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Self-Esteem Scores In a study of a group of women science majors who remained in their profession and a group who left their profession within a few months of graduation, the researchers collected the data shown here on a self-esteem questionnaire. At \(\alpha=0.05,\) can it be concluded that there is a difference in the self-esteem scores of the two groups? Use the \(P\) -value method. $$ \begin{array}{ll}{\text { Leavers }} & {\text { Stayers }} \\\ {\bar{X}_{1}=3.05} & {\bar{X}_{2}=2.96} \\ {\sigma_{1}=0.75} & {\sigma_{2}=0.75} \\ {n_{1}=103} & {n_{2}=225}\end{array} $$

Short Answer

Expert verified
There is no significant difference in self-esteem scores between the two groups at \(\alpha=0.05\).

Step by step solution

01

State the Hypotheses

To determine if there is a difference in self-esteem scores, we set up our null hypothesis (\(H_0\)) and alternative hypothesis (\(H_1\)). The null hypothesis, \(H_0\), states that there is no difference in self-esteem scores between the two groups: \(\mu_1 = \mu_2\). The alternative hypothesis, \(H_1\), states that there is a difference: \(\mu_1 eq \mu_2\).
02

Calculate the Test Statistic

We use the formula for the z-test statistic for two independent samples: \[ z = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \] Plugging in the values: \[ z = \frac{3.05 - 2.96}{\sqrt{\frac{0.75^2}{103} + \frac{0.75^2}{225}}} \] \[ = \frac{0.09}{\sqrt{\frac{0.5625}{103} + \frac{0.5625}{225}}} \approx \frac{0.09}{\sqrt{0.005468 + 0.0025}} \approx \frac{0.09}{\sqrt{0.007968}} \approx \frac{0.09}{0.0893} \approx 1.008 \]
03

Determine the P-value

Using a standard normal distribution table, we find the p-value associated with a z-score of 1.008. The two-tailed p-value for \(|z| > 1.008\) is approximately 0.3134. This value represents the probability of observing a difference at least as extreme as the one found.
04

Compare P-value to the Significance Level

We compare the p-value (0.3134) to the significance level \(\alpha = 0.05\). Since 0.3134 is greater than 0.05, we do not reject the null hypothesis.
05

State the Conclusion

Since the p-value is greater than the significance level, there is insufficient evidence to conclude that there is a significant difference in self-esteem scores between the two groups of women. We fail to reject the null hypothesis.

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Key Concepts

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

Understanding the Z-test
The z-test is a statistical test used to determine if there is a significant difference between the means of two populations. In our example with the self-esteem scores, the z-test helps us decide whether the leavers and stayers differ significantly in their mean scores.

The test relies on certain assumptions:
  • The samples are independent of each other.
  • The sample sizes are sufficiently large (usually above 30).
  • The data is drawn from a normally distributed population.
The equation used for the z-test compares the difference between sample means relative to the variability of scores within the samples. A higher z-value indicates a more significant difference between the groups.

Using the formula for our case: \[ z = \frac{3.05 - 2.96}{\sqrt{\frac{0.75^2}{103} + \frac{0.75^2}{225}}} \]This computation helps us determine the likelihood of observing the difference between the two groups just by chance, under the null hypothesis.
The P-value Method
The p-value method is a popular approach in hypothesis testing to assess the strength of our findings. After calculating the z-test statistic, we look up the corresponding p-value. This value tells us the probability of obtaining test results at least as extreme as those observed, assuming the null hypothesis is true.

In our example, we found that the p-value was approximately 0.3134. This is the chance of witnessing such a difference, or one more extreme, if there truly was no difference in self-esteem scores between the two groups.

A smaller p-value indicates stronger evidence against the null hypothesis. It's important to note that the p-value does not measure the probability that the null hypothesis is true, rather it measures the probability of observing the data if the null hypothesis were true.
Explaining Significance Level
The significance level, denoted by \(\alpha\), is the threshold we set for deciding whether to reject the null hypothesis. In many fields, a common significance level is 0.05, meaning we are willing to accept a 5% chance of rejecting the null hypothesis when it is true (also known as a Type I error).

In our exercise with the self-esteem scores, \(\alpha = 0.05\) was chosen. When we compare this to our calculated p-value of 0.3134, we find that the p-value is greater. This relationship tells us that the evidence isn't strong enough to conclude a significant difference in self-esteem between leavers and stayers.
  • If the p-value is less than or equal to \(\alpha,\) we reject the null hypothesis.
  • If the p-value is greater than \(\alpha,\) we fail to reject the null hypothesis.
This comparison helps balance the risks of making incorrect conclusions in hypothesis testing.
Understanding Null Hypothesis
The null hypothesis, often denoted as \(H_0\), is a statement suggesting that there is no effect or no difference between groups, and any observed difference is due to sampling or experimental error. In our self-esteem score exercise, the null hypothesis posits that there is no difference in self-esteem scores between the leavers and stayers.

The null hypothesis is crucial as it provides a baseline to compare our findings against. Hypothesis testing involves calculating the probability of observing data as unusual as ours, assuming the null hypothesis is correct.
  • If this probability is low, we have grounds to reject the null hypothesis.
  • If it’s not low, we do not have sufficient evidence to reject it, implying the null hypothesis might be true.
In our case, since the p-value exceeded the significance level, we failed to reject the null hypothesis, suggesting a lack of evidence for a difference in means between the two groups.

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Most popular questions from this chapter

What are the two different degrees of freedom associated with the \(F\) distribution?

For Exercises 7 through \(27,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Bullying Bullying is a problem at any age but especially for students aged 12 to 18 . A study showed that \(7.2 \%\) of all students in this age bracket reported that bullied at school during the past six months with 6 th grade having the highest incidence at \(13.9 \%\) and 12 th grade the lowest at \(2.2 \%\). To see if there is a difference between public and private schools, 200 students were randomly selected from each. At the 0.05 level of significance, can a difference be concluded? $$ \begin{array}{ccc}{} & {\text { Private }} & {\text { Public }} \\ \hline \text { Sample size } & {200} & {200} \\ {\text { No. bullied }} & {13} & {16}\end{array} $$

For Exercises 2 through \(12,\) perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Reducing Errors in Spelling A ninth-grade teacher wishes to see if a new spelling program will reduce the spelling errors in his students' writing. The number of spelling errors made by the students in a five-page report before the program is shown. Then the number of spelling errors made by students in a five-page report after the program is shown. At \(\alpha=0.05,\) did the program work? $$ \begin{array}{lllllll}{\text { Before }} & {8} & {3} & {10} & {5} & {9} & {11} & {12} \\ \hline \text { After } & {6} & {4} & {8} & {1} & {4} & {7} & {11}\end{array} $$

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Home Prices A real estate agent compares the selling prices of randomly selected homes in two municipalities in southwestern Pennsylvania to see if there is a difference. The results of the study are shown. Is there enough evidence to reject the claim that the average cost of a home in both locations is the same? Use \(\alpha=0.01 .\) $$ \begin{array}{cc}{\text { Scott }} & {\text { Ligonier }} \\ \hline \overline{X_{1}=\$ 93,430^{*}} & {\bar{X}_{2}=\$ 98,043^{*}} \\\ {\sigma_{1}=\$ 5602} & {\sigma_{2}=\$ 4731} \\ {n_{1}=35} & {n_{2}=40}\end{array} $$

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