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Use a t-distribution to find a confidence interval for the difference in means \(\mu_{1}-\mu_{2}\) using the relevant sample results from paired data. 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, and that differences are computed using \(d=x_{1}-x_{2}\). A \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=556.9, s_{d}=\) \(143.6, n_{d}=100\)

Short Answer

Expert verified
The best estimate for the difference of means ( \(\mu_{1}-\mu_{2}\) ) is 556.9, the margin of error is 27.8936, and the 90% confidence interval for the difference in means is (529.0064, 584.7936)

Step by step solution

01

Identify given values

From the problem, the mean difference ( \(\bar{x}_{d}\) ) is 556.9, the standard deviation of the differences ( \(s_{d}\)) is 143.6, and the sample size ( \(n_{d}\) ) is 100. The confidence level is 90%.
02

Calculate the t-value

We find the t-value that corresponds to our desired confidence level and degrees of freedom. Degrees of freedom, in this case, equals \( n_{d} - 1\), which is 99. For a 90% confidence level, the t-value (two-tailed) for 99 degrees of freedom is approximately 1.660.
03

Calculate the margin of error

The formula for the margin of error \( E \) with a t-distribution is \( E = t \cdot \frac{s_{d}}{\sqrt{n_{d}}} \). Substituting with provided values, \( E = 1.660 \cdot \frac{143.6}{\sqrt{100}} \) which results in 27.8936.
04

Determine the confidence interval

Our confidence interval is determined as \( \bar{x}_d ± E \), thus our interval is \( 556.9 ± 27.8936 \)

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Key Concepts

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

Understanding t-distribution
When diving into statistics, the concept of the t-distribution is essential, especially when working with smaller sample sizes or when the population standard deviation is unknown. The t-distribution, or Student's t-distribution, is a type of probability distribution that is symmetric and bell-shaped, much like the standard normal distribution, yet with heavier tails.

Heavier tails indicate a higher probability of values further from the mean, which is crucial when dealing with uncertainty in estimates from smaller samples. As the sample size increases, the t-distribution approaches the normal distribution. The t-distribution comes into play when calculating confidence intervals or conducting hypothesis testing for the mean population value when the sample size is below 30 or the population variance is unknown. In such scenarios, the t-distribution provides more reliable and adjusted intervals and test results than the normal distribution.
Paired Sample t-Test Explained
The paired sample t-test, also known as the dependent t-test, is a powerful statistical procedure used to compare the means from two related groups. These groups are 'paired' because they are somehow related or matched.

Examples of such pairings include before-and-after observations on the same subjects, or matched subjects in different conditions. In the paired sample t-test, we calculate the difference \(d=x_{1}-x_{2}\) for each matched pair of observations and then analyze these differences using the t-test. This method helps to control for variability between subjects, as the comparison is based only on the changes within each pair, not between different pairs. As a result, this test can be more sensitive in detecting differences than two independent samples t-tests.
Calculating the Margin of Error
The concept of margin of error plays a pivotal role in understanding the precision of an estimate. It is the range above and below the observed statistic value that is believed to contain, with a certain level of confidence, the true population parameter.

Mathematically, the margin of error (\(E\)) can be calculated in context of a t-distribution as \( E = t \cdot \frac{s_{d}}{\sqrt{n_{d}}} \), where \(t\) is the t-distribution critical value associated with our desired confidence level, \(s_d\) is the standard deviation of the differences observed in our paired sample, and \(n_d\) is the number of differences or pairings. Higher confidence levels or larger variabilities result in a wider margin of error, suggesting less precision, while a larger sample size reduces the margin of error, indicating a more precise estimate.

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Most popular questions from this chapter

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.

Using Data 5.1 on page \(375,\) we find a significant difference in the proportion of fruit flies surviving after 13 days between those eating organic potatoes and those eating conventional (not organic) potatoes. ask you to conduct a hypothesis test using additional data from this study. \(^{40}\) In every case, we are testing $$\begin{array}{ll}H_{0}: & p_{o}=p_{c} \\\H_{a}: & p_{o}>p_{c}\end{array}$$ where \(p_{o}\) and \(p_{c}\) represent the proportion of fruit flies alive at the end of the given time frame of those eating organic food and those eating conventional food, respectively. Also, in every case, we have \(n_{1}=n_{2}=500 .\) Show all remaining details in the test, using a \(5 \%\) significance level. Effect of Organic Bananas after 15 Days After 15 days, 345 of the 500 fruit flies eating organic bananas are still alive, while 320 of the 500 eating conventional bananas are still alive.

Use a t-distribution and the given matched pair sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distribution of the differences is relatively normal. Assume that differences are computed using \(d=x_{1}-x_{2}\). Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}>\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lllllllllll} \hline \text { Situation } & 1 & 125 & 156 & 132 & 175 & 153 & 148 & 180 & 135 & 168 & 157 \\ \text { Situation } & 2 & 120 & 145 & 142 & 150 & 160 & 148 & 160 & 142 & 162 & 150 \\ \hline \end{array} $$

A data collection method is described to investigate a difference in means. In each case, determine which data analysis method is more appropriate: paired data difference in means or difference in means with two separate groups. To study the effect of sitting with a laptop computer on one's lap on scrotal temperature, 29 men have their scrotal temperature tested before and then after sitting with a laptop for one hour.

When we want \(95 \%\) confidence and use the conservative estimate of \(p=0.5,\) we can use the simple formula \(n=1 /(M E)^{2}\) to estimate the sample size needed for a given margin of error ME. In Exercises 6.40 to 6.43, use this formula to determine the sample size needed for the given margin of error. A margin of error of 0.01

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