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Close Confidants and Social Networking Sites Exercise 6.93 introduces a study \(^{48}\) in which 2006 randomly selected US adults (age 18 or older) were asked to give the number of people in the last six months "with whom you discussed matters that are important to you." The average number of close confidants for the full sample was \(2.2 .\) In addition, the study asked participants whether or not they had a profile on a social networking site. For the 947 participants using a social networking site, the average number of close confidants was 2.5 with a standard deviation of 1.4 , and for the other 1059 participants who do not use a social networking site, the average was 1.9 with a standard deviation of \(1.3 .\) Find and interpret a \(90 \%\) confidence interval for the difference in means between the two groups.

Short Answer

Expert verified
The 90% confidence interval for the difference in means between the group using social networking sites and the group not using social networking sites is given by (x̄1 - x̄2) - ME to (x̄1 - x̄2) + ME. Substitute the calculated values of x̄1, x̄2 and ME in the formula.

Step by step solution

01

Identify the data

Identify the sample sizes, means, and standard deviations for each group. For the group using a social networking site, the sample size (n1) is 947, the mean (x̄1) is 2.5, and the standard deviation (s1) is 1.4. For the group not using a social networking site, the sample size (n2) is 1059, the mean (x̄2) is 1.9, and the standard deviation (s2) is 1.3.
02

Calculate the standard error

The standard error (SE) of the difference in sample means can be calculated using the formula: SE = sqrt[ (s1²/n1) + (s2²/n2) ] = sqrt[ (1.4²/947) + (1.3²/1059) ].
03

Find the critical value

For a 90% confidence level, the critical value for t is the t-score with 90% of the distribution's area lying between -t and +t from a t-distribution table. With large sample sizes, t approximates the z-score, which for 90% confidence interval is 1.645. You can also use a calculator or statistics software to find the exact t value.
04

Calculate the margin of error

The Margin of error (ME) can be calculated using the formula: ME = t * SE. Substitute the calculated values of t and SE in the formula.
05

Find the confidence interval

The confidence interval is the difference in sample means (x̄1 - x̄2) plus and minus the margin of error. So, calculate (x̄1 - x̄2) - ME and (x̄1 - x̄2) + ME.

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Key Concepts

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

Standard Error
In the realm of statistics, the Standard Error (SE) is a measure that tells us how far we can expect a sample statistic to deviate from the true population parameter. For example, when dealing with differences between two sample means, as in the textbook exercise, the standard error helps gauge the variability or uncertainty of this difference.

In the exercise, students calculated the standard error of the difference in sample means between two groups. To be more precise, the formula used is: \[\text{SE} = \sqrt{\left(\frac{s1^2}{n1}\right) + \left(\frac{s2^2}{n2}\right)}\]where \(s1\) and \(s2\) represent standard deviations of the two samples, and \(n1\) and \(n2\) are the respective sample sizes. Simplifying complex formulas and demystifying them is key. By understanding that the standard error essentially provides an estimate of the 'standard' distance between our estimated difference and the true difference in populations, students can better grasp the concept.
Margin of Error
Delving into the concept of Margin of Error (ME), we uncover the range within which we expect the true parameter to lie, given the sample statistics. It's a cushion that accounts for sampling variability. In the exercise, the margin of error is an essential piece of the confidence interval puzzle, as it defines the width of the interval around the sample means' difference.

To calculate it, the formula used is: \[\text{ME} = t \times \text{SE}\]This incorporates the standard error (SE) and multiplies it by a critical value from the t-distribution, labeled as \(t\). This critical value is essentially a multiplier that ensures the confidence interval spans a range that includes the true mean difference with a specific level of confidence, in this case, 90%. It's crucial to convey that the margin of error reflects how much we would expect our estimate to vary if we were to take multiple samples.
T-distribution
Moving on to the t-distribution, this is a probability distribution that estimates the population parameters when the sample size is small and/or the population standard deviation is unknown. It resembles the normal distribution but has heavier tails, meaning it predicts more variability. In the context of the exercise, it's used to find the critical value (\(t\)) that helps define the edges of our confidence interval.

Students must understand that as sample sizes increase, the t-distribution becomes more like the normal distribution. This concept, often represented as a bell-shaped curve, is critical when making inferences about a population based on sample data. We use this distribution to capture the uncertainty of the mean estimate, which then helps calculate the margin of error, and the overall confidence interval.
Sample Means Difference
Finally, let's touch upon the Sample Means Difference. In studies comparing two groups, researchers often investigate whether there's a significant difference between the groups' averages. This exercise illustrates such a scenario, comparing the number of close confidants between social media users and non-users.

The difference is computed simply as \[(x̄1 - x̄2)\]A positive result indicates that the first group has a higher mean, while a negative means the second group does. But how sure can we be that this sample difference reflects the true difference in populations? Confidence intervals calculated from margins of error and standard errors allow us to state this difference with a level of certainty. It's important for students to not just calculate these values but to interpret them—understanding that a confidence interval that does not contain zero suggests a statistically significant difference between the groups.

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Most popular questions from this chapter

A survey is planned to estimate the proportion of voters who support a proposed gun control law. The estimate should be within a margin of error of \(\pm 2 \%\) with \(95 \%\) confidence, and we do not have any prior knowledge about the proportion who might support the law. How many people need to be included in the sample?

Effect of Splitting the Bill Exercise 2.153 on page 105 describes a study to compare the cost of restaurant meals when people pay individually versus splitting the bill as a group. In the experiment half of the people were told they would each be responsible for individual meal costs and the other half were told the cost would be split equally among the six people at the table. The 24 people paying individually had a mean cost of 37.29 Israeli shekels with a standard deviation of 12.54 , while the 24 people splitting the bill had a higher mean cost of 50.92 Israeli shekels with a standard deviation of 14.33. The raw data can be found in SplitBill and both distributions are reasonably bell-shaped. Use this information to find and interpret a \(95 \%\) confidence interval for the difference in mean meal cost between these two situations.

Who Watches More TV: Males or Females? The dataset StudentSurvey has information from males and females on the number of hours spent watching television in a typical week. Computer output of descriptive statistics for the number of hours spent watching TV, broken down by gender, is given: \(\begin{array}{l}\text { Descriptive Statistics: TV } \\ \text { Variable } & \text { Gender } & \mathrm{N} & \text { Mean } & \text { StDev } \\ \text { TV } & \mathrm{F} & 169 & 5.237 & 4.100 \\ & \mathrm{M} & 192 & 7.620 & 6.427 \\\ \text { Minimum } & \mathrm{Q} 1 & \text { Median } & \mathrm{Q} 3 & \text { Maximum } \\ & 0.000 & 2.500 & 4.000 & 6.000 & 20.000 \\ & 0.000 & 3.000 & 5.000 & 10.000 & 40.000\end{array}\) (a) In the sample, which group watches more TV, on average? By how much? (b) Use the summary statistics to compute a \(99 \%\) confidence interval for the difference in mean number of hours spent watching TV. Be sure to define any parameters you are estimating. (c) Compare the answer from part (c) to the confidence interval given in the following computer output for the same data: \(\begin{array}{l}\text { Two-sample T for TV } \\ \text { Gender } & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \mathrm{F} & 169 & 5.24 & 4.10 & 0.32 \\ \mathrm{M} & 192 & 7.62 & 6.43 & 0.46 \\ \text { Difference } & =\mathrm{mu}(\mathrm{F})-\mathrm{mu}(\mathrm{M}) & & \end{array}\) Estimate for for difference: -2.383 r difference: (-3.836,-0.930) \(99 \% \mathrm{Cl}\) for (d) Interpret the confidence interval in context.

In Exercises 6.203 and \(6.204,\) use Stat Key or other technology to generate a bootstrap distribution of sample differences in means and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample standard deviations as estimates of the population standard deviations. Difference in mean commuting time (in minutes) between commuters in Atlanta and commuters in St. Louis, using \(n_{1}=500, \bar{x}_{1}=29.11,\) and \(s_{1}=20.72\) for Atlanta and \(n_{2}=500, \bar{x}_{2}=21.97,\) and \(s_{2}=14.23\) for St. Louis

In Exercises 6.188 to 6.191 , use the t-distribution to find a confidence interval for a difference in means \(\mu_{1}-\mu_{2}\) given the relevant sample results. Give the best estimate for \(\mu_{1}-\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed. A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=75.2, s_{1}=10.7, n_{1}=30\) and \(\bar{x}_{2}=69.0, s_{2}=8.3, n_{2}=20 .\)

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