}}$$
03

Calculate the z-statistic

Perform calculations: $$z = \frac{-6}{\frac{29.7}{\sqrt{50}}} = \frac{-6}{4.2} \approx -1.43$$ So, the z-statistic is approximately -1.43.
04

Determine if the z-statistic is unusual

Compare the absolute value of the z-statistic to the standard value of 2 to determine if it is unusual: $$|-1.43| < 2$$ Since the absolute value of the z-statistic is less than 2, we can conclude that this is not an unusual value of z.

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Key Concepts

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

Sample Mean
The sample mean, symbolized as \(\bar{x}\), is a critical value that represents the average of a subset of the population being studied. In our exercise, the sample mean was calculated to be 149 based on the responses of 50 randomly chosen Facebook users.

Understanding the sample mean is essential because it serves as an estimator of the population mean. To calculate it, you simply sum up all the values in the sample and then divide the sum by the number of observations in the sample. It should be noted that while the sample mean can give a good estimation of the population mean, it is not necessarily an exact match due to the variability inherent in sampling.

The accuracy of the sample mean as an estimator of the population mean increases with the sample size. Larger samples tend to give a mean value that is closer to the population mean, which illustrates the concept of the Law of Large Numbers. The reliability of conclusions drawn from the sample mean can be assessed through statistical tests such as the calculation of a z-statistic.
Population Mean
The population mean, represented by the Greek letter \(\mu\), refers to the average of all individuals or elements in the entire group we aim to analyze or understand. It is a parameter that signifies the central point of the distribution of the entire population.

The term might sound similar to the sample mean, but the key difference lies in the scope: the population mean considers every single member of the group without exception. In our specific exercise scenario, the reported population mean of Facebook friends is 155.

Obtaining the exact population mean can be challenging or even impractical when dealing with large populations, which is why researchers often rely on samples to estimate this value. It is the true average we are trying to estimate or get close to by collecting and analyzing sample data.
Sample Standard Deviation
The sample standard deviation (commonly denoted as \(s\)) is a measure of the amount of variability or spread in a sample data set. It quantifies how much the individual data points differ from the sample mean.

In our exercise, the sample standard deviation is 29.7, which suggests that on average, the number of friends that each individual in the sample has deviates from the sample mean (149) by about 29.7. Calculating the standard deviation involves determining the average squared deviation of each data point from the sample mean, taking the square root of that average.

A higher standard deviation indicates that the data points are more spread out from the mean, whereas a lower standard deviation implies that they are clustered closer to the mean. Importantly, while the standard deviation provides insights into the variability within a sample, it does not impact the sample mean itself.
Statistical Significance
The concept of statistical significance relates to the likelihood that the result obtained from a data sample can be generalized to the population at large. It indicates whether any observed differences or relationships in the data are meaningful or if they could simply be due to random chance.

In terms of the z-statistic, absolute values of z greater than 2 typically indicate statistical significance, suggesting that the sample mean is different from the population mean to a degree that is unlikely to occur by random chance. In the given exercise, the computed z-value of approximately -1.43 does not exceed the standard threshold of 2, indicating that the difference between our sample mean and population mean is not statistically significant.

This means that if there is a true difference in the average number of Facebook friends, the results from this specific sample do not sufficiently show it. It's important to consider that 'not significant' does not imply 'no difference'; rather, it points to insufficient evidence to conclude that a difference exists based on the sample data.

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Most popular questions from this chapter

In a comparison of the mean 1 -month weight losses for women aged \(20-30\) years, these sample data were obtained for each of two diets: $$ \begin{array}{lcc} \hline & \text { Diet I } & \text { Diet II } \\ \hline \text { Sample Size } n & 40 & 40 \\ \text { Sample Mean } \bar{x}(\mathrm{~kg}) & 4.5 & 3.6 \\ \text { Sample Variance } s^{2} & 0.89 & 1.18 \\ \hline \end{array} $$ Do the data provide sufficient evidence to indicate that diet I produces a greater mean weight loss than diet II? Use \(\alpha=.05 .\)

A company wants to implement a flextime schedule so that workers can schedule their own work hours, but it needs a minimum mean of 7 hours per day per assembly worker in order to operate effectively. A random sample of 80 workers was asked to submit a tentative flextime schedule. If the mean number of hours per day for Monday was 6.7 hours and the standard deviation was 2.7 hours, do the data provide sufficient evidence to indicate that the mean number of hours worked per day on Mondays, for all of the company's assemblers, will be less than 7 hours? Test using \(\alpha=.05 .\)

Independent random samples were selected from two binomial populations, with sample sizes and the number of successes given. State the null and alternative hypotheses to test for a difference in the two population proportions. Calculate the necessary test statistic, the p-value, and draw the appropriate conclusions with \(\alpha=.01 .\) $$ \begin{array}{lcc} \hline & {\text { Population }} \\ & 1 & 2 \\ \hline \text { Sample Size } & 500 & 500 \\ \text { Number of Successes } & 120 & 147 \\ \hline \end{array} $$

Independent random samples of 280 and 350 observations were selected from binomial populations 1 and 2, respectively. Sample 1 had 132 successes, and sample 2 had 178 successes. Do the data present sufficient evidence to indicate that the proportion of successes in population 1 is smaller than the proportion in population 2 ? Use one of the two methods of testing presented in this section, and explain your conclusions.

Does Mars, Inc. use the same proportion of red M\&M'S in its plain and peanut varieties? Random samples of plain and peanut M\&M'S provide the following sample data: $$ \begin{array}{lcc} \hline & \text { Plain } & \text { Peanut } \\ \hline \text { Sample Size } & 56 & 32 \\ \text { Number of Red M\&M'S } & 12 & 8 \end{array} $$ a. Use a test of hypothesis to determine whether there is a significant difference in the proportions of red candies for the two types of M\&M'S. Let \(\alpha=.05 .\) b. Calculate a \(95 \%\) confidence interval for the difference in the proportion of red candies for the two types of M\&M'S. Does this interval confirm your results in part a?

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