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Consider the hypothesis test \(H_{0}: \mu_{1}=\mu_{2}\) against \(H_{1}: \mu_{1}=\mu_{2} .\) Suppose that sample sizes \(n_{1}=15\) and \(n_{2}=15,\) that \(\bar{x}_{1}=6.2\) and \(\bar{x}_{2}=7.8,\) and that \(s_{1}^{2}=4\) and \(s_{2}^{2}=6.25 .\) Assume that \(\sigma_{1}^{2}=\sigma_{2}^{2}\) and that the data are drawn from normal distributions. Use \(\alpha=0.05\) (a) Test the hypothesis and find the \(P\) -value. (b) Explain how the test could be conducted with a confidence interval. (c) What is the power of the test in part (a) if \(\mu_{1}\) is 3 units less than \(\mu_{2}\) ? (d) Assume that sample sizes are equal. What sample size should be used to obtain \(\beta=0.05\) if \(\mu_{1}\) is 2.5 units less than \(\mu_{2}\) ? Assume that \(\alpha=0.05 .\)

Short Answer

Expert verified
Reject H0 with a p-value of 0.021. Confidence interval does not contain 0. Power is high; n=20 per group for \( \beta=0.05 \).

Step by step solution

01

Formulate Hypotheses

We start with the null hypothesis (H0: \( \mu_1 = \mu_2 \)) and the alternative hypothesis (H1: \( \mu_1 eq \mu_2 \)). This is a two-tailed test as we are checking for equality.
02

Calculate Test Statistic

Calculate the pooled standard deviation \( s_p \) using the formula: \[ s_p = \sqrt{\frac{(n_1-1) s_1^2 + (n_2-1) s_2^2}{n_1+n_2-2}} \] Plug in the values: \[ s_p = \sqrt{\frac{14 \times 4 + 14 \times 6.25}{28}} = \sqrt{5.125} = 2.2638 \] Now, calculate the test statistic \( t \): \[ t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \times \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} = \frac{6.2 - 7.8}{2.2638 \times \sqrt{\frac{1}{15} + \frac{1}{15}}} = -2.454 \]
03

Determine P-Value and Conclusion

For \( t = -2.454 \), with degrees of freedom \( df = n_1 + n_2 - 2 = 28 \), find the p-value using a t-distribution table or calculator. Since it's a two-tailed test, we double the p-value. Given \( \alpha = 0.05 \), if p-value < 0.05, reject H0. Here, the p-value is approximately 0.021, so we reject H0 at the 5% significance level.
04

Use Confidence Interval to Test Hypothesis

Construct a 95% confidence interval for \( \mu_1 - \mu_2 \) using: \[ (\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2, df} \times s_p \times \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} \] Plug in values: \[ (6.2 - 7.8) \pm 2.048 \times 2.2638 \times \sqrt{\frac{1}{15} + \frac{1}{15}} = -1.6 \pm 2.048 \times 1.161 \] \[ CI = (-4.00, 0.80) \] Since 0 is not within the interval, reject H0.
05

Calculate Power of the Test

To find the power, use: \[ \beta = P(\text{Fail to reject H0 when } \mu_1 - \mu_2 = -3) \] Convert this using the non-centrality parameter and find the power from tables or software. Using \( \delta = \frac{-3}{2.2638 \sqrt{2/15}} \approx -3.944 \), and looking up tables, power=1-\beta is high.
06

Determine Sample Size for Given Power

To find \( n \) with \( \beta=0.05 \), set \[ n = \frac{(z_{\alpha/2} + z_\beta)^2 \times (s_p^2)}{(\mu_1 - \mu_2)^2} \] Assuming new \( \mu_1 - \mu_2 = -2.5 \), calculate: \[ n = \frac{(1.96 + 1.645)^2 \times (5.125)}{2.5^2} \approx 19 \] Thus, 20 per group would be sufficient.

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Key Concepts

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

Two-tailed Test
In statistics, a two-tailed test is used when we want to determine if there are any differences between two means when we have no specific direction for our hypothesis. This means we are testing for the possibility of the relationship in both directions. The null hypothesis ( H鈧 ) states that there's no effect or difference, specifically, that the means are equal ( 渭鈧 = 渭鈧 ). On the other hand, the alternative hypothesis ( H鈧 ) we set up is non-directional and posits that the means are not equal ( 渭鈧 鈮 渭鈧 ).
A common scenario for a two-tailed test is when we are comparing the means of two independent samples. Instead of predicting which mean will be greater or less, we just want to know if they are different. This test is often visualized with a normal curve distribution, marking the critical regions (extremes on both tails of the curve), representing the areas where the result would be considered unlikely if the null hypothesis is true. This helps to determine if we should reject the null hypothesis, based on the variability and the size of difference in our sample means.
P-value Calculation
The P-value is a crucial part of statistical hypothesis testing. It helps you understand the strength of the results you have found. When you conduct a hypothesis test, the P-value is computed based on your data and helps determine whether you can reject the null hypothesis. The smaller the P-value, the stronger the evidence against the null hypothesis.
For a two-tailed test, once you have computed the test statistic (like a t-value), you find the P-value by checking how extreme your test statistic is, assuming the null hypothesis is true. This is done through a t-distribution table or software.
  • If your P-value is less than your chosen significance level (伪), you reject the null hypothesis. A common 伪 level used is 5% (0.05).
  • In the example given, the P-value was found to be approximately 0.021. Since this P-value is less than 0.05, it tells us there is a statistically significant difference between the two groups' means.
This evidence allows researchers to statistically reject the null hypothesis, accepting the alternative hypothesis instead.
Confidence Interval
A confidence interval gives a range of values for an unknown parameter and is used in hypothesis testing to understand where a true mean difference might lie. In the context of comparing two means, a 95% confidence interval is constructed to estimate the range in which the difference between the two population means might fall.
The formula for the confidence interval when dealing with two means is:\[(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2, df} \times s_p \times \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\]This range tells us that if the interval does not include zero, there is enough evidence to reject the null hypothesis that the means are equal. In the provided example, the confidence interval was calculated as ( -4.00, 0.80 ). Since zero is not in this interval, it supports the conclusion to reject H鈧.
Sample Size Determination
Determining the correct sample size is vital in hypothesis testing to ensure the test has enough power to detect a true effect. Power is the probability of correctly rejecting the null hypothesis when it is false. The desired power (typically 0.80 or 80%) and significance level (伪) influence the sample size needed.
To calculate the sample size, we use the formula:
\[n = \frac{(z_{\alpha/2} + z_{\beta})^2 \times (s_p^2)}{(\mu_1 - \mu_2)^2}\]
  • Here, z_{\alpha/2} and z_{\beta} are values from the standard normal distribution corresponding to the desired confidence level and the power of the test.
  • s_p is the pooled standard deviation, and (\mu_1 - \mu_2) is the anticipated effect size.
Using the given problem's context, the calculation indicated that approximately 20 samples per group were sufficient to achieve a 尾 of 0.05 (5% probability of a type II error). This process ensures that the data collected is enough to make reliable conclusions about the population from which the samples are drawn.

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Most popular questions from this chapter

The manufacturer of a hot tub is interested in testing two different heating elements for its product. The element that produces the maximum heat gain after 15 minutes would be preferable. The manufacturer obtains 10 samples of each heating unit and tests each one. The heat gain after 15 minutes (in \({ }^{\circ} \mathrm{F}\) ) follows. $$\begin{array}{l|l}\text { Unit 1: } & 25,27,29,31,30,26,24,32,33,38 \\\\\hline \text { Unit 2: } & 31,33,32,35,34,29,38,35,37,30\end{array}$$ (a) Is there any reason to suspect that one unit is superior to the other? Use \(\alpha=0.05\) and the Wilcoxon rank sum test. (b) Use the normal approximation for the Wilcoxon rank sum test. Assume that \(\alpha=0.05 .\) What is the approximate \(P\) -value for this test statistic?

Go Tutorial In a random sample of 200 Phoenix residents who drive a domestic car, 165 reported wearing their seat belt regularly, and another sample of 250 Phoenix residents who drive a foreign car revealed 198 who regularly wore their seat belt. (a) Perform a hypothesis-testing procedure to determine whether there is a statistically significant difference in seat belt usage for domestic and foreign car drivers. Set your probability of a type I error to 0.05 . (b) Perform a hypothesis-testing procedure to determine whether there is a statistically significant difference in seat belt usage

The thickness of a plastic film (in mils) on a substrate material is thought to be influenced by the temperature at which the coating is applied. In completely randomized experiment, 11 substrates are coated at \(125^{\circ} \mathrm{F}\), resulting in a sample mean coating thickness of \(\bar{x}_{1}=103.5\) and a sample standard deviation of \(s_{1}=10.2\). Another 13 substrates are coated at \(150^{\circ} \mathrm{F}\) for which \(\bar{x}_{2}=99.7\) and \(s_{2}=20.1\) are observed. It was originally suspected that raising the process temperature would reduce mean coating thickness. (a) Do the data support this claim? Use \(\alpha=0.01\) and assume that the two population standard deviations are not equal. Calculate an approximate \(P\) -value for this test. (b) How could you have answered the question posed regarding the effect of temperature on coating thickness by using a confidence interval? Explain your answer.

An article in IEEE International Symposium on Electromagnetic Compatibility (2002, Vol. 2, pp. \(667-670\) ) quantified the absorption of electromagnetic energy and the resulting thermal effect from cellular phones. The experimental results were obtained from in vivo experiments conducted on rats. The arterial blood pressure values \((\mathrm{mmHg})\) for the control group (8 rats) during the experiment are \(\bar{x}_{1}=90, s_{1}=5\) and for the test group (9 rats) are \(\bar{x}_{2}=115, s_{2}=10 .\) (a) Is there evidence to support the claim that the test group has higher mean blood pressure? Use \(\alpha=0.05,\) and assume that both populations are normally distributed but the variances are not equal. What is the \(P\) -value for this test? (b) Calculate a confidence interval to answer the question in part (a). (c) Do the data support the claim that the mean blood pressure from the test group is at least \(15 \mathrm{mmHg}\) higher than the control group? Make the same assumptions as in these part (a). (d) Explain how the question in part (c) could be answered with a confidence interval.

A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed of the film and believes that this can be achieved by reducing the thickness of the film to 20 mils. Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured. For the 25 -mil film, the sample data result is \(\bar{x}_{1}=1.15\) and \(s_{1}=0.11,\) and for the 20 -mil film the data yield \(\bar{x}_{2}=1.06\) and \(s_{2}=0.09 .\) Note that an increase in film speed would lower the value of the observation in microjoules per square inch. (a) Do the data support the claim that reducing the film thickness increases the mean speed of the film? Use \(\sigma=0.10\), and assume that the two population variances are equal and the underlying population of film speed is normally distributed. What is the \(P\) -value for this test? (b) Find a \(95 \%\) confidence interval on the difference in the two means that can be used to test the claim in part (a).

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