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Independent random samples of 36 and 45 observations are drawn from two quantitative populations, 1 and \(2,\) respectively. The sample data summary is shown here: $$\begin{array}{lcc} & \text { Sample 1 } & \text { Sample 2 } \\\\\hline \text { Sample Size } & 36 & 45 \\\\\text { Sample Mean } & 1.24 & 1.31 \\\\\text { SampleVariance } & .0560 & .0540\end{array}$$ Do the data present sufficient evidence to indicate that the mean for population 1 is smaller than the mean for population 2 ? Use one of the two methods of testing presented in this section, and explain your conclusions

Short Answer

Expert verified
In an exercise with two independent random samples of size 36 and 45 from two quantitative populations, the goal is to perform a hypothesis test to determine if the population means are different. The steps involved in a two-sample t-test include stating the null and alternative hypotheses, computing the test statistic, calculating the degrees of freedom, determining the critical t-value, and making a decision on the null hypothesis. Use a t-distribution table or an online calculator to help with the calculations.

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis is that the population means are equal, i.e., \(\mu_1 = \mu_2\). The alternative hypothesis is that the population means are not equal, i.e., \(\mu_1 \neq \mu_2\).
02

Compute the test statistic

We'll calculate the test statistic for a two-sample t-test, which is given by: \(t = \frac{(\overline{x}_1 - \overline{x}_2)}{\sqrt{\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}}}\), where \(\overline{x}_1\), \(s^2_1\), and \(n_1\) are the sample mean, sample variance, and sample size of population 1, respectively. Similarly, \(\overline{x}_2\), \(s^2_2\), and \(n_2\) are the sample mean, sample variance, and sample size of population 2, respectively. Using the given data, \(t = \frac{(1.24 - 1.31)}{\sqrt{\frac{0.0560}{36} + \frac{0.0540}{45}}}\).
03

Calculate the degrees of freedom

The degrees of freedom can be calculated using the following formula for two-sample t-test: $$\text{df} = \frac{(\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2})^2}{\frac{(\frac{s^2_1}{n_1})^2}{n_1 - 1} + \frac{(\frac{s^2_2}{n_2})^2}{n_2 - 1}}$$ Using the given data, \(\text{df} = \frac{(\frac{0.0560}{36} + \frac{0.0540}{45})^2}{\frac{(\frac{0.0560}{36})^2}{36 - 1} + \frac{(\frac{0.0540}{45})^2}{45 - 1}}\).
04

Determine the critical t-value

Using a standard t-distribution table or an online calculator, we can determine the critical t-value corresponding to our chosen significance level (e.g., \(\alpha = 0.05\)) and the degrees of freedom calculated in step 3.
05

Make a decision on the null hypothesis

Compare the calculated test statistic (from step 2) with the critical t-value (from step 4). If the absolute value of the test statistic is greater than the critical t-value, we reject the null hypothesis, which means there's evidence to support that the population means are not equal. If the absolute value of the test statistic is less than or equal to the critical t-value, we fail to reject the null hypothesis, which means there's not enough evidence to support that the population means are not equal. Remember to use a t-distribution table or an online calculator to help you with calculations in steps 2 and 3.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a method used in statistics to decide if there is enough evidence to reject a null hypothesis. It is a critical tool to determine if the observed data can disprove an initial assumption or belief about a population parameter. In this exercise, our null hypothesis states that the two population means are equal: \(\mu_1 = \mu_2\). The alternative hypothesis, which is what we test against, is that the means are not equal: \(\mu_1 eq \mu_2\).

The two-sample t-test is often used when comparing the means from two different groups to determine if they differ significantly. By calculating a test statistic and comparing it to a critical value, we can make conclusions about our hypotheses.

  • The **test statistic** is calculated based on the sample data.
  • The **null hypothesis** is rejected if the test statistic is larger in magnitude than the critical value.
  • If the test statistic is within the range of the critical value, the null hypothesis is not rejected.
Understanding how to perform a hypothesis test is crucial for drawing meaningful conclusions from data, especially in fields such as business, healthcare, and social sciences.
Degrees of Freedom
Degrees of Freedom (df) refer to the number of independent values or quantities that can vary in an analysis without breaking any constraints. They are crucial in many statistical calculations, including t-tests, where they affect the distribution of the test statistic.

In a two-sample t-test, the degrees of freedom combine the sample sizes and variances from both groups. Calculating the degrees of freedom precisely can be complex. It is generally determined using the formula involving variances and sample sizes:

\[\text{df} = \frac{\left(\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}\right)^2}{\frac{(\frac{s^2_1}{n_1})^2}{n_1 - 1} + \frac{(\frac{s^2_2}{n_2})^2}{n_2 - 1}}\]

Where \(s_1^2\) and \(s_2^2\) are the sample variances, and \(n_1\) and \(n_2\) are the sample sizes for the two populations, respectively.

  • A higher number of degrees of freedom often leads to a more reliable test result.
  • As the degrees of freedom increase, the t-distribution becomes more similar to a normal distribution.
Ensure you use the correct formula and double-check calculations, as an incorrect value can lead to flawed conclusions.
Significance Level
The significance level, denoted by \(\alpha\), is the probability of rejecting the null hypothesis when it is actually true. It represents the threshold for how much risk you are willing to take of making a Type I error. In practice, this is usually set at common levels such as 0.05, 0.01, or 0.10, which correspond to 5%, 1%, or 10% risk respectively.

Setting the right significance level is crucial because it directly influences the decision-making process in hypothesis testing.

  • **Lower \(\alpha\) values (e.g., 0.01):** Stricter criteria, reducing the chance of a Type I error, but increasing the risk of a Type II error.
  • **Higher \(\alpha\) values (e.g., 0.10):** Relaxed criteria, increasing the chance of a Type I error.
In the context of this exercise, choosing a significance level like 0.05 means that we are accepting a 5% risk that our conclusion that the populations means are not equal is incorrect.

Understanding significance levels helps you interpret the results of a hypothesis test properly and make informed decisions based on data.

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Most popular questions from this chapter

PCBs Polychlorinated biphenyls (PCBs) have been found to be dangerously high in some game birds found along the marshlands of the southeastern coast of the United States. The Federal Drug Administration (FDA) considers a concentration of PCBs higher than 5 parts per million (ppm) in these game birds to be dangerous for human consumption. A sample of 38 game birds produced an average of 7.2 ppm with a standard deviation of \(6.2 \mathrm{ppm}\). Is there sufficient evidence to indicate that the mean ppm of \(\mathrm{PCBs}\) in the population of game birds exceeds the FDA's recommended limit of 5 ppm? Use \(\alpha=.01\)

A random sample of \(n=1400\) observations from a binomial population produced \(x=529\) a. If your research hypothesis is that \(p\) differs from .4 what hypotheses should you test? b. Calculate the test statistic and its \(p\) -value. Use the \(p\) -value to evaluate the statistical significance of the results at the \(1 \%\) level. c. Do the data provide sufficient evidence to indicate that \(p\) is different from \(.4 ?\)

Suppose you wish to detect a difference between \(\mu_{1}\) and \(\mu_{2}\) (either \(\mu_{1}>\mu_{2}\) or \(\mu_{1}<\mu_{2}\) ) and, instead of running a two-tailed test using \(\alpha=.05,\) you use the following test procedure. You wait until you have collected the sample data and have calculated \(\bar{x}_{1}\) and \(\bar{x}_{2}\) If \(\bar{x}_{1}\) is larger than \(\bar{x}_{2},\) you choose the alternative hypothesis \(H_{\mathrm{a}}: \mu_{1}>\mu_{2}\) and run a one-tailed test placing \(\alpha_{1}=.05\) in the upper tail of the \(z\) distribution. If, on the other hand, \(\bar{x}_{2}\) is larger than \(\bar{x}_{1},\) you reverse the procedure and run a one-tailed test, placing \(\alpha_{2}=.05\) in the lower tail of the \(z\) distribution. If you use this procedure and if \(\mu_{1}\) actually equals \(\mu_{2},\) what is the probability \(\alpha\) that you will conclude that \(\mu_{1}\) is not equal to \(\mu_{2}\) (i.e., what is the probability \(\alpha\) that you will incorrectly reject \(H_{0}\) when \(H_{0}\) is true)? This exercise demonstrates why statistical tests should be formulated prior to observing the data.

Female Models In a study to assess various effects of using a female model in automobile advertising, 100 men were shown photographs of two automobiles matched for price, color, and size, but of different makes. One of the automobiles was shown with a female model to 50 of the men (group \(\mathrm{A}\) ), and both automobiles were shown without the model to th other 50 men (group B). In group A, the automobile shown with the model was judged as more expensive by 37 men; in group \(B\), the same automobile was judged as the more expensive by 23 men. Do these results indicate that using a female model influences the perceived cost of an automobile? Use a one-tailec test with \(\alpha=.05 .\)

A random sample of \(n=1000\) observations from a binomial population produced \(x=279 .\) a. If your research hypothesis is that \(p\) is less than .3 , what should you choose for your alternative hypothesis? Your null hypothesis? b. What is the critical value that determines the rejection region for your test with \(\alpha=.05 ?\) c. Do the data provide sufficient evidence to indicate that \(p\) is less than .3 ? Use a \(5 \%\) significance level.

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