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Chapter 10: Problem 62

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?

Short Answer

Expert verified
Whether the mean length of girls' names is significantly longer than that of boys' names is determined by comparing the t-value and t-critical. If the t-value is greater or equal to the t-critical, then we reject the null hypothesis and conclude that the mean length of girls' names is longer. Otherwise, we cannot reject the null hypothesis and the hypothesis that the mean length of girls' names is longer does not hold.

Step by step solution

01

Identifying null and alternate hypothesis

Firstly, the null hypothesis \(H_0\) is that the mean length of boys' names and girls' names is the same, while the alternate hypothesis \(H_1\) is that the mean length of girls' names is greater than boys' names.
02

Calculating the t-value

The t-value, a test statistic that measures the size of the difference relative to the variation in the data, is calculated with the following formula: \[t = \frac{\bar{x}_G - \bar{x}_B}{\sqrt{\frac{s^2_G}{n_G}+\frac{s^2_B}{n_B}}}\],where \( \bar{x}_G = 6.133, s_G = 1.456, n_G = 30\) are the sample mean, standard deviation, and size for girls' names, and \( \bar{x}_B = 5.767, s_B = 1.87, n_B = 30\) are the sample mean, standard deviation, and size for boys' names.
03

Calculating the t-critical value

The critical value is obtained from the t-distribution table (Student's t-table). The relevant degrees of freedom is calculated as: \(df = n_B + n_G - 2 = 30 + 30 - 2 = 58\). Given our pre-set significance level of 0.05 and considering \(H_1\), we'll use a one-tailed test. Therefore, consulting the table, the t-critical value which corresponds to the cumulative probability of 0.95 (1 - 0.05) and dof = 58 is roughly 1.671.
04

Deciding on the hypothesis

If the t-value is greater than or equal to the critical value, then reject the null hypothesis, meaning the study supports the idea that girls' names are longer. Otherwise, don't reject the null hypothesis and conclude that the results are statistically insignificant.

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Key Concepts

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

Null and Alternate Hypotheses
In statistical hypothesis testing, we start by defining two competing hypotheses: the null hypothesis (denoted as \( H_0 \)) and the alternate hypothesis (denoted as \( H_1 \)). These hypotheses provide a framework for making decisions based on data.
  • The null hypothesis \( H_0 \) represents a statement of no effect or no difference. In our case, \( H_0 \): The mean length of boys' names is equal to that of girls' names.
  • The alternate hypothesis \( H_1 \) suggests that there is an effect or a difference. Here, \( H_1 \): The mean length of girls' names is greater than that of boys' names.
Deciding which hypothesis to support depends on the data outcome. If the data suggests a significant difference, the null hypothesis is rejected in favor of the alternate. By clearly outlining \( H_0 \) and \( H_1 \), we can methodically approach hypothesis testing and ensure the conclusion drawn is based on statistical evidence.
T-Distribution
The t-distribution is a fundamental tool in hypothesis testing, particularly when working with smaller sample sizes. It allows us to determine how far the sample mean is expected to be from the population mean. The distribution is symmetric and bell-shaped, similar to the standard normal distribution, but with heavier tails.
Some aspects to understand about the t-distribution:
  • The shape of the t-distribution is defined by the degrees of freedom (df), which depends on the sample size. More degrees of freedom imply the t-distribution becomes more like the standard normal distribution.
  • In cases where the sample size is small or the population standard deviation is unknown, the t-distribution is more appropriate than the normal distribution.
  • It provides critical values against which calculated test statistics (like the t-value) can be compared to decide on hypotheses.
In this exercise, we used the t-distribution to determine the t-critical value for 58 degrees of freedom and a significance level of 0.05.
Significance Level
The significance level, denoted as \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. It reflects the level of risk we are willing to take for making a Type I error in hypothesis testing.
  • Common significance levels are 0.01, 0.05, or 0.10. In our scenario, a significance level of 0.05 was selected, meaning there is a 5% risk we are willing to accept for falsely rejecting the null hypothesis.
  • It serves as a threshold for decision-making. If the p-value (or calculated t-value) is less than or equal to \( \alpha \), we reject \( H_0 \).
  • By setting a significance level, we can make statistically informed decisions about our hypotheses. In this case, it determines whether the difference in name lengths is deemed statistically significant.
The choice of significance level aids in maintaining a balance between being too cautious (leading to false negative results) and too liberal (leading to false positive results) in inference.
Sample Mean
The sample mean, denoted as \( \bar{x} \), is the arithmetic average of a data set and provides an estimate of the population mean. It plays a critical role in hypothesis testing as it is one of the key statistics used to compare different groups.
Key points about the sample mean include:
  • The sample mean serves as a point estimate of the population mean. It helps determine how representative a sample is of the population.
  • In the exercise, \( \bar{x}_G = 6.133 \) is the sample mean for girls' names and \( \bar{x}_B = 5.767 \) for boys' names. This initial comparison hinted that girls' names might be longer on average.
  • When conducting the t-test, the difference between the sample means is used to assess whether the observed difference is statistically significant.
Understanding the sample mean informs our hypothesis tests and, alongside other statistical metrics, helps us draw meaningful conclusions from sample data.

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Most popular questions from this chapter

To test the effect of a physical-fitness \- course on one's physical ability, the number of sit-ups that a person could do in 1 minute, both before and after the course, was recorded. Ten randomly selected participants scored as shown in the following table. Can you \- conclude that a significant amount of improvement took place? Use \(\alpha=0.01\) and assume normality.$$\begin{array}{lllllllllll}\text { Before } & 29 & 22 & 25 & 29 & 26 & 24 & 31 & 46 & 34 & 28 \\\\\text { After } & 30 & 26 & 25 & 35 & 33 & 36 & 32 & 54 & 50 & 43 \\\\\hline\end{array}$$a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

10.30 Does a content title help a reader comprehend a piece of writing? Twenty-six participants were given an article to read without a title. They then rated themselves one their comprehension of the information on a scale from 1 to \(10,\) where 10 was complete comprehension. The same 26 participants were then given the article again, this time withe an appropriate title, and asked to rate their comprehension The resulting summarized data were given as \(\bar{d}=4.76\) and \(s_{d}=2.33,\) where \(d=\) rating with title - rating without title Comprehension was generally higher on the second reading than on the first by an average of 3.2 on this scale. Does this sample provide sufficient evidence that a content title does make a difference with respect to comprehension? Use \(\alpha=0.05\)

Was used to complete a \(t\) -test of the difference between two means using the following two independent samples. $$\begin{array}{lllllllll}\hline \text { Sample 1 } & 33.7 & 21.6 & 32.1 & 38.2 & 33.2 & 35.9 & 34.1 & 39.8 \\\& 23.5 & 21.2 & 23.3 & 18.9 & 30.3 & & & \\\\\hline \text { Sample 2 } & 28.0 & 59.9 & 22.3 & 43.3 & 43.6 & 24.1 & 6.9 & 14.1 \\\& 30.2 & 3.1 & 13.9 & 19.7 & 16.6 & 13.8 & 62.1 & 28.1 \\\\\hline\end{array}$$ a. Assuming normality, verify the results (two sample means and standard deviations, and the calculated \(t \star)\) by calculating the values yourself. b. Use Table 7 in Appendix \(B\) to verify the \(p\) -value based on the calculated df. c. Find the \(p\) -value using the smaller number of degrees of freedom. Compare the two \(p\) -values.

In trying to estimate the amount of growth that took place in the trees recently planted by the County Parks Commission, 36 trees were randomly selected from the 4000 planted. The heights of these trees were measured and recorded. One year later, another set of 42 trees was randomly selected and measured. Do the two sets of data (36 heights, 42 heights) represent dependent or independent samples? Explain.

Use a computer to demonstrate the truth of the statement describing the sampling distribution of \(\bar{x}_{1}-\bar{x}_{2}\) Use two theoretical normal populations: \(N_{1}(100,20)\) and \(N_{2}(120,20)\) a. To get acquainted with the two theoretical populations, randomly select a very large sample from each. Generate 2000 data values, calculate mean and standard deviation, and construct a histogram using class boundaries that are multiples of one-half of a standard deviation (10) starting at the mean for each population. b. If samples of size 8 are randomly selected from each population, what do you expect the distribution of \(\bar{x}_{1}-\bar{x}_{2}\) to be like (shape of distribution, mean, standard error \() ?\) c. Randomly draw a sample of size 8 from each population, and find the mean of each sample. Find the difference between the sample means. Repeat 99 more times. d. The set of \(100\left(\bar{x}_{1}-\bar{x}_{2}\right)\) values forms an empirical sampling distribution of \(\bar{x}_{1}-\bar{x}_{2} .\) Describe the empirical distribution: shape (histogram), mean, and standard error. (Use class boundaries that are multiples of standard error from mean for easy comparison to the expected.) e. Using the information found in parts a-d, verify the statement about the \(\bar{x}_{1}-\bar{x}_{2}\) sampling distribution made on page 496 f. Repeat the experiment a few times and compare the results.
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