Question

A paired-difference experiment was conducted using \(n=10\) pairs of observations. a. Test the null hypothesis \(H_{0}:\left(\mu_{1}-\mu_{2}\right)=0\) against \(H_{\mathrm{a}}:\left(\mu_{1}-\mu_{2}\right) \neq 0\) for \(\alpha=.05, \bar{d}=.3,\) and \(s_{d}^{2}=\) .16. Give the approximate \(p\) -value for the test. b. Find a \(95 \%\) confidence interval for \(\left(\mu_{1}-\mu_{2}\right)\). c. How many pairs of observations do you need if you want to estimate \(\left(\mu_{1}-\mu_{2}\right)\) correct to within .1 with probability equal to \(.95 ?\)

Short answer

Find a \(95 \%\) confidence interval for \(\left(\mu_{1}-\mu_{2}\right)\). Determine the required number of observation pairs to estimate \(\left(\mu_{1}-\mu_{2}\right)\) correctly to within 0.1 with a probability of 0.95. **Answer:** The approximate p-value for the test is 0.0424. The \(95 \%\) confidence interval for \(\left(\mu_{1}-\mu_{2}\right)\) is approximately (0.0459, 0.5541). The required number of observation pairs to estimate \(\left(\mu_{1}-\mu_{2}\right)\) correctly to within 0.1 with a probability of 0.95 is 21 pairs.

Step-by-step solution

Perform the t-test

To test the null hypothesis \(H_{0}:\left(\mu_{1}-\mu_{2}\right)=0\) versus the alternative hypothesis \(H_{\mathrm{a}}:\left(\mu_{1}-\mu_{2}\right) \neq 0\), we use the t-test. The test statistic is given by: $$ t=\frac{\bar{d}-\Delta_{0}}{\frac{s_{d}}{\sqrt{n}}} $$ Substitute the given values of \(\bar{d}=0.3,\Delta_0=0,s_d^2=0.16\) (which gives \(s_d=\sqrt{0.16} =0.4\)) and \(n=10\) into the formula to calculate the test statistic.

Calculate the test statistic

We plug in the values into the formula to get: $$ t=\frac{0.3-0}{\frac{0.4}{\sqrt{10}}}=\frac{0.3}{0.4/\sqrt{10}} \approx 2.3717 $$ Our test statistic, t, is approximately 2.3717. Since the alternative hypothesis \(H_{\mathrm{a}}:\left(\mu_{1}-\mu_{2}\right) \neq 0\) is a two-tailed test, we have to calculate the two-tailed p-value.

Calculate the p-value

To calculate the p-value, we need to find the probability that a t-distributed random variable with 9 degrees of freedom (which is equal to \(n-1\)) would be greater than the calculated test statistic in absolute value. Using a t-distribution table or a calculator, we find that the p-value is approximately \(2(1 - P(T_9 > 2.3717)) \approx 0.0424\). **b. Find \(95 \%\) Confidence Interval**

Identify the formula

A \(95 \%\) confidence interval for \(\left(\mu_{1}-\mu_{2}\right)\) can be calculated using the following formula: $$ CI = \left(\bar{d} - t_{\alpha/2}\frac{s_{d}}{\sqrt{n}}, \bar{d} + t_{\alpha/2}\frac{s_{d}}{\sqrt{n}}\right) $$ We already have values for \(\bar{d}=0.3\) and \(s_{d}=0.4\). We need the critical t-value \(t_{\alpha/2}\) for a two-tailed test with 9 degrees of freedom and a 0.025 significance level on each tail.

Find critical t-value

Using a t-distribution table or a calculator, we find the critical t-value \(t_{\alpha/2} \approx 2.2622\).

Calculate confidence interval

Plug in the values into the formula to calculate the confidence interval: $$ CI=\left(0.3 - 2.2622\frac{0.4}{\sqrt{10}}, 0.3 + 2.2622\frac{0.4}{\sqrt{10}}\right) \approx (0.0459, 0.5541) $$ The \(95 \%\) confidence interval for \(\left(\mu_{1}-\mu_{2}\right)\) is approximately (0.0459, 0.5541). **c. Determine required number of observation pairs**

Identify the formula

The sample size needed to estimate \(\left(\mu_{1}-\mu_{2}\right)\) correct to within 0.1 with probability equal to 0.95 is given by: $$ n = \left(\frac{t_{\alpha/2}s_d}{E}\right)^2 $$ Here, \(t_{\alpha/2}\) is the critical t-value for a two-tailed test with a 0.025 significance level on each tail (same as in part b), \(s_d=0.4\) and \(E=0.1\).

Calculate the required number of pairs

Plug in the values to calculate the required number of pairs: $$ n = \left(\frac{2.2622 \times 0.4}{0.1}\right)^2 \approx 20.4368 $$ Since we cannot have a fraction of an observation, we round up the value to the nearest whole number, which is 21. Thus, 21 pairs of observations are required if we want to estimate \(\left(\mu_{1}-\mu_{2}\right)\) correct to within 0.1 with probability equal to 0.95.

Key concepts

Hypothesis Testing

Hypothesis testing is a statistical method used to make inferences about population parameters based on sample data. It involves setting up two competing hypotheses: the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)). The null hypothesis typically suggests no effect or no difference, whereas the alternative implies some effect or difference.

In the context of a paired difference test, we are interested in examining if there is a difference between two related groups. For example, this could involve testing if a new treatment is more effective than an old one within the same group of subjects. In this exercise, we conduct a t-test to determine if the difference in means between two paired samples is significant.

The t-test formula for paired samples is given by:\[ t = \frac{\bar{d} - \Delta_0}{s_d/\sqrt{n}}\]where \(\bar{d}\) is the mean of differences, \(\Delta_0\) is the hypothesized difference (often 0), \(s_d\) is the standard deviation of differences, and \(n\) is the number of pairs. A key step in hypothesis testing is calculating the test statistic, which is then used to find the p-value. The p-value indicates the probability of observing the test results under the assumption that the null hypothesis is true. If the p-value is smaller than the pre-set significance level (\(\alpha = 0.05\)), we reject the null hypothesis in favor of the alternative hypothesis.

Confidence Interval

A confidence interval provides a range of values within which we expect the true population parameter to lie, with a certain level of confidence. A 95% confidence interval implies that if we were to repeat the study many times, approximately 95% of the calculated intervals would contain the true difference in means.

To calculate a confidence interval for a paired difference, we use:\[CI = \left(\bar{d} - t_{\alpha/2}\frac{s_{d}}{\sqrt{n}}, \bar{d} + t_{\alpha/2}\frac{s_{d}}{\sqrt{n}}\right)\]Here, \(t_{\alpha/2}\) is the critical value from the t-distribution associated with the desired confidence level, and \(n-1\) is used as the degrees of freedom. In our exercise, for 9 degrees of freedom and a significance level of 0.025 in each tail, the critical t-value is found to be approximately 2.2622.

Plugging in the given values, \(\bar{d} = 0.3\) and \(s_{d} = 0.4\), the calculated confidence interval is approximately (0.0459, 0.5541). This means we are 95% confident that the actual difference in means lies within this interval. A confidence interval not including zero supports the hypothesis of a significant difference between the paired samples.

Sample Size Calculation

Calculating the appropriate sample size is crucial in designing an experiment to ensure the study has sufficient power to detect a true effect. The sample size required depends on several factors, including the desired precision of the estimate, confidence level, and variability in the data.

To determine the number of paired observations needed for a certain precision, we use the formula:\[n = \left(\frac{t_{\alpha/2}s_d}{E}\right)^2\]where \(E\) represents the margin of error, and \(t_{\alpha/2}\) is the critical t-value based on the desired confidence level. In this exercise, we wish to estimate the difference within an error margin of 0.1 with a 95% confidence level. The standard deviation \(s_{d}\) is given as 0.4, and the critical t-value is 2.2622.

Plugging these values into the formula, we find that we need approximately 20.4368 pairs. Rounding up gives us 21 pairs. This ensures that our estimates are precise enough to meet the chosen level of confidence and accuracy, highlighting the importance of planning adequately to achieve reliable results in hypothesis testing.