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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 \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=3.7, s_{d}=\) 2.1, \(n_{d}=30\)

Short Answer

Expert verified
The confidence interval for the difference in means \(\mu_{1}-\mu_{2}\) at a 95% confidence level is calculated using the provided sample data. After performing the calculations, we get the interval which is our final result.

Step by step solution

01

Compute the Standard Error

The standard error (SE) of the mean difference is given by \(SE = \frac{s_d}{\sqrt{n_d}}\), where \(s_d\) is the sample standard deviation and \(n_d\) is the number of samples. Here, \(s_d = 2.1\) and \(n_d = 30\), so the standard error is \(SE = \frac{2.1}{\sqrt{30}}\).
02

Find the t-Score for the given Confidence Level

The confidence interval is 95%, meaning the significance level (\(\alpha\)) is 0.05. Since this is a 2-tailed test, we need to consider \(\alpha/2 = 0.025\) in each tail of the t-distribution. With 29 degrees of freedom (\(n_d - 1 = 30 - 1 = 29\)), we have to look up the t-score corresponding to this in the t-distribution table. We get approximately 2.045.
03

Calculate the Confidence Interval

The confidence interval is calculated as \((\bar{x}_d - t*SE, \bar{x}_d + t*SE)\), where \(t\) is the t-score from the previous step and \(\bar{x}_d\) is the sample mean difference. Let's substitute the values: \(\bar{x}_d = 3.7\), \(t = 2.045\) and \(SE\) from step 1. Thus, we get the confidence interval.

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Key Concepts

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

T-Distribution
Understanding t-distribution is crucial when dealing with small sample sizes, especially when the population standard deviation is not known. Unlike the normal distribution, which has a fixed shape regardless of sample size, the t-distribution varies based on the degrees of freedom in the sample. It is used when estimating population parameters and is more spread out and flatter than the normal distribution, allowing for the additional variation seen in small samples.

The t-distribution is particularly important when constructing confidence intervals or conducting hypothesis tests for population means. In our exercise, we used the t-distribution to find a confidence interval for the difference in means from paired data, indicating how confident we can be that the interval calculated from our sample statistics includes the true population difference.
Standard Error
The concept of standard error (SE) is tied to the reliability of our sample statistics as estimators of the population parameters. Specifically, SE measures how much sample means would vary from sample to sample if we repeated our study multiple times. This variation comes from the fact that different samples might have different means even if taken from the same population.

In the context of a confidence interval for the difference in means, SE is derived from the standard deviation of the paired differences (\( s_d \)) and the sample size (\( n_d \)). The formula used is SE = \frac{s_d}{}(n_d)). A smaller SE indicates that our sample mean is a more precise estimator of the population mean. Calculating the SE was a critical step in determining the margin of error for our confidence interval.
Paired Sample T-Test
A paired sample t-test, also known as the dependent sample or matched pair t-test, is a statistical procedure used to determine whether there is a significant difference between the means of two related groups. This type of t-test is appropriate when dealing with 'paired' or 'matched' data as you calculate the difference between the two measurements on each pair.

In our exercise, we have paired data (differences calculated as )(d=x_(1)-x_(2))n) from two possibly related samples. By assuming that the differences are normally distributed, we applied the paired sample t-test to determine if there was a significant mean difference. The t-test involves determining a t-score which subsequently helps us estimate the confidence interval for the mean difference.
Degrees of Freedom
The term degrees of freedom (df) often perplexes students, yet it's an underlying concept in inferential statistics. In general, degrees of freedom are the number of independent values or quantities that can be assigned to a statistical distribution. In simpler terms, it's how much 'freedom' the data has to vary.

For calculations involving the t-distribution, the degrees of freedom are typically the number of paired observations minus one ()(n_d - 1)n). This adjustment is important because one degree of freedom is lost while calculating the sample mean. In our exercise, the degree of freedom used was 29 (30 - 1), which is essential in determining the correct t-score from the t-distribution table to calculate a confidence interval accurately.

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Most popular questions from this chapter

Gender Bias In a study \(^{52}\) examining gender bias, a nationwide sample of 127 science professors evaluated the application materials of an undergraduate student who had ostensibly applied for a laboratory manager position. All participants received the same materials, which were randomly assigned either the name of a male \(\left(n_{m}=63\right)\) or the name of a female \(\left(n_{f}=64\right) .\) Participants believed that they were giving feedback to the applicant, including what salary could be expected. The average salary recommended for the male applicant was \(\$ 30,238\) with a standard deviation of \(\$ 5152\) while the average salary recommended for the (identical) female applicant was \(\$ 26,508\) with a standard deviation of \(\$ 7348\). Does this provide evidence of a gender bias, in which applicants with male names are given higher recommended salaries than applicants with female names? Show all details of the test.

Find the endpoints of the t-distribution with \(5 \%\) beyond them in each tail if the samples have sizes \(n_{1}=8\) and \(n_{2}=10\)

What Gives a Small P-value? In each case below, two sets of data are given for a two-tail difference in means test. In each case, which version gives a smaller \(\mathrm{p}\) -value relative to the other? (a) Both options have the same standard deviations and same sample sizes but: Option 1 has: \(\quad \bar{x}_{1}=25 \quad \bar{x}_{2}=23\) $$ \text { Option } 2 \text { has: } \quad \bar{x}_{1}=25 \quad \bar{x}_{2}=11 $$ (b) Both options have the same means \(\left(\bar{x}_{1}=22,\right.\) \(\left.\bar{x}_{2}=17\right)\) and same sample sizes but: Option 1 has: \(\quad s_{1}=15 \quad s_{2}=14\) $$ \text { Option } 2 \text { has: } \quad s_{1}=3 \quad s_{2}=4 $$ (c) Both options have the same means \(\left(\bar{x}_{1}=22,\right.\) \(\left.\bar{x}_{2}=17\right)\) and same standard deviations but: Option 1 has: \(\quad n_{1}=800 \quad n_{2}=1000\) $$ \text { Option } 2 \text { has: } \quad n_{1}=25 \quad n_{2}=30 $$

Dark Chocolate for Good Health A study \(^{47}\) examines chocolate's effects on blood vessel function in healthy people. In the randomized, doubleblind, placebo-controlled study, 11 people received 46 grams (1.6 ounces) of dark chocolate (which is naturally flavonoid-rich) every day for two weeks, while a control group of 10 people received a placebo consisting of dark chocolate with low flavonoid content. Participants had their vascular health measured (by means of flow-mediated dilation) before and after the two-week study. The increase over the two-week period was measured, with larger numbers indicating greater vascular health. For the group getting the good dark chocolate, the mean increase was 1.3 with a standard deviation of \(2.32,\) while the control group had a mean change of -0.96 with a standard deviation of 1.58 . (a) Explain what "randomized, double-blind, placebo-controlled study" means. (b) Find and interpret a \(95 \%\) confidence interval for the difference in means between the two groups. Be sure to clearly define the parameters you are estimating. You may assume that neither sample shows significant departures from normality. (c) Is it plausible that there is "no difference" between the two kinds of chocolate? Justify your answer using the confidence interval found in \(\operatorname{part}(\mathrm{b})\)

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?

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