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Assume that the populations are normally distributed. (a) Test whether \(\mu_{1}<\mu_{2}\) at the \(\alpha=0.05\) level of significance for the given sample data. (b) Construct a \(95 \%\) confidence interval about \(\mu_{1}-\mu_{2}\). $$ \begin{array}{lcc} & \text { Sample 1 } & \text { Sample 2 } \\ \hline n & 40 & 32 \\ \hline \bar{x} & 94.2 & 115.2 \\ \hline s & 15.9 & 23.0 \\ \hline \end{array} $$

Short Answer

Expert verified
Conduct the t-test and confidence interval calculations using the given formulas. Compare the test statistic with the critical value to test the hypothesis and compute the confidence interval.

Step by step solution

01

- State Hypotheses for Part (a)

Define the null and alternative hypotheses. The null hypothesis is that there is no difference in the population means, i.e., \(H_0: \mu_1 = \mu_2\).The alternative hypothesis is that \( \mu_1 < \mu_2 \). Thus, \( H_a: \mu_1 < \mu_2 \).
02

- Calculate the Test Statistic for Part (a)

To test the hypothesis, use the formula for the test statistic for two independent means: \[ t = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] Substituting the given values: \( \bar{x}_1 = 94.2, \quad \bar{x}_2 = 115.2, \quad s_1 = 15.9, \quad s_2 = 23.0, \quad n_1 = 40, \quad n_2 = 32 \) \[ t = \frac{94.2 - 115.2}{\sqrt{\frac{(15.9)^2}{40} + \frac{(23.0)^2}{32}}} \] Compute the value.
03

- Determine Degrees of Freedom and Critical Value for Part (a)

Using the formula for degrees of freedom: \[ df = \frac{ \left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2 }{ \frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1} } \] Approximate the degrees of freedom, find the critical value of \( t \) for \( \alpha = 0.05 \) for a one-tailed test.
04

- Make a Decision for Part (a)

Compare the test statistic value with the critical value. If the test statistic is less than the critical value, reject the null hypothesis; otherwise, fail to reject the null hypothesis.
05

- Construct the Confidence Interval for Part (b)

The formula for the confidence interval about \( \mu_1 - \mu_2 \) is given by: \[ (\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2, df} \cdot \sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } \] Using the previously calculated degrees of freedom and the critical value for the \ 95 \% \ confidence level, substitute the known values.

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Key Concepts

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

Confidence Intervals
Confidence intervals provide a range of values that likely contain the population parameter. They help in estimating the unknown population parameters.

For example, a 95% confidence interval means that if we were to repeat our sample many times, 95% of the intervals calculated from those samples would contain the true population mean difference (\mu_1 - \mu_2).

Calculating a confidence interval involves:
  • Determining the sample statistic, such as the difference between means \(\bar{x}_1 - \bar{x}_2\).
  • Finding the appropriate critical value based on the desired confidence level (like \alpha=0.05 for 95\%).
  • Computing the margin of error using the standard error and the critical value.
For our exercise:
- The formula used is: \[(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2, df} \cdot \sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }\]
- Plugging in the relevant values provides the confidence interval range.

This interval gives us an estimate with high certainty about the difference between the two population means.
t-test
The t-test examines if the means of two groups are significantly different from each other. It is widely used when the population standard deviations are unknown and the sample sizes are small.

There are several types of t-tests, but for this exercise, we're using an independent samples t-test. This test compares the means from two different groups (Sample 1 and Sample 2).

The formula for the t-test statistic used in our example is:
\[ t = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]

This formula accounts for the mean difference, the variance within each sample, and the sample sizes. After calculating the t-value, you compare it with the critical t-value from t-distribution tables.

It assesses whether the observed difference in sample means significantly deviates from zero, indicating a genuine difference in population means.
Statistical Significance
Statistical significance helps us understand whether our results are likely due to chance or if they reflect a real effect in the population.

In our example, we test whether \(\mu_1 < \mu_2\) at \alpha=0.05 significance level. Here, \alpha represents the probability of rejecting the null hypothesis when it is true (Type I error).

To determine significance:
  • Set up null and alternative hypotheses (\(H_0: \mu_1 = \mu_2\) and \(H_a: \mu_1 < \mu_2\)).
  • Calculate the t-test statistic and degrees of freedom.
  • Find the critical value from the t-distribution.
  • Compare the test statistic with the critical value.
If the test statistic falls in the rejection region (less than the critical value), we reject the null hypothesis, indicating a statistically significant difference.

In our case, if the test result shows \(\mu_1 < \mu_2\) significantly, it implies the population from Sample 1 has a lower mean than Sample 2.
Degrees of Freedom
Degrees of freedom (df) are essential in hypothesis testing and confidence interval calculations. They refer to the number of independent values that can vary in an analysis.

Degrees of freedom in a t-test for two independent samples are calculated using:
\[ df = \frac{ \left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2 }{ \frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1} } \]

This accounts for sample variances and sizes, providing a more precise t-distribution critical value.

In our example, degrees of freedom helps us determine the critical t-value needed for hypothesis testing and confidence interval construction.

The higher the degrees of freedom, the closer the t-distribution resembles the normal distribution. This provides more reliable results for inference about population parameters.

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Most popular questions from this chapter

A quality-control engineer wants to find out whether or not a new machine that fills bottles with liquid has less variability than the machine currently in use. The engineer calibrates each machine to fill bottles with 16 ounces of a liquid. After running each machine for 5 hours, she randomly selects 15 filled bottles from each machine and measures their contents. She obtains the following results: $$ \begin{array}{ccc|ccc}& {\text { Old Machine }} & && {\text { New Machine }} \\ \hline 16.01 & 16.04 & 15.96 & 16.02 & 15.96 & 16.05 \\ \hline 16.00 & 16.07 & 15.89 & 15.95 & 15.99 & 16.02 \\ \hline 16.04 & 16.05 & 15.91 & 16.00 & 15.97 & 16.03 \\ \hline 16.10 & 16.01 & 16.00 & 16.06 & 16.05 & 15.94 \\ \hline 15.92 & 16.16 & 15.92 & 16.08 & 15.96 & 15.95 \\ \hline \end{array} $$ (a) Is the variability in the new machine less than that of the old machine at the \(\alpha=0.05\) level of significance? Note: Normal probability plots indicate that the data are normally distributed. (b) Draw boxplots of each data set to confirm the results of part (a) visually.

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.

In June \(2014,\) the Gallup organization surveyed 1134 American adults and found that 298 had confidence in public schools. In June \(2013,\) the Gallup organization had surveyed 1134 American adults and found that 360 had confidence in public schools. Suppose that a newspaper article has a headline that reads, "Confidence in Public Schools Deteriorates." Is this an accurate headline? Why?

Perform the appropriate hypothesis test. A random sample of size \(n=13\) obtained from a population that is normally distributed results in a sample mean of 45.3 and sample standard deviation of \(12.4 .\) An independent sample of size \(n=18\) obtained from a population that is normally distributed results in a sample mean of 52.1 and sample standard deviation of 14.7 . Does this constitute sufficient evidence to conclude that the population means differ at the \(\alpha=0.05\) level of significance?

Aluminum Bottles The aluminum bottle, first introduced in 1991 by CCL Container for mainly personal and household items such as lotions, has become popular with beverage manufacturers. Besides being lightweight and requiring less packaging, the aluminum bottle is reported to cool faster and stay cold longer than typical glass bottles. A small brewery tests this claim and obtains the following information regarding the time (in minutes) required to chill a bottle of beer from room temperature \(\left(75^{\circ} \mathrm{F}\right)\) to serving temperature \(\left(45^{\circ} \mathrm{F}\right) .\) Construct and interpret a \(90 \%\) confidence interval for the mean difference in cooling time for clear glass versus aluminum. $$ \begin{array}{lcc} & \text { Clear Glass } & \text { Aluminum } \\ \hline \text { Sample size } & 42 & 35 \\ \hline \begin{array}{l} \text { Mean time to } \\ \text { chill (minutes) } \end{array} & 133.8 & 92.4 \\ \hline \begin{array}{l} \text { Sample standard } \\ \text { deviation (minutes) } \end{array} & 9.9 & 7.3 \\ \hline \end{array} $$

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