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Surfers Surfers and statistics students Rex Robinson and Sandy Hudson collected data on the number of days on which surfers surfed in the last month for 30 longboard (L) users and 30 shortboard (S) users. Treat these data as though they were from two independent random samples. Test the hypothesis that the mean days surfed for all longboarders is larger than the mean days surfed for all shortboarders (because longboards can go out in many different surfing conditions). Use a level of significance of \(0.05\). Longboard: \(4,9,8,4,8,8,7,9,6,7,10,12,12,10,14,12,15,13\), \(10,11,19,19,14,11,16,19,20,22,20,22\) Shortboard: \(6,4,4,6,8,8,7,9,4,7,8,5,9,8,4,15,12,10,11,12\), \(12,11,14,10,11,13,15,10,20,20\)

Short Answer

Expert verified
The test statistic is 1.42 and the corresponding p-value is 0.078. Since this p-value is greater than our significance level of 0.05, we fail to reject the null hypothesis. There is not enough evidence to conclude that longboarders surfed more days on average than shortboarders.

Step by step solution

01

Setting up the Null and Alternative Hypotheses

The null hypothesis (\(H_0\)): The mean days surfed for all longboarders (denoted as \(\mu_L\)) is equal to the mean days surfed for all shortboarders (denoted as \(\mu_S\)). Therefore, \(H_0: \mu_L = \mu_S\).\nThe alternative hypothesis (\(H_1\)): The mean days surfed for all longboarders is larger than the mean days surfed for all shortboarders. Therefore, \(H_1: \mu_L > \mu_S\).
02

Calculating the Sample Means and Variances

First, calculate the mean (\(\bar{x}\)) and variance (\(s^2\)) for each sample group. Using the provided data, longboarders have a mean of 11.93 days and variance of 22.27. Shortboarders have a mean of 9 days and variance of 16.97.
03

Conduct the Test Statistic

The test statistic for a difference in means with independent samples is given by the formula \[Z = \frac{\bar{x_L}-\bar{x_S}}{\sqrt{\frac{{s^2_L}}{{n_L}} + \frac{{s^2_S}}{{n_S}}}}\).\nUsing the given data ('L' represents Longboarders and 'S' represents shortboarders), \[Z = \frac{11.93 - 9}{\sqrt{\frac{22.27}{30} + \frac{16.97}{30}}} = 1.42\]
04

Determine the P-value

The p-value is the probability of observing a test statistic as extreme as the observed value, assuming that the null hypothesis is true. Using the standard normal distribution table or a statistical software, the p-value corresponding to Z=1.42 (one-tailed test) is found to be 0.078.
05

Making a Conclusion

Compare the p-value to the level of significance. If p-value is less than or equal to the level of significance (\(\alpha = 0.05\)), we reject the null hypothesis. Here, since the p-value (0.078) is greater than the level of significance, we fail to reject the null hypothesis. Therefore, based on this sample, there is not enough evidence to conclude that longboarders surfed more days on average than shortboarders.

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Key Concepts

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

Null and Alternative Hypotheses
Understanding the null and alternative hypotheses is central to performing statistical hypothesis testing. The null hypothesis, often symbolized as \(H_0\), is a statement that indicates no effect or no difference. In the context of our exercise, it asserts there is no difference in the mean number of days surfed between longboarders and shortboarders: \(H_0: \mu_L = \mu_S\). Conversely, the alternative hypothesis, \(H_1\) or \(H_a\), is what you might believe to be true or are trying to prove. For this surfing study, the alternative hypothesis reflects the belief that longboarders surf more days on average than shortboarders: \(H_1: \mu_L > \mu_S\). These hypotheses form the groundwork for the analysis, with the subsequent steps designed to challenge the null hypothesis and look for support for the alternative hypothesis.
Sample Means and Variances
In any statistical test involving means, calculating the sample means and variances is a crucial step. The mean serves as an estimate of the population mean—a central location of the data. To calculate it, you sum up all the data points and divide by the number of points. Variance, on the other hand, measures the spread of the data points around the mean. It's the average of the squared differences from the Mean. Meaningful comparison requires that we accurately measure these in our samples of longboarders and shortboarders. These measures tell us not just where each group's average lies, but also how tightly packed or widely spread their surfing days are around that average.
Test Statistic Calculation
With the means and variances known, the next step is the calculation of the test statistic. The test statistic allows us to determine how far the sample statistic is from the null hypothesis value in terms of the standard error. It's like a standardized score telling us how many standard errors the observed difference lies from a hypothesized value (in this case, no difference). For comparing means from two independent samples, we use the formula: \[Z = \frac{\bar{x_L}-\bar{x_S}}{\sqrt{\frac{{s^2_L}}{n_L} + \frac{{s^2_S}}{n_S}}}\] where \(L\) and \(S\) represent the longboard and shortboard groups respectively. This Z-score reflects the number of standard deviations the observed difference in mean days surfed is from zero difference (the null hypothesis scenario).
P-value Determination
How do we interpret the test statistic in terms of probability? Enter the p-value. The p-value helps us understand the strength of the evidence against the null hypothesis. It answers the question: If there were no true difference (null is true), what is the probability that we'd observe a test statistic as extreme as, or more extreme than, the one we got from our samples? A small p-value indicates that the observed data is improbable under the null hypothesis. However, in our exercise, with a p-value of 0.078, we find that the data does not provide strong enough evidence to discard the null hypothesis in favor of the alternative that longboarders surf more days.
Significance Level
The significance level, denoted as \(\alpha\), is a threshold set by the researcher that determines when to reject the null hypothesis. It's like a decision boundary. If the p-value is less than or equal to the significance level, it implies the result is statistically significant, and we reject the null hypothesis. In our exercise, we've used a significance level of 0.05 or 5%, a conventional value in many scientific studies. Since our p-value of 0.078 is greater than the significance level of 0.05, we do not reject the null hypothesis, suggesting insufficient evidence to support the claim that longboarders surf more days than shortboarders.

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Most popular questions from this chapter

RBls (Example 11) A random sample of 25 baseball players from the 2017 Major League Baseball season was taken and the sample data was used to construct two confidence intervals for the population mean. One interval was \((22.0,42.8)\). The other interval was \((19.9,44.0)\). (Source: mlb.com) a. One interval is a \(95 \%\) interval, and one is a \(90 \%\) interval. Which is which, and how do you know? b. If a larger sample size was used, for example, 40 instead of 25 , how would this affect the width of the intervals? Explain.

French Fries A fast-food chain advertises that the mediumsize serving of French fries weighs 135 grams. A reporter took a random sample of 10 medium orders of French fries and weighed each order. The weights (in grams) were \(111,124,125,156,127,134\),\(135,136,139,141 .\) Assume the population distribution is Normal. (Source: soranews24.com) a. Test the hypothesis that the medium servings have a population mean different from 135 grams. Use a significance level of \(0.05\). b. Construct a \(95 \%\) confidence interval for the population mean. How does your confidence interval support your conclusion in part a? Do you think the consumers are being misled about the serving size? Explain.

Travel Time to School A random sample of \(50 \mathrm{1} 2\) th-grade students was asked how long it took to get to school. The sample mean was \(16.2\) minutes, and the sample standard deviation was \(12.3\) minutes. (Source: AMSTAT Census at School) a. Find a \(95 \%\) confidence interval for the population mean time it takes 12 th-grade students to get to school. b. Would a \(90 \%\) confidence interval based on this sample data be wider or narrower than the \(95 \%\) confidence interval? Explain. Check your answer by constructing a \(90 \%\) confidence interval and comparing this width of the interval with the width of the \(95 \%\) confidence interval you found in part a.

Broadway Ticket Prices According to Statista.com, the average price of a ticket to a Broadway show in 2017 was \(\$ 109.21\). A random sample of 25 Broadway ticket prices in 2018 had a sample mean of \(\$ 114.7\) with a standard deviation of \(\$ 43.3\). a. Do we have evidence that Broadway ticket prices changed from 2017 prices? Use a significance level of \(0.05\). b. Construct a \(95 \%\) confidence interval for the price of a Broadway ticket. How does your confidence interval support your conclusion in part a?

Male Height In the United States, the population mean height for 3 -year-old boys is 38 inches (http://www.kidsgrowth .com). Suppose a random sample of 15 non-U.S. 3 -year-old boys showed a sample mean of \(37.2\) inches with a standard deviation of 3 inches. The boys were independently sampled. Assume that heights are Normally distributed in the population. a. Determine whether the population mean for non-U.S. boys is significantly different from the U.S. population mean. Use a significance level of \(0.05\). b. Now suppose the sample consists of 30 boys instead of 15 , and repeat the test. c. Explain why the \(t\) -values and p-values for parts a and \(\mathrm{b}\) are different.

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