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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 \(13-15 .\) What is the probability of observing a value of \(z\) greater than 1.43 or less than -1.43 ? What might you conclude about the average number of Facebook friends?

Short Answer

Expert verified
Based on the given data and calculated probability (15.28%), there seems to be no significant difference between the reported average number of Facebook friends (155) and the observed average number of Facebook friends (149) for the sample of 50 Facebook users.

Step by step solution

01

Calculate the standard error

The standard error is calculated using the formula: $$ SE = \frac{s}{\sqrt{n}} $$ Where \(s\) is the standard deviation and \(n\) is the sample size. In this case, \(s = 29.7\) and \(n = 50\). So the standard error is: $$ SE = \frac{29.7}{\sqrt{50}} \approx 4.204 $$
02

Calculate the z-score

The z-score is calculated using the formula: $$ z = \frac{\bar{x} - \mu}{SE} $$ Where \(z\) is the z-score, \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, and \(SE\) is the standard error. In this case, \(\bar{x} = 149\), \(\mu = 155\), and \(SE \approx 4.204\). Plugging these values into the formula, we get: $$ z = \frac{149 - 155}{4.204} \approx -1.43 $$
03

Refer to the z-table

Now we want to find the probability of observing a value of \(z\) greater than 1.43 or less than -1.43. To do this, we can refer to the z-table. For \(z = 1.43\), we find that the probability is 0.9236; However, the z-table gives us the probability of observing a value of \(z\) less than 1.43, to find the probability of observing a value greater than 1.43, we need to subtract this value from 1: $$ P(z > 1.43) = 1 - P(z < 1.43) = 1 - 0.9236 = 0.0764 $$ Since the z-score is symmetric around 0, the probability of observing a value of \(z\) less than -1.43 is equal to the probability of observing a value greater than 1.13, which is 0.0764.
04

Calculate the probability and draw a conclusion

To find the probability of observing a value of \(z\) greater than 1.43 or less than -1.43, we should sum up the probabilities found in step 3: $$ P(z > 1.43 \text{ or } z < -1.43) = P(z > 1.43) + P(z < -1.43) = 0.0764 + 0.0764 = 0.1528 $$ The probability of observing a value of \(z\) greater than 1.43 or less than -1.43 is 0.1528 or 15.28%. This means there is a 15.28% chance of observing a sample mean that is more extreme than the one we have if the true population mean is indeed 155. This probability is not very low, so we might conclude that there is no significant difference between the reported average number of Facebook friends (155) and the observed average number of Facebook friends (149) for the sample of 50 Facebook users.

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Key Concepts

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

Standard Error
Understanding the standard error (SE) is crucial for interpreting the variability within an experiment or study. The standard error is a statistical term that measures the amount of variation or dispersion of a set of sample means relative to the true population mean.

When calculating the SE, as seen in our exercise example, you use the formula: \[\begin{equation} SE = \frac{s}{\sqrt{n}} \end{equation}\]where \(s\) is the sample standard deviation and \(n\) is the number of observations in the sample. The SE gives us an estimation of how much the computed sample mean (\(\overline{x}\)) deviates from the actual population mean (\(\mu\)). A smaller SE indicates that the sample mean is likely to be close to the population mean, thus providing a more precise estimate of the population mean.

When assessing results, such as in hypothesis testing, the standard error serves as a fundamental component because it allows us to determine how much credibility to give to the sample mean when making inferences about the true population mean.
Sample Mean
The sample mean (\(\overline{x}\)) is simply the average of all measurements in a sample. It's a central value used to summarize a dataset and a fundamental aspect of descriptive statistics.

In the context of our Facebook friends example, the sample mean refers to the average number of friends from the chosen 50 Facebook users. It was calculated to be \(\overline{x}=149\). This number is an estimate of the population mean, which can vary from sample to sample. The accuracy of this estimate will depend on the sample size and the variability of the data within the population, which are reflected in the standard error. The concept of simple mean is essential as it forms the basis for comparison when performing hypothesis testing or calculating z-scores.
Population Mean
The population mean (\(\mu\)), sometimes referred to as the expected value, is the average of all measurements across the entire population. In theory, it represents the true average, but in most practical situations, it's often impossible to calculate because we can't measure every single individual in a population.

In our exercise, the reported population mean is 155 friends. This is the value we are testing our sample against to see if there's a significant difference or if it's reasonable to consider that our sample mean could come from a population with this mean. The population mean is a fixed value, unlike the sample mean, which might change from one sample to another due to random variation.
Normal Distribution
The concept of normal distribution, or the bell curve, is a foundational pillar in statistics. It's a continuous probability distribution that is symmetrical and the mean, median, and mode of the distribution are identical.

A key feature of the normal distribution is that it allows statisticians to make inferences about population parameters using sample data. Most z-score calculations assume that the distribution of sample means is normally distributed when the sample size is sufficiently large, due to the Central Limit Theorem. In our exercise involving the average number of Facebook friends, the fact that our calculations are based on the z-score implies an underlying assumption of normal distribution for the sampling distribution of the mean.
Hypothesis Testing
Hypothesis testing is a method used in statistics to test a hypothesis about a parameter in a population using sample data. It is a systematic procedure that uses evidence from a sample to draw conclusions about the population.

In the final steps of our example, after calculating the standard error and z-score, we moved on to the hypothesis testing phase. We have a null hypothesis (H0) that assumes there is no difference between the population mean and the sample mean. The alternative hypothesis (H1) would argue that a significant difference does exist. By calculating the probability of observing a z-score as extreme or more extreme than our sample's z-score, we determine whether we can reject the null hypothesis.

In this case, with a probability of 15.28% being a fairly large value, we do not have strong enough evidence to reject the null hypothesis. Hence, we might conclude that the observed difference in the mean number of Facebook friends (155 in the population versus 149 in the sample) is not statistically significant, and could be due to chance variation within normal sampling error.

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Most popular questions from this chapter

A random sample of \(n=35\) observations from a quantitative population produced a mean \(\bar{x}=2.4\) and a standard deviation of \(s=.29 .\) Your research objective is to show that the population mean \(\mu\) exceeds 2.3. Use this information to answer the questions. Calculate \(\beta=P\left(\right.\) accept \(H_{0}\) when \(\left.\mu=2.4\right)\)

In a Pew Research report concerning the rise of automation in the United States, \(56 \%\) of the participants indicated that they would not ride in a driverless car, and \(87 \%\) favor a requirement of having a human in the driver's seat in case of an emergency. \({ }^{14}\) Suppose that the number of participants was \(n=500\). Is there sufficient evidence to conclude that more than a simple majority of Americans would not ride in a driverless car? a. Use a formal test of hypothesis with \(\alpha=.05\) to determine whether more than \(50 \%\) of Americans would not ride in a driverless car. b. Use the \(p\) -value approach. Do the two approaches lead to the same conclusion?

The braking ability was compared for two 2018 automobile models. Random samples of 64 automobiles were tested for each type. The recorded measurement was the distance (in meters) required to stop when the brakes were applied at 80 kilometers per hour. These are the computed sample means and variances: \begin{tabular}{ll} \hline Model I & Model II \\ \hline \(\bar{x}_{1}=36.0\) & \(\bar{x}_{2}=33.2\) \\ \(s_{1}^{2}=9.48\) & \(s_{2}^{2}=8.09\) \\ \hline \end{tabular} Do the data provide sufficient evidence to indicate a difference between the mean stopping distances for the two models?

What is the power of a test and how is it related to \(\beta ?\)

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 \(\alpha=.01 .\) $$ \begin{array}{lcc} \hline & {\text { Population }} \\ & 1 & 2 \\ \hline \text { Sample Size } & 800 & 640 \\ \text { Number of Successes } & 337 & 374 \\ \hline \end{array} $$

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