/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 One could reason that high schoo... [FREE SOLUTION] | 91Ó°ÊÓ

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One could reason that high school seniors would have more money issues than high school juniors. Seniors foresee expenses for college as well as their senior trip and prom. So does this mean that they work more than their junior classmates? Christine, a senior at HFL High School, randomly collected the following data (recorded in hours/week) from students that work: $$\begin{array}{llll}\hline \text { Seniors } & n_{s}=17 & \bar{x}_{s}=16.4 & s_{s}=10.48 \\\\\text { Juniors } & n_{i}=20 & \bar{x}_{i}=18.405 & s_{i}=9.69 \\\\\hline\end{array}$$ Assuming that work hours are normally distributed, do these data suggest that there is a significant difference between the average number of hours that HFL seniors and juniors work per week? Use \(\alpha=0.10\).

Short Answer

Expert verified
The short answer will depend on the values obtained for t and P-value, and how they compare with the significance level of 0.10. You will either reject or fail to reject the null hypothesis, and conclude accordingly whether or not there is a statistically significant difference between mean working hours of high school seniors and juniors.

Step by step solution

01

State the Null and Alternative Hypotheses

The null hypothesis \(H_0\) states that the mean working hours of juniors and seniors are equal. It can be written as: \(H_0 : \mu_s = \mu_j\), where \(\mu_s\) and \(\mu_j\) are the mean working hours of seniors and juniors respectively. The alternative hypothesis \(H_a\) states that the mean working hours are not equal. It can be written as: \(H_a : \mu_s \neq \mu_j\).
02

Compute the Test Statistic

In this two-sample t-test, when the sample size \(n > 30\), the t-score is calculated as \[ t = \frac{\bar{x}_s - \bar{x}_j}{\sqrt{\frac{s_s^2}{n_s} + \frac{s_j^2}{n_j}}}\], where \(\bar{x}_s\) and \(\bar{x}_j\) are the sample means, \(s_s\) and \(s_j\) are the sample standard deviations, and \(n_s\) and \(n_j\) are the sample sizes. Substituting the given values into this equation, we obtain \(t = \frac{16.4 - 18.405}{\sqrt{\frac{(10.48)^2}{17} + \frac{(9.69)^2}{20}}}\). Find the resulting value.
03

Find the P-value

Find the two-tailed P-value associated with the calculated t-value by looking it up in the t-distribution table or using statistical software. Remember, because we have a two-sided test, the P-value is twice the area that is beyond this t-value. Use the appropriate degrees of freedom which is minimum of \(n_s - 1 = 17 - 1 = 16\) and \(n_j - 1 = 20 - 1 = 19\), so \(\text{df} = 16\).
04

Make a Decision and Interpret the Results

Compare the P-value with the significance level \(\alpha = 0.10\). If the P-value is equal to or less than \(\alpha\), reject the null hypothesis, which would suggest that there is statistically significant evidence to say that the mean hours worked per week by seniors and juniors are different. If the P-value is greater than \(\alpha\), do not reject the null hypothesis, which would suggest the difference is not statistically significant. Summarize the results in the context of the problem.

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Key Concepts

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

Two-Sample t-Test
The two-sample t-test is a statistical method used to determine if there is a significant difference between the means of two independent groups. In our high school example, this test helps us analyze whether the average working hours of seniors and juniors are different.

We use the sample data: means, standard deviations, and sizes from each group to compute a test statistic called the t-score. The formula for this is given by:
  • \[ t = \frac{\bar{x}_s - \bar{x}_i}{\sqrt{\frac{s_s^2}{n_s} + \frac{s_j^2}{n_j}}} \]
Here's what everything represents:
  • \(\bar{x}_s\) and \(\bar{x}_i\) are the means of seniors and juniors respectively.
  • \(s_s\) and \(s_i\) are their respective standard deviations.
  • \(n_s\) and \(n_i\) are their sample sizes.
Once the t-score is calculated, it is compared against a t-distribution to determine the likelihood of observing such a score under the null hypothesis. This process is crucial when sample sizes are small or the standard deviations of groups could be unequal.

Understanding this concept is important as it allows us to make informed decisions in daily research and statistics.
Significance Level (Alpha)
Significance level, often denoted as \( \alpha \), is a critical component in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true. In simpler terms, it helps determine the line we draw between observing a rare event versus a true effect.

In the given exercise, the significance level is set to \( \alpha = 0.10 \). This means that we are allowing a 10% chance of falsely concluding that there is a difference in the mean working hours between seniors and juniors.

Significance levels are chosen arbitrarily, although common levels are 0.05 and 0.01. The level you choose can affect how "convinced" we are by the data: lower significance levels mean stricter evidence requirements. Here is a quick guide:
  • Higher \( \alpha \): Acknowledges wider margin for error, more false positives.
  • Lower \( \alpha \): Requires stronger evidence to reject the null hypothesis.
In practical terms, understanding the chosen \( \alpha \) level helps in interpreting whether results are significant or if repeated testing would tend to yield similar conclusions.
Null and Alternative Hypotheses
Hypotheses are foundational to any statistical test. They define the framework of what you're trying to explore or conclude. In hypothesis testing, we usually construct two different hypotheses: the null hypothesis \(H_0\) and the alternative hypothesis \(H_a\).

The null hypothesis (\(H_0\)) suggests no effect or no difference. It serves as a starting point and holds that any observed differences arise due to sampling variability. For our exercise, the null hypothesis is:
  • \(H_0 : \mu_s = \mu_j \) — meaning the average hours worked by seniors and juniors are equal.

The alternative hypothesis (\(H_a\)) proposes that there is a true effect, and the differences are not due to random chance. For the high school example, it is:
  • \(H_a : \mu_s eq \mu_j \) — meaning the average hours worked by seniors and juniors are not equal.

These hypotheses create the basis for statistical analysis, supplying a structured approach to determine if the observed data supports a specific claim. Testing these can lead to conclusions drawn about populations that are otherwise impractical to observe entirely. By setting clear hypotheses, researchers can drive data-driven decisions with robust confidence.

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Most popular questions from this chapter

Ten recently diagnosed diabetics were tested to determine whether an educational program would be effective in increasing their knowledge of diabetes. They were given a test, before and after the educational program, concerning self-care aspects of diabetes. The scores on the test were as follows: $$\begin{array}{lcccccccccc}\text { Patient } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \text { Before } & 75 & 62 & 67 & 70 & 55 & 59 & 60 & 64 & 72 & 59 \\\\\text { Afler } & 77 & 65 & 68 & 72 & 62 & 61 & 60 & 67 & 75 & 68 \\\\\hline\end{array}$$The following MINITAB output may be used to determine whether the scores improved as a result of the program. Verify the values shown on the output [mean difference(MEAN), standard deviation of the difference (STDEV), standard error of the difference (SE MEAN), \(t \star\) (T-Value), and \(p\) -value] by calculating the values yourself.

When a hypothesis test is two-tailed and Excel is used to calculate the \(p\) -value, what additional step must be taken?

State the null hypothesis, \(H_{o}\), and the alternative hypothesis, \(H_{a},\) that would be used to test the following claims: a. The variances of populations \(A\) and \(B\) are not equal. b. The standard deviation of population I is larger than the standard deviation of population II. c. The ratio of the variances for populations \(A\) and \(B\) is different from 1. d. The variability within population \(\mathrm{C}\) is less than the variability within population D.

Is having a long, more complex first name more dignified for a girl? Are girls' names longer than boys' names? With current names like "Alexandra," "Madeleine," and "Savannah," it certainly appears so. To test this theory,random samples of seventh-grade girls and boys were taken. Let \(x\) be the number of letters in each seventh grader's first name.$$\begin{array}{llll}\hline \text { Boys' Names } & n=30 & \bar{x}=5.767 & s=1.870 \\\\\text { Girls' Names } & n=30 & \bar{x}=6.133 & s=1.456 \\\\\hline\end{array}$$ At the 0.05 level of significance, do the data support the contention that the mean length of girls' names is longer than the mean length of boys' names?

A research project was undertaken to evaluate the amount of force needed to elicit a designated response on equipment made from two distinct designs: the existing design and an improved design. The expectation was that new design equipment would require less force than the current equipment. Fifty units of each design were tested and the required force recorded. A lower force level and reduced variability are both considered desirable. a. Describe both sets of data using means, standard deviations, and histograms. b. Check the assumptions for comparing the variances and means of two independent samples. Describe your findings. c. Will one-or two-tailed tests be appropriate for testing the expectations for the new design? Why? d. Is there significant evidence to show that the new design has reduced the variability in the required force? Use \(\alpha=0.05\) e. Is there significant evidence to show that the new design has reduced the mean amount of force? Use \(\alpha=0.05\) f. Did the new design live up to expectations? Explain.

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