/*! 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 53 Who talks more鈥攎en or women? R... [FREE SOLUTION] | 91影视

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Who talks more鈥攎en or women? Researchers equipped random samples of 56 male and 56 female students from a large university with a small device that secretly records sound for a random 30 seconds during each 12.5-minute period over two days. Then they counted the number of words spoken by each subject during each recording period and, from this, estimated how many words per day each subject speaks. The female estimates had a mean of 16,177 words per day with a standard deviation of 7520 words per day. For the male estimates, the mean was 16,569 and the standard deviation was 9108. (a) Do these data provide convincing evidence of a difference in the average number of words spoken in a day by male and female students at this university? Carry out an appropriate test to support your answer. (b) Interpret the P-value from part (a) in the context of this study.

Short Answer

Expert verified
Perform a two-sample t-test to compare the means. Interpret the p-value based on your significance level.

Step by step solution

01

Define the Hypotheses

The first step in hypothesis testing is to define the null and alternative hypotheses. For this problem:- Null Hypothesis (\(H_0\)): There is no difference in the average number of words spoken per day between males and females, i.e., \( \mu_1 = \mu_2 \).- Alternative Hypothesis (\(H_a\)): There is a difference in the average number of words spoken per day between males and females, i.e., \( \mu_1 eq \mu_2 \).
02

Choose the Right Test

Since we are comparing the means from two independent samples (male and female students) and we have the standard deviations, we will use a two-sample t-test for means.
03

Calculate the Test Statistic

The test statistic for a two-sample t-test is given by:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]where:- \( \bar{x}_1 = 16,177 \) is the mean for females,- \( \bar{x}_2 = 16,569 \) is the mean for males,- \( s_1 = 7,520 \) is the standard deviation for females,- \( s_2 = 9,108 \) is the standard deviation for males,- \( n_1 = 56 \) and \( n_2 = 56 \) are the sample sizes.Substitute these values into the formula to calculate \( t \).
04

Compute the Degrees of Freedom

The degrees of freedom for a two-sample t-test can be calculated using the formula:\[df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}\]This value can be used appropriately to find the critical value and p-value from t-distribution tables or software.
05

Determine the P-value

Using the calculated test statistic and degrees of freedom, determine the p-value. This can be done using statistical software or t-distribution tables.
06

Make a Decision

Compare the p-value to the significance level (often \( \alpha = 0.05 \)). If the p-value is less than \( \alpha \), reject the null hypothesis. Otherwise, fail to reject the null hypothesis.
07

P-value Interpretation

In the context of this study, the p-value indicates the probability of observing a difference in the word counts as extreme as what was observed, if there was actually no real difference between males and females in terms of average words spoken.

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Key Concepts

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

Two-Sample T-Test
When researchers want to compare the means of two separate groups, such as males and females, a two-sample t-test is a helpful tool. This test examines whether there is a statistically significant difference between the two group means. In our exercise, we're looking at the average number of words spoken per day by male and female students. Assume we have two sets of estimates鈥攐ne that includes the mean and standard deviation of words spoken by females, and similarly for males.

The two-sample t-test takes into account the difference between these means, as well as the variability (standard deviations) and sample sizes of both groups. By computing a test statistic from this information, we can then determine the likelihood that this observed difference might have happened by random chance if, in reality, there was no difference in word count between the two groups.
P-Value Interpretation
The p-value plays a critical role in hypothesis testing. It helps us understand the probability of finding an effect as extreme or more extreme than the one in our sample data if the null hypothesis is actually true. In simpler terms, it tells us how "surprised" we should be to see our data if there's really no difference between the groups being compared.

In the word count study, after calculating the test statistic and obtaining a p-value, we interpret this value within the context of our significance level (commonly set at 0.05). A small p-value (typically 鈮 0.05) indicates that the observed difference in word counts between males and females is unlikely to have occurred under the null hypothesis鈥攕uggesting that a real difference may exist. Conversely, a p-value larger than 0.05 suggests insufficient evidence to claim a difference.
Statistical Significance
Statistical significance refers to the likelihood that a result from data analysis is not due to random chance. When performing a hypothesis test, we often set a significance level, denoted by \( \alpha \), such as 0.05, which serves as a threshold for making decisions.

In the context of our study, if the p-value calculated from the t-test is less than \( \alpha = 0.05 \), we say the result is statistically significant. This means the evidence is strong enough to conclude that there is a meaningful difference in average words spoken by males and females. However, it is essential to remember that statistical significance does not measure the size of the effect or its practical importance鈥攐nly that an effect exists.
Null and Alternative Hypotheses
In hypothesis testing, we start by formulating two competing hypotheses: the null hypothesis \( (H_0) \) and the alternative hypothesis \( (H_a) \). The null hypothesis is generally a statement of no effect or no difference鈥攊n this case, it suggests that there is no difference in the average number of words spoken per day between males and females.

Conversely, the alternative hypothesis posits that a difference does exist between the groups. Here, it states that the averages are not equal. After defining these hypotheses, statistical tests are conducted to determine which is more likely to be true given the data. The conclusion drawn provides insights on whether observed differences, if any, are statistically significant and not just a product of random sampling variability.

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Most popular questions from this chapter

Information online \((8.2,10.1)\) A random digit dialing sample of 2092 adults found that 1318 used the Internet. \(^{45}\) Of the users, 1041 said that they expect businesses to have Web sites that give product information; 294 of the 774 nonusers said this. (a) Construct and interpret a 95% confidence interval for the proportion of all adults who use the Internet. (b) Construct and interpret a 95% confidence interval to compare the proportions of users and nonusers who expect businesses to have Web sites.

Did the random assignment work? A large clinical trial of the effect of diet on breast cancer assigned women at random to either a normal diet or a low-fat diet. To check that the random assignment did produce comparable groups, we can compare the two groups at the start of the study. Ask if there is a family history of breast cancer: 3396 of the \(19,541\) women in the low-fat group and 4929 of the \(29,294\) women in the control group said "Yes."\(^{15}\) If the random assignment worked well, there should not be a significant difference in the proportions with a family history of breast cancer. (a) How significant is the observed difference? Carry out an appropriate test to help answer this question. (b) Describe a Type I and a Type II error in this setting. Which is more serious? Explain.

Computer gaming Do experienced computer game players earn higher scores when they play with someone present to cheer them on or when they play alone? Fifty teenagers who are experienced at playing a particular computer game have volunteered for a study. We randomly assign 25 of them to play the game alone and the other 25 to play the game with a supporter present. Each player鈥檚 score is recorded. (a) Is this a problem about comparing means or comparing proportions? Explain. (b) What type of study design is being used to produce data?

Fear of crime The elderly fear crime more than younger people, even though they are less likely to be victims of crime. One study recruited separate random samples of 56 black women and 63 black men over the age of 65 from Atlantic City, New Jersey. Of the women, 27 said they 鈥渇elt vulnerable鈥 to crime; 46 of the men said this.\(^{12}\) (a) Construct and interpret a 90% confidence interval for the difference in population proportions (men minus women). (b) Does your interval from part (a) give convincing evidence of a difference between the population proportions? Explain.

What鈥檚 wrong? A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state鈥檚 driver鈥檚 license exam. An incoming class of 100 students is randomly assigned to two groups, each of size 50. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, 30 of Instructor A鈥檚 students and 22 of Instructor B鈥檚 students pass the state exam. Do these results give convincing evidence that Instructor A is more effective? Min Jae carried out the significance test shown below to answer this question. Unfortunately, he made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one. State: I want to perform a test of $$H_{0} : p_{1}-p_{2}=0$$ $$H_{a} : p_{1}-p_{2}>0$$ where \(p_{1}=\) the proportion of Instructor A's students that passed the state exam and \(p_{2}=\) the proportion of Instructor B's students that passed the state exam. Since no significance level was stated, I'll use \(\sigma=0.05\) Plan: If conditions are met, I'll do a two-sample \(z\) test for comparing two proportions. \(\bullet\) Random The data came from two random samples of 50 students. \(\bullet\) Normal The counts of successes and failures in the two groups - \(30,20,22\) , and \(28-\) are all at least \(10 .\) \(\bullet\) Independent There are at least 1000 students who take this driving school's class. Do: From the data, \(\hat{p}_{1}=\frac{20}{50}=0.40\) and \(\hat{p}_{2}=\frac{30}{50}=0.60 .\) So the pooled proportion of successes is $$\hat{p}_{C}=\frac{22+30}{50+50}=0.52$$ \(\bullet\) Test statistic $$z=\frac{(0.40-0.60)-0}{\sqrt{\frac{0.52(0.48)}{100}+\frac{0.52(0.48)}{100}}}=-2.83$$ Conclude: The P-value, \(0.9977,\) is greater than \(\alpha=\) \(0.05,\) so we fail to reject the null hypothesis. There is not convincing evidence that Instructor A's pass rate is higher than Instructor B's.

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