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Chapter 8: Problem 83

Pew Research conducts polls on social media use. In \(2012,66 \%\) of those surveyed reported using Facebook. In 2018 , \(76 \%\) reported using Facebook. a. Assume that both polls used samples of 100 people. Do a test to see whether the proportion of people who reported using Facebook was significantly different in 2012 and 2018 using a \(0.01\) significance level. b. Repeat the problem, now assuming the sample sizes were both 1500 . (The actual survey size in 2018 was \(1785 .\).) c. Comment on the effect of different sample sizes on the p-value and on the conclusion.

Short Answer

Expert verified
The procedure, calculates and compares Z-values with the critical value for both 100 samples and 1500 samples from 2012 and 2018. After this, it is seen how the p-value changes with different sample sizes.

Step by step solution

01

Setting up the Hypothesis

The null Hypothesis, \(H_0\), is that the proportion of people who used Facebook in 2012 and 2018 are equal; and the Alternate Hypothesis \(H_a\) is that the proportions are not equal.
02

Calculating the Test Statistic for 100 samples

We will calculate the test statistic with \[Z = \frac{(\hat{p}_1-\hat{p}_2) - 0}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1}+\frac{1}{n_2})}}\]\nHere, we have \(\hat{p}_1\) as the proportion in 2018 which is 0.76, \(\hat{p}_2\) is the proportion in 2012 which is 0.66. \(n_1\) and \(n_2\) is 100 both for years 2012 and 2018. And the pooled sample proportion \(\hat{p}\) is \(\frac{\hat{p}_1n_1 + \hat{p}_2n_2}{n_1+n_2}\) which we have to calculate.
03

Critical Value and Conclusion for 100 samples

For the 0.01 significance level, the critical value is obtained from the Z-table, which is approximately 2.33 and 2.33. If the calculated Z value is less than -2.33 or greater than 2.33, we reject the null hypothesis.
04

Calculating for 1500 samples

We will repeat step 2 and 3 with 1500 as the shared \(n_1\) and \(n_2\) values.
05

Effect of Different Samples Sizes

By comparing the p-value (the probability the result occurred by chance) from the test results with a 100-person sample and the 1500-person sample, we can see how the sample size affects the p-value.

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Key Concepts

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

Sample Size Effect
Sample size plays a crucial role in statistical hypothesis testing. A larger sample size provides more reliable results and reduces the margin of error. This can lead to:
  • Increased precision in estimating population parameters.
  • Lower p-values, making it easier to detect significant differences when they exist.
When the sample size increases, the standard error decreases, making the test statistic (in this case, the Z-value) larger for the same difference in proportions. This can result in a more definitive conclusion as it helps to distinguish actual effects from random noise in the data.
Significance Level
The significance level, often denoted as \(\alpha\), serves as a threshold for determining whether the null hypothesis should be rejected. A common choice is 0.05, but in this exercise, we use 0.01, indicating a more stringent criterion.
  • A smaller significance level means fewer false positives but increases the chance of missing a true effect.
  • Setting the significance level involves balancing the risk of Type I errors (rejecting a true null hypothesis) and sensitivity to true differences.
In practice, choosing an appropriate level depends on the context and consequences of potential decisions based on the test results.
P-Value
The p-value is a measure that helps you understand the probability of observing a result as extreme as the test result, assuming the null hypothesis is true. It helps in deciding whether or not to reject the null hypothesis:
  • A small p-value (typically \( \leq 0.05 \) or \( \leq 0.01 \)) indicates strong evidence against the null hypothesis.
  • A larger p-value suggests that the observed data are consistent with the null hypothesis.
The p-value varies with changes in sample size. A larger sample usually results in a smaller p-value, which can provide clearer insights into the differences being tested.
Z-Test
A Z-test is used to determine if there's a significant difference between sample and population means or between two sample means when the standard deviation is known:
  • It is useful when sample sizes are large (typically \(n \geq 30\)).
  • The formula \( Z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \) is applied to calculate the test statistic for proportions.
For this exercise, conducting a Z-test allows for the comparison of Facebook usage proportions from two different years. It provides a way to interpret if the observed differences are statistically significant given the sample data.

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Most popular questions from this chapter

According to the Bureau of Labor Statistics, \(10.1 \%\) of Americans are self- employed. A researcher wants to determine if the self-employment rate in a certain area is different. She takes a random sample of 500 working residents from the area and finds that 62 are self-employed. a. Test the hypothesis that the proportion of self-employed workers in this area is different from \(10.1 \%\). Use a \(0.05\) significance level. b. After conducting the hypothesis test, a further question one might ask. "What proportion of workers in this area are self-employed?" Use the sample data to find a \(95 \%\) confidence interval for the proportion of workers who are self-employed in the area from which the sample was drawn. How does this confidence interval support the hypothesis test conclusion?

A community college used enrollment records of all students and reported that that the percentage of the student population identifying as female in 2010 was \(54 \%\) whereas the proportion identifying as female in 2018 was \(52 \%\). Would it be appropriate to use this information for a hypothesis test to determine if the proportion of students identifying as female at this college had declined? Explain.

The National Association for Law Placement estimated that \(86.7 \%\) of law school graduates in 2015 found employment. An economist thinks the current employment rate for law school graduates is different from the 2015 rate. Pick the correct pair of hypotheses the economist could use to test this claim. \(\begin{array}{ll}\text { i. } \mathrm{H}_{0}: p \neq 0.867 & \text { ii. } \mathrm{H}_{0}: p=0.867 \\ \mathrm{H}_{\mathrm{a}}: p=0.867 & \mathrm{H}_{\mathrm{a}}: p>0.867 \\ \text { iii. } \mathrm{H}_{0}: p=0.867 & \text { iv. } \mathrm{H}_{0}: p=0.867 \\\ \mathrm{H}_{\mathrm{a}^{\circ}}=p<0.867 & \mathrm{H}_{\mathrm{a}}: p \neq 0.867\end{array}\)

When comparing two sample proportions with a two-sided alternative hypothesis, all other factors being equal, will you get a smaller p-value if the sample proportions are close together or if they are far apart? Explain.

No-Carb Diet A weight-loss diet claims that it causes weight loss by eliminating carbohydrates (breads and starches) from the diet. To test this claim, researchers randomly assign overweight subjects to two groups. Both groups eat the same amount of calories, but one group eats almost no carbs, and the other group includes carbs in their meals. After 2 months, the researchers test the claim that the no-carb diet is better than the usual diet. They record the proportion of each group that lost more than \(5 \%\) of their initial weight. They then announce that they failed to reject the null hypothesis. Which of the following are valid interpretations of the researchers' findings? a. There were no significant differences in effectiveness between the no-carb diet and the carb diet. b. The no-carb diet and the carb diet were equally effective. c. The researchers did not see enough evidence to conclude that the no-carb diet was more effective. d. The no-carb diet was less effective than the carb diet.
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