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Chapter 1: Problem 105

SSHA scores Here are the scores on the Survey of Study Habits and Attitudes (SSHA) for 18 first-year college women: \(\begin{array}{rrrrrrrrrr}{154} & {109} & {137} & {115} & {152} & {140} & {154} & {178} & {101} \\ {103} & {126} & {126} & {137} & {165} & {165} & {129} & {200} & {148} \\ \hline\end{array}\) and for 20 first-year college men: \(\begin{array}{rrrrrr}{108} & {140} & {114} & {91} & {180} & {115} & {126} \\\ {92} & {169} & {146} & {109} & {132} & {75} & {88} \\ {113} & {151} & {70} & {115} & {187} & {104} \\ \hline\end{array}\) Do these data support the belief that women have better study habits and attitudes toward learning than men? (Note that high scores indicate good study habits and attitudes toward learning.) Follow the four-step process.

Short Answer

Expert verified
Women have better study habits than men according to this data, assuming the t-test rejects the null hypothesis.

Step by step solution

01

Define the Hypotheses

We need to set up our null and alternative hypotheses. The null hypothesis (H鈧�) is that the mean scores for women and men are equal (i.e., no difference in study habits and attitudes), while the alternative hypothesis (H鈧�) is that the mean score for women is higher than that for men:\[H_0: \mu_w = \mu_m\]\[H_1: \mu_w > \mu_m\]where \(\mu_w\) is the population mean score for women, and \(\mu_m\) is the population mean score for men.
02

Calculate Sample Statistics

To test these hypotheses, first calculate the sample mean and standard deviation for both groups. For women:Mean \( \bar{x}_w = \frac{154 + 109 + 137 + 115 + 152 + 140 + 154 + 178 + 101 + 103 + 126 + 126 + 137 + 165 + 165 + 129 + 200 + 148}{18} \approx 140.94 \)For men:Mean \( \bar{x}_m = \frac{108 + 140 + 114 + 91 + 180 + 115 + 126 + 92 + 169 + 146 + 109 + 132 + 75 + 88 + 113 + 151 + 70 + 115 + 187 + 104}{20} \approx 119.55 \) Calculate the standard deviations as well (we assume they are needed for further calculations).
03

Perform the Test

Since we have independent samples and possibly unequal variances, use a two-sample t-test for the means. Calculate the test statistic using the formula:\[ t = \frac{\bar{x}_w - \bar{x}_m}{\sqrt{\frac{s_w^2}{n_w} + \frac{s_m^2}{n_m}}} \]where \(s_w\) and \(s_m\) are the sample standard deviations for women and men respectively, and \(n_w\) and \(n_m\) are the sample sizes.Use a t-distribution table or calculator to find the critical value at a significance level (often \(\alpha = 0.05\)). If the calculated t-value is greater than the critical value, reject the null hypothesis.
04

Conclusion

Compare the t-statistic with the critical value. If the t-statistic is greater, we conclude that there is significant evidence to reject the null hypothesis in favor of the alternative hypothesis. This would suggest that women indeed have better study habits and attitudes towards learning than men, according to the given data. If the t-statistic is not greater, we fail to reject the null hypothesis, meaning we do not have enough evidence to support the claim that women have better study habits than men.

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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 whether there is a significant difference between the means of two independent groups. In the context of the exercise, we're comparing the average study habits and attitudes of first-year college women to those of men.

Here's how it works:
  • We collect a sample from each group (in this case, scores of women and men).
  • Calculate the means and standard deviations for each sample.
  • Use the two-sample t-test formula to compute the test statistic, which measures the size of the difference relative to the variation in our data.
The assumption here is that both samples come from populations that follow a normal distribution, and they should ideally have equal variances. If variances differ significantly, we use a modified version of the test known as Welch's t-test. With our calculated t-statistic, we compare it against a critical value from the t-distribution table to determine whether the difference between group means is statistically significant.
Null Hypothesis
The null hypothesis is a foundational concept in hypothesis testing, and it's critical to understand its role. It represents a statement of no effect or no difference and serves as a default or starting assumption. In our exercise, the null hypothesis (\(H_0\)) claims that there are no differences in the study habits and attitudes between the genders.

Formally, it is expressed as \(H_0: \mu_w = \mu_m\), where \(\mu_w\) and \(\mu_m\) are the population mean scores for women and men, respectively. The null hypothesis posits that any observed differences in our sample data occur by chance.

When conducting hypothesis testing, rejecting the null hypothesis is key to establishing that there is a significant difference between the groups under study.
Alternative Hypothesis
The alternative hypothesis is what researchers aim to provide evidence for in their experiments. Unlike the null hypothesis, it suggests that there is a meaningful difference or effect. In the scenario of comparing study habits between genders, the alternative hypothesis (\(H_1\)) claims that women have better study habits and attitudes than men.

Mathematically, it is noted as \(H_1: \mu_w > \mu_m\). This one-tailed hypothesis specifically suggests that the mean score of women (\(\mu_w\)) is greater than that of men (\(\mu_m\)).

If the data provides enough evidence against the null hypothesis (meaning the t-statistic is significantly large), we reject the null hypothesis in favor of the alternative hypothesis. This step is vital for concluding that such a difference in study habits likely exists.
Significance Level
The significance level, often denoted by \(\alpha\), is a critical threshold in hypothesis testing. It is the probability of rejecting the null hypothesis when it is actually true (Type I error). Commonly set at 0.05, the significance level determines how conservative or liberal a test is.

For the exercise, the significance level might be at 5% (\(0.05\)), meaning that there is a 5% risk of concluding that there is a difference in study habits between women and men when no such difference exists in the population.
  • If the p-value, calculated from the test statistics, is less than \(\alpha\), the result is statistically significant, leading us to reject the null hypothesis.
  • Conversely, if the p-value is greater than \(\alpha\), we fail to reject the null hypothesis, indicating insufficient evidence to support the alternative hypothesis.
The significance level guides decisions in statistical testing, ensuring results are reliable and not due to random chance.

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Most popular questions from this chapter

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