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Chapter 11: Problem 2

Assume that the populations are normally distributed. (a) Test whether \(\mu_{1} \neq \mu_{2}\) at the \(\alpha=0.05\) level of significance for the given sample data. (b) Construct a \(95 \%\) confidence interval about \(\mu_{1}-\mu_{2}\). $$ \begin{array}{ccc} & \text { Sample } \mathbf{1} & \text { Sample } 2 \\ \hline n & 20 & 20 \\ \hline \bar{x} & 111 & 104 \\ \hline s & 8.6 & 9.2 \\ \hline \end{array} $$

Short Answer

Expert verified
Reject the null hypothesis (\( \mu_1 \eq \mu_2 \)). The 95% confidence interval for \( \mu_1 - \mu_2 \) is (1.3, 12.7).

Step by step solution

01

State the hypotheses

Formulate the null and alternative hypotheses. The null hypothesis: \( H_{0}: \mu_{1} = \mu_{2} \) and the alternative hypothesis: \( H_{1}: \mu_{1} \eq \mu_{2} \)
02

Calculate the test statistic

Use the formula for the t-test statistic: \[ t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] Here, \( \bar{x}_1 = 111 \), \( \bar{x}_2 = 104 \), \( s_1 = 8.6 \), \( s_2 = 9.2 \), and \( n_1 = n_2 = 20 \)
03

Plug in the values and solve

Plug in the given values into the t-test formula: \[ t = \frac{(111 - 104)}{\sqrt{\frac{8.6^2}{20} + \frac{9.2^2}{20}}} = \frac{7}{\sqrt{3.69 + 4.225}} = \frac{7}{\sqrt{7.915}} = \frac{7}{2.814} \approx 2.49 \]
04

Determine the critical value

Identify the critical value. Since the alternative hypothesis is two-tailed, use the t-distribution with \( df = n_1 + n_2 - 2 = 20 + 20 - 2 = 38 \) and \( \alpha = 0.05 \). The critical t-value is approximately \( t_{critical} \approx \pm 2.024 \)
05

Compare and make a decision

Compare the test statistic to the critical value. \( t = 2.49 \) is greater than \( t_{critical} \), so we reject the null hypothesis. There is sufficient evidence to conclude that \( \mu_{1} \eq \mu_{2} \)
06

Find the confidence interval

Use the confidence interval formula: \[ ( \bar{x}_1 - \bar{x}_2 ) \pm t_{critical} \sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } \] Plug in the values: \[(111 - 104) \pm 2.024 \sqrt{\frac{8.6^2}{20} + \frac{9.2^2}{20}} = 7 \pm 2.024 \sqrt{7.915} \approx 7 \pm 5.7 \] Thus, the 95% confidence interval is \[ (1.3, 12.7) \]

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Key Concepts

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

t-test
Let's start by understanding the t-test, a key concept in hypothesis testing. The t-test is used to determine if there is a significant difference between the means of two groups. It helps assess whether the observed differences are real or just due to random variations. This test is particularly useful when dealing with small sample sizes and when the population standard deviation is unknown.
The formula for the t-test statistic is:
\[ t = \frac{(\bar{x}_1 - \bar{x}_2) - (\text{difference in population means})}{\text{SE(m)}} \]
where SE(m) is the standard error of the mean difference, calculated as:
\[ \text{SE(m)} = \frac{\text{s}_1^2}{n1} + \frac{\text{s}_2^2}{n2} \]
After calculating the t-value, it is compared against a critical value from the t-distribution table to decide if we can reject the null hypothesis.
confidence interval
Next, let's look at confidence intervals. A confidence interval gives a range of values within which we expect the true population parameter to fall. It is associated with a confidence level, usually expressed as 95%, meaning we are 95% confident that the interval contains the true mean difference.
To construct a 95% confidence interval for the difference between two means consider the following formula:
\[ ( \bar{x}_1 - \bar{x}_2 ) \text{ 卤 } t_{critical} \times SE(m) \]
In our step-by-step solution, we used the sample data to calculate the margin of error and then added and subtracted this from the difference in sample means to find the interval.
sample mean comparison
When comparing sample means, it's crucial to understand the role of variability within your data. Differences in the sample means can suggest differences in the population means, but we must determine whether these differences are statistically significant.
Using the t-test and confidence intervals, we can examine if the difference in sample means is likely due to random chance. The larger the t-value and the narrower the confidence interval, the more confident we are that the sample mean differences reflect a true difference in the populations being compared.
In our example, the difference between Sample 1 and Sample 2 was tested and statistically analyzed, revealing that the observed difference is significant at the 0.05 level.

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Most popular questions from this chapter

Stock fund managers are investment professionals who decide which stocks should be part of a portfolio. In a recent article in the Wall Street Journal ( \(N\) ot \(a\) Stock-Picker's Market, WSJ January 25,2014 ), the performance of stock fund managers was considered based on dispersion in the market. In the stock market, risk is measured by the standard deviation rate of return of stock (dispersion). When dispersion is low, then the rate of return of the stocks that make up the market are not as spread out. That is, the return on Company \(X\) is close to that of \(Y\) is close to that of \(Z,\) and so on. When dispersion is high, then the rate of return of stocks is more spread out; meaning some stocks outperform others by a substantial amount. Since \(1991,\) the dispersion of stocks has been about \(7.1 \% .\) In some years, the dispersion is higher (such as 2001 when dispersion was \(10 \%)\), and in some years it is lower (such as 2013 when dispersion was \(5 \%)\). So, in 2001 , stock fund managers would argue, one needed to have more investment advice in order to identify the stock market winners, whereas in 2013 , since dispersion was low, virtually all stocks ended up with returns near the mean, so investment advice was not as valuable. (a) Suppose you want to design a study to determine whether the proportion of fund managers who outperform the market in low-dispersion years is less than the proportion of fund managers who outperform the market in high-dispersion years. What would be the response variable in this study? What is the explanatory variable in this study? (b) What or who are the individuals in this study? (c) To what population does this study apply? (d) What would be the null and alternative hypothesis? (e) Suppose this study was conducted and the data yielded a \(P\) -value of \(0.083 .\) Explain what this result suggests.

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