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Find the following confidence intervals for \(\mu_{\mathrm{d}}\), assuming that the populations of paired differences are normally distributed. a. \(n=12, \bar{d}=17.5, \quad s_{d}=6.3, \quad\) confidence level \(=99 \%\) b. \(n=27, \quad \bar{d}=55.9, \quad s_{d}=14.7\), confidence level \(=95 \%\) c. \(n=16, \bar{d}=29.3, s_{d}=8.3, \quad\) confidence level \(=90 \%\)

Short Answer

Expert verified
The confidence intervals are: a. [15.105, 19.895] b. [51.42, 60.38] c. [26.76, 31.84]

Step by step solution

01

Calculate t-score

Use the values given in the exercise and apply them to the t-distribution equation. For \(n=12\) and the confidence level of \(99\% \), the degree of freedom is \(12-1=11\). The one-tailed t-score for a confidence level \(99\% \) with a degree of freedom of \(11\) is approximately \(3.106\). Repeat this step for each given scenario.
02

Compute Confidence Interval for Scenario a

Apply the values to the confidence interval formula. Given \(\bar{d}=17.5\), \(s_{d}=6.3\), \(n=12\) and \(t=3.106\), the confidence interval will be \((17.5 \pm 3.106 \times \frac{6.3}{\sqrt{12}})\), which results in an interval of \([15.105, 19.895]\).
03

Compute Confidence Intervals for Other Scenarios

Repeat step 2 for the other scenarios. Given: - Scenario b: \(\bar{d}=55.9\), \(s_{d}=14.7\), \(n=27\) and \(t\) for \(95\% \) confidence with \(26\) degrees of freedom is approximately \(2.056\). The interval computes to \([51.42, 60.38] \). - Scenario c: \(\bar{d}=29.3\), \(s_{d}=8.3\), \(n=16\) and \(t\) for \(90\% \) confidence with \(15\) degrees of freedom is approximately \(1.753\). The interval calculates to \([26.76, 31.84] \).

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Key Concepts

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

Understanding the t-distribution
The t-distribution, also known as Student's t-distribution, is a fundamental concept in statistics. It's used when estimating the means of a normally distributed population in situations where the sample size is small. Unlike the normal distribution, the t-distribution is wider and has heavier tails, which means there's more room for variability. This characteristic ensures that confidence intervals are wider when based on small samples.

The shape of the t-distribution depends on the degrees of freedom. As the degrees of freedom increase, the t-distribution approaches the normal distribution. In practice, the t-distribution is used to calculate confidence intervals for population means when the population standard deviation is unknown. You can find the appropriate t-value by referring to t-tables, which list t-values corresponding to different confidence levels and degrees of freedom.
  • Use t-distribution when sample size is small.
  • Applicable for normally distributed populations.
  • Wider intervals allow for uncertainties in smaller samples.
The Role of Degrees of Freedom
Degrees of freedom generally refer to the number of values in a calculation that are free to vary. In the context of confidence intervals, the degrees of freedom (df) are calculated as the sample size minus one ( -1"). So, for example, if you have a sample size of 12, the degrees of freedom are 11.

This concept is crucial for determining the t-value used in creating the confidence interval. The degrees of freedom affect the shape of the t-distribution. Fewer degrees of freedom result in a distribution with fatter tails and more variability, thus requiring a higher t-value to maintain a given level of confidence.
  • Calculate as n-1 for a single sample.
  • Influences the shape of the t-distribution.
  • More degrees indicate the distribution becomes closer to normal.
Standard Deviation and Its Importance
Standard deviation ( ext{sd}") is a measure of how spread out the numbers in a data set are. In the context of confidence intervals, it's crucial because it provides insight into the variability of differences in paired data. This measurement indicates whether the differences are consistently near the mean or scattered over a wide range.

A smaller standard deviation means the data points are close to the mean, and consequently, the confidence interval will be narrower, indicating more precision. Conversely, a larger standard deviation indicates more variability among the differences, resulting in a wider confidence interval.

In paired samples, standard deviation is specifically used for calculating the standard error, which is then multiplied by the t-value to find the confidence interval.
  • Indicates data variability.
  • Smaller value means tighter confidence interval.
  • Crucial for calculating standard error in paired differences.
Paired Differences Explained
Paired differences come into play when data is collected in pairs. For example, you might be measuring before-and-after effects on the same subjects. Rather than analyzing each group's data separately, the differences between each pair are considered.

By focusing on these differences, paired sample analysis accounts for individual variability. This approach often results in a more precise estimate as it neutralizes factors that consistently affect paired observations.

In confidence intervals, the mean of these differences is calculated, along with the standard deviation of differences. These values are used to determine if the average change is significantly different from zero. This analysis helps in understanding improvements or changes due to an intervention or time.
  • Used in before-and-after studies.
  • Neutralizes consistent individual variability.
  • Focuses on differences, improving result precision.

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Most popular questions from this chapter

The following information was obtained from two independent samples selected from two populations with unknown but equal standard deviations. $$ \begin{array}{lll} n_{1}=55 & \bar{x}_{1}=90.40 & s_{1}=11.60 \\ n_{2}=50 & \bar{x}_{2}=86.30 & s_{2}=10.25 \end{array} $$ a. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\)

A November 2011 Gallup poll asked American adults about their views of healthcare and the healthcare system in the United States. Although feelings about the quality of healthcare were positive, the same cannot be said about the quality of the healthcare system. According to this study, \(29 \%\) of Independents and \(27 \%\) of Democrats rated the healthcare system as being excellent or good (www.gallup.com/poll/ \(150788 /\) Americans-Maintain- Negative-View-Healthcare-Coverage.aspx). Suppose that these results were based on samples of 1200 Independents and 1300 Democrats. a. Let \(p_{1}\) and \(p_{2}\) be the proportions of all Independents and all Democrats, respectively, who will rate the healthcare system as being excellent or good. Construct a \(97 \%\) confidence interval for \(p_{1}-p_{2}\) b. Using a \(1 \%\) significance level, can you conclude that \(p_{1}\) is different from \(p_{2}\) ? Use both the critical-value and the \(p\) -value approaches.

According to Pew Research Center surveys, \(79 \%\) of U.S. adults were using the Internet in January 2011 and \(83 \%\) were using it in January 2012 (USA TODAY, January 26,2012 ). Suppose that these percentages are based on random samples of 1800 U.S. adults in January 2011 and 1900 in January \(2012 .\) a. Let \(p_{1}\) and \(p_{2}\) be the proportions of all U.S. adults who were using the Internet in January 2011 and January 2012, respectively. Construct a \(98 \%\) confidence interval for \(p_{1}-p_{2}\) b. Using a \(1 \%\) significance level, can you conclude that \(p_{1}\) is lower than \(p_{2}\) ? Use both the criticalvalue and the \(p\) -value approaches.

A mail-order company has two warehouses, one on the West Coast and the second on the East Coast. The company's policy is to mail all orders placed with it within 72 hours. The company's quality control department checks quite often whether or not this policy is maintained at the two warehouses. A recently taken sample of 400 orders placed with the warehouse on the West Coast showed that 364 of them were mailed within 72 hours. Another sample of 300 orders placed with the warehouse on the East Coast showed that 279 of them were mailed within 72 hours. a. Construct a \(97 \%\) confidence interval for the difference between the proportions of all orders placed at the two warehouses that are mailed within 72 hours. b. Using a \(2.5 \%\) significance level, can you conclude that the proportion of all orders placed at the warehouse on the West Coast that are mailed within 72 hours is lower than the corresponding proportion for the warehouse on the East Coast?

The global recession has led more and more people to move in with relatives, which has resulted in a large number of multigenerational households. An October 2011 Pew Research Center poll showed that \(11.5 \%\) of people living in multigenerational households were living below the poverty level, and 14.6\% of people living in other types of households were living below the poverty level (www. pewsocialtrends.org/201 1/10/03/fighting-poverty-in-a-bad- cconomy-americans-move-in-with-relatives/? sre-pre-headline). Suppose that these results were based on samples of 1000 people living in multigenerational households and 2000 people living in other types of households. a. Let \(p_{1}\) be the proportion of all people in multigenerational households who live below the poverty level and \(p_{2}\) be the proportion of all people in other types of households who live below the poverty level. Construct a 98\% confidence interval for \(p_{1}-p_{2}\) - b. Using a \(2.5 \%\) significance level, can you conclude that \(p_{1}\) is less than \(p_{2}\) ? Use the critical-value approach. c. Repeat part b using the \(p\) -value approach.

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