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Chapter 16: Problem 626

An investigator tested two samples, one of boys and one of girls. He wanted to know whether the results for girls are significantly more variable than for boys. Test for this at the \(5 \%\) level of significance using the following sample data. \begin{tabular}{|c|c|c|} \cline { 2 - 3 } \multicolumn{1}{c|} {} & Sample Size & \(\mathrm{S}^{2}\) \\ \hline Boys & 8 & \(50.21\) \\ \hline Girls & 9 & \(147.62\) \\ \hline \end{tabular}

Short Answer

Expert verified
We conducted an F-test to compare the variances of the given samples of boys and girls. We calculated the F-statistic to be \(2.94\) and found the critical F-value to be \(3.68\) at a \(5\%\) level of significance. Since the F-statistic (\(2.94\)) was less than the critical F-value (\(3.68\)), we failed to reject the null hypothesis. This led us to conclude that there is no sufficient evidence to support the claim that the results for girls are significantly more variable than for boys at a \(5\%\) level of significance.

Step by step solution

01

Identify the null and alternate hypotheses

We want to test if the results for girls are significantly more variable than for boys. So, we set the null hypothesis (H_0) and the alternate hypothesis (H_a) as follows: \(H_0: \sigma_1^2 \leq \sigma_2^2\) \(H_a: \sigma_1^2 > \sigma_2^2\) where \(\sigma_1^2\) is the population variance for boys, and \(\sigma_2^2\) is the population variance for girls.
02

Calculate the F-statistic

To calculate the F-statistic, we use the given sample variances and the formula: \(F = \frac{\mathrm{S}_2^2}{\mathrm{S}_1^2}\) Where \(F\) is the F-statistic, and \(\mathrm{S}_1^2\) and \(\mathrm{S}_2^2\) are the sample variances for boys and girls, respectively. We plug in the given values: \(F = \frac{147.62}{50.21} \approx 2.94\)
03

Determine the critical F-value

To determine the critical F-value, we need the degrees of freedom for both samples, as well as the desired level of significance. The degrees of freedom for each sample are: \(df_1 = n_1 - 1 = 8 - 1 = 7\) \(df_2 = n_2 - 1 = 9 - 1 = 8\) The level of significance is given as \(5\%\), or \(0.05\). Using an F-table or statistical software, we find the critical F-value at a \(5\%\) level of significance for \(df_1 = 7\) and \(df_2 = 8\) is approximately \(3.68\).
04

Compare the F-statistic to the critical F-value

Now, we compare the calculated F-statistic (\(2.94\)) to the critical F-value (\(3.68\)): Since \(2.94 < 3.68\), we fail to reject the null hypothesis.
05

Interpret the results

Because we failed to reject the null hypothesis, we conclude that, at a \(5\%\) level of significance, there is no sufficient evidence to support the claim that the results for girls are significantly more variable than for boys.

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Key Concepts

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

F-test
The F-test is a statistical test used to compare two variances. When an investigator wants to determine if one sample is more variable than another, the F-test comes into play. In the context of the given problem, we're comparing the variability between boys and girls.

The F-statistic is calculated by taking the ratio of the two sample variances. Specifically, we divide the variance of the girls' sample by the variance of the boys' sample. The resulting F-statistic is then compared to a critical value obtained from statistical tables.

If the computed F-statistic is greater than the critical value, it suggests significant variability between the two samples. However, if it's lower, like in our case with an F-statistic of 2.94 and a critical value of 3.68, it indicates no significant difference.
Variability
Variability refers to how spread out the data points are in a dataset. When we talk about sample variance, it鈥檚 a measure of this spread in numbers. In simple terms, it tells us how much the numbers in a group differ from the average number of that group.

In the problem, the variance of boys' scores was lower at 50.21 compared to girls' 147.62. This suggests that girls' scores fluctuate more widely from their average compared to boys' scores, hinting at greater variability among girls. Understanding this concept is crucial since the primary focus of the problem is to assess if one gender shows more variation in scores than the other.

Higher variability means more diverse outcomes, while lower variability implies the results are clustered closer to the mean.
Level of significance
The level of significance in hypothesis testing is a threshold set by the investigator to determine when to reject the null hypothesis. It represents the probability of making a Type I error, which occurs when the null hypothesis is incorrectly rejected.

In our problem, a 5% level of significance was chosen. This means there's a 5% risk of concluding that the girls' scores are more variable than the boys' scores when they actually aren't.

Some key points about the level of significance:
  • A common choice is 5%, but sometimes 1% or 10% levels are used, depending on the context.
  • The smaller the level of significance, the stricter the criteria to reject the null hypothesis.
  • It helps keep the balance between sensitivity to detect an effect and the risk of error.
Degrees of freedom
Degrees of freedom (df) play a vital role in statistical tests like the F-test. They represent the number of values in the final calculation of a statistic that are free to vary. In simple words, it鈥檚 a count of information used for estimating parameters.

For each sample, degrees of freedom are calculated as the sample size minus one. Therefore:
  • Boys: Sample size = 8; so, degrees of freedom = 8 - 1 = 7.
  • Girls: Sample size = 9; so, degrees of freedom = 9 - 1 = 8.

In testing, degrees of freedom help in determining the critical F-value from statistical tables. Without the correct degrees of freedom, finding an appropriate critical value would be inaccurate, potentially leading to incorrect conclusions about the variability in samples.

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Most popular questions from this chapter

In testing a hypothesis concerned with the value of a population mean, first the level of significance to be used in the test is specified and then the regions of acceptance and rejection for evaluating the obtained sample mean are determined. If the 1 percent level of significance is used, indicate the percentages of sample means in each of the areas of the normal curve, assuming that the population hypothesis is correct, and the test is two-tailed.

The standard deviation of a particular dimension of a metal component is small enough so that it is satisfactory in subsequent assembly. A new supplier of metal plate is under consideration and will be preferred if the standard deviation of his product is not larger, because the cost of his product is lower than that of the present supplier. A sample of 100 items from each supplier is obtained. The results are as follows: New supplier: \(\mathrm{S}_{1}^{2}=.0058\) Old supplier: \(\mathrm{S}_{2}{ }^{2}=.0041\) Should the new supplier's metal plates be purchased? Test at the \(5 \%\) level of significance.

Consider the probability distribution function $$ \mathrm{f}(\mathrm{x} ; \theta)=(1 / \theta) \mathrm{e}^{-\mathrm{x} / \theta} \quad 0<\mathrm{x}<\infty $$ $$ =0 $$ elsewhere. It is desired to test the hypothesis \(\mathrm{H}_{0}: \theta=2\) against alternate hypothesis \(\mathrm{H}_{1}: \theta>2\). Suppose a random sample \(\mathrm{X}_{1}, \mathrm{X}_{2}\) is used and the critical region is \(\mathrm{X}_{1}+\mathrm{X}_{2} \geq 9.5\) Calculate an expression for the power function, \(\mathrm{K}\left(\theta_{1}\right)\), for all \(\theta_{1}>2\), and specifically for \(\theta_{1}=4\).

Suppose we have a binomial distribution for which \(\mathrm{H}_{0}\) is that the probability of success on a single trial, \(\mathrm{p}=1 / 2\). Suppose also that \(\mathrm{H}_{1}\) is \(\mathrm{p}=2 / 3\). Show how the power of the normal approximation to the binomial test increases as n increases by finding the critical value, \(K\), and the type II error, \(\beta\), for each of the following values of n: \(36,64,100,144\), and 196 . For which of these values of \(\mathrm{N}\) does \(\beta\) first fall as low as \(.5 ?\) Use \(\alpha=.01\).

Suppose we have a sample with \(\mathrm{n}=10\) and \(\underline{\mathrm{Y}}=50 . \mathrm{We}\) wish to test \(\mathrm{H}_{0}: \mu=47\) against \(\mathrm{H}_{1}: \mu \neq 47 .\) We would like to know the probability given \(\mu=47\) of observing a random sample from the population with \(\underline{Y}=50 .\) We will reject \(\mathrm{H}_{0}\) if the probability is less than \(\alpha=.05\) that \(\mathrm{H}_{0}\) is true when \(\underline{Y}=50\). Assume for this sample that \(\sum(\mathrm{Y}-\mathrm{Y})^{2}=99.2250 .\) Calculate the estimated population variance, the estimated \(\sigma^{2} \underline{\mathrm{Y}}\) and the estimated slandered error of the mean. Then calculate the t statistic and determine whether \(\mathrm{H}_{0}\) can be rejected. Also, suppose \(\alpha=.01\). Within what limits may \(\underline{Y}\) vary without our having to reject \(\mathrm{H}_{0}\) ?
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