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Chapter 6: Problem 270

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 95% confidence interval for \(\mu_{1}-\mu_{2}\) lies between the two values calculated in Step 3.

Step by step solution

01

Calculate the Standard Error

First, calculate the standard error (SE) of the difference scores. The standard error is calculated as the standard deviation of the difference scores \(s_{d}\) divided by the square root of the size of the sample \(n_{d}\). This is computed as: \(SE = s_{d}/\sqrt{n_{d}} = 2.1/\sqrt{30}\).
02

Identify the Critical Value from the t-distribution

Next, identify the critical value (t*) from the t-distribution for a 95% confidence interval. Because the degrees of freedom (df) for a sample size of 30 is 29 (df= n-1), we find t* using df = 29. Depending on the source, the value could slightly vary but t* is approximately 2.045 for a 95% confidence interval with 29 degrees of freedom.
03

Calculate the Confidence Interval

Now, calculate the confidence interval using the formula: Confidence interval = \(\bar{x}_{d} \pm (t* \times SE)\). Substituting known values, we get: Confidence interval = \(3.7 \pm (2.045 \times SE)\). The two values obtained represent the lower and upper limits of the 95% confidence interval for \(\mu_{1}-\mu_{2}\).

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Key Concepts

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

Paired Difference
In statistical analysis, paired difference refers to the comparison of two related samples or groups. Often, these samples consist of measurements taken on the same subjects under different conditions or times. The primary intention of analyzing paired differences is to evaluate if there is a significant change from one condition to another.

To compute paired differences:
  • Identify pairs: Each subject should have two measurements, one for each condition being compared.
  • Compute the difference: Subtract the second measurement from the first for each subject. This generates the difference scores.
  • Analyze the average of these differences using statistical tests.
The paired difference approach reduces variability because it accounts for the innate characteristics of individual subjects that could affect outcomes. It removes between-subject variability and focuses on changes attributable to conditions.
Standard Error
The standard error (SE) is a critical statistic in hypothesis testing and confidence interval estimation. It represents the variability of a sample statistic, in this case, the mean of the paired differences, estimates the expected fluctuation if the sampling process is repeated.

To calculate the standard error of the paired differences:
  • Determine the standard deviation of the paired differences, denoted as \(s_d\).
  • Divide the standard deviation by the square root of the sample size \(n_d\). This formula is expressed as \(SE = \frac{s_d}{\sqrt{n_d}}\).
The result is a smaller standard error when compared to the standard deviation, this usually indicates less variance and thus a more precise mean estimate from the sample results.
t-distribution
The t-distribution is pivotal in conducting hypothesis tests and constructing confidence intervals when dealing with small sample sizes or unknown population variances. It resembles a normal distribution but has thicker tails, allowing for more variability. This feature makes it apt when data variability is higher or when the sample size is limited.

Key Points about the t-distribution:
  • The shape of the t-distribution changes based on its degrees of freedom \(df\), which is equal to the sample size minus one (\(n - 1\)).
  • As the sample size increases, the t-distribution increasingly resembles the normal distribution.
  • Critical values (like \(t^*\)), necessary for establishing confidence intervals, are derived from this distribution.
Using the t-distribution helps address unknown variances and provides more reliable results for smaller samples, as evidenced in determining the critical value for a confidence interval with 29 degrees of freedom in this exercise.

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Most popular questions from this chapter

Use StatKey or other technology to generate a bootstrap distribution of sample 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 deviation as an estimate of the population standard deviation. Mean commute time in Atlanta, in minutes, using the data in CommuteAtlanta with \(n=500\), \(\bar{x}=29.11,\) and \(s=20.72\)

Random samples of the given sizes are drawn from populations with the given means and standard deviations. For each scenario: (a) Find the mean and standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 300 from Population 1 with mean 75 and standard deviation 18 and samples of size 500 from Population 2 with mean 83 and standard deviation 22

Exercise 4.86 on page 263 introduces a matched pairs study in which 47 participants had cell phones put on their ears and then had their brain glucose metabolism (a measure of brain activity) measured under two conditions: with one cell phone turned on for 50 minutes (the "on" condition) and with both cell phones off (the "off" condition). Brain glucose metabolism is measured in \(\mu \mathrm{mol} / 100 \mathrm{~g}\) per minute, and the differences of the metabolism rate in the on condition minus the metabolism rate in the off condition were computed for all participants. The mean of the differences was 2.4 with a standard deviation of \(6.3 .\) Find and interpret a \(95 \%\) confidence interval for the effect size of the cell phone waves on mean brain metabolism rate.

Refer to a study on hormone replacement therapy. Until 2002 , hormone replacement therapy (HRT), taking hormones to replace those the body no longer makes after menopause, was commonly prescribed to post-menopausal women. However, in 2002 the results of a large clinical trial \(^{56}\) were published, causing most doctors to stop prescribing it and most women to stop using it, impacting the health of millions of women around the world. In the experiment, 8506 women were randomized to take HRT and 8102 were randomized to take a placebo. Table 6.16 shows the observed counts for several conditions over the five years of the study. (Note: The planned duration was 8.5 years. If Exercises 6.205 through 6.208 are done correctly, you will notice that several of the p-values are just below \(0.05 .\) The study was terminated as soon as HRT was shown to significantly increase risk (using a significance level of \(\alpha=0.05)\), because at that point it was unethical to continue forcing women to take HRT). Does HRT influence the chance of a woman getting cancer of any kind? $$ \begin{array}{lcc} \hline \text { Condition } & \text { HRT Group } & \text { Placebo Group } \\ \hline \text { Cardiovascular Disease } & 164 & 122 \\ \text { Invasive Breast Cancer } & 166 & 124 \\ \text { Cancer (all) } & 502 & 458 \\ \text { Fractures } & 650 & 788 \\ \hline \end{array} $$

Standard Error from a Formula and a Bootstrap Distribution In Exercises 6.19 to \(6.22,\) use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of lie detector trials in which the technology misses a lie, with \(n=48\) and \(\hat{p}=0.354\)
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