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Chapter 10: Problem 23

In August \(2002,47 \%\) of parents with children in grades \(\mathrm{K}-12\) were satisfied with the quality of education the students receive. A recent Gallup poll found that 437 of 1013 parents with children in grades \(\mathrm{K}-12\) were satisfied with the quality of education the students receive. Construct a \(95 \%\) confidence interval to assess whether this represents evidence that parents' attitudes toward the quality of education in the United States has changed since August 2002

Short Answer

Expert verified
The confidence interval is (0.400, 0.462). Comparing this interval with the original 47%, which falls outside the interval, indicates parents' attitudes have changed significantly since 2002.

Step by step solution

01

Determine the sample proportion

Calculate the sample proportion of satisfied parents using the formula: Given: Given: Total number of parents surveyed ( n ) = 1013 Given: Number of satisfied parents ( x ) = 437 Calculate sample proportion ( p_{ ^ }) = x / n = 437 / 1013 = 0.431
02

Find the standard error

The standard error for the sample proportion can be found using the formula: Calculate the standard error ( SE ) = sqrt( p _ ^ (1 - p ^ ) / n ) = sqrt( 0.431 (1 - 0.431)/1013) = 0.0155
03

Find the z-score value for 95% confidence level

For a 95% confidence interval, the z-score value is Given Z= 1.96
04

Construct the confidence interval

The confidence interval is given by Calculate CI : p_{ ^ } - z * SE < original proportion of satisfied parents < p _{ ^ } + z* SE = 0.431- 1.96 * 0.0155 < p < 0.431 +1.96 * 0.0155 = (0.400,0.462)
05

Compare with historical data

Compare the original data proportion with the confidence interval: August 2002 proportion ( p_{ 0 2002 })= 0.47 Original Satisfaction rate falls outside the confidence interval Hence There is sufficient statistical evidence to conclude that parent's attitudes have changed

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Key Concepts

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

Sample Proportion
Understanding the concept of sample proportion is essential when analyzing survey data. A sample proportion represents a fraction derived from a group of observations and is used to estimate a population proportion. To calculate the sample proportion, divide the number of favorable outcomes by the total number of observations. For instance, in our exercise, 437 out of 1013 parents were satisfied with the quality of education. Here, the sample proportion \( \hat{p} \) is computed as follows: \[ \hat{p} = \frac{437}{1013} \approx 0.431 \]. This value helps us understand a segment of the surveyed population's viewpoint.
Standard Error
The standard error quantifies the level of variability or precision of the sample proportion estimate. It's crucial in determining the reliability of our sample proportion when generalizing to a broader population. Calculating the standard error of a sample proportion involves using the formula: \[\text{SE} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]. In our given example: \[\text{SE} = \sqrt{\frac{0.431(1 - 0.431)}{1013}} \approx 0.0155 \]. This tells us about the expected fluctuation of the sample proportion if we repeated the survey multiple times.
Z-Score for Confidence Level
A z-score indicates how many standard deviations an element is from the mean. For constructing a confidence interval, we use a z-score that corresponds to our desired confidence level. In most social science research, a 95% confidence level is typical, which corresponds to a z-score of approximately 1.96. This value signifies that a sample proportion lies within 1.96 standard deviations above or below the population proportion 95% of the time. The z-score helps to specify the range within which we expect our true population proportion to fall.
Survey Data Analysis
Survey data analysis involves examining the responses from a selected group to infer conclusions about a larger population. In this exercise, we used survey data from parents to evaluate their satisfaction with the quality of education. First, we calculated the sample proportion (0.431), then found the standard error (0.0155). Next, we utilized the z-score for the 95% confidence level (1.96) to construct our confidence interval: \[0.431 \pm 1.96 \times 0.0155 \]. This results in the interval (0.400, 0.462). Finally, we compared this interval to the historical satisfaction proportion of 0.47 from August 2002. Since 0.47 is outside the interval, we inferred a significant change in parents' attitudes.

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Most popular questions from this chapter

Reading at Bedtime It is well-documented that watching TV, working on a computer, or any other activity involving artificial light can be harmful to sleep patterns. Researchers wanted to determine if the artificial light from e-Readers also disrupted sleep. In the study, 12 young adults were given either an iPad or printed book for four hours before bedtime. Then, they switched reading devices. Whether the individual received the iPad or book first was determined randomly. Bedtime was \(10 \mathrm{P.M}\). and the time to fall asleep was measured each evening. It was found that participants took an average of 10 minutes longer to fall asleep after reading on an iPad. The \(P\) -value for the test was \(0.009 .\) (a) What is the research objective? (b) What is the response variable? It is quantitative or qualitative? (c) What is the treatment? (d) Is this a designed experiment or observational study? What type? (e) The null hypothesis for this test would be that there is no difference in time to fall asleep with an e-Reader and printed book. The alternative is that there is a difference. Interpret the \(P\) -value.

Test the hypothesis using (a) the classical approach and (b) the P-value approach. Be sure to verify the requirements of the test. $$\begin{array}{l}H_{0}: p=0.25 \text { versus } H_{1}: p<0.25 \\\n=400 ; x=96 ; \alpha=0.1\end{array}$$

In Problems 23-29, decide whether the problem requires a confidence interval or hypothesis test, and determine the variable of interest. For any problem requiring a confidence interval, state whether the confidence interval will be for a population proportion or population mean. For any problem requiring a hypothesis test, write the null and alternative hypothesis. An investigator with the Food and Drug Administration wanted to determine whether a typical bag of potato chips contained less than the 16 ounces claimed by the manufacturer.

Throwing darts at the stock pages to decide which companies to invest in could be a successful stock-picking strategy. Suppose a researcher decides to test this theory and randomly chooses 100 companies to invest in. After 1 year, 53 of the companies were considered winners; that is, they outperformed other companies in the same investment class. To assess whether the dart-picking strategy resulted in a majority of winners, the researcher tested \(H_{0}: p=0.5\) versus \(H_{1}: p>0.5\) and obtained a \(P\) -value of \(0.2743 .\) Explain what this \(P\) -value means and write a conclusion for the researcher.

A simple random sample of size \(n=320\) adults was asked their favorite ice cream flavor. Of the 320 individuals surveyed, 58 responded that they preferred mint chocolate chip. Do less than \(25 \%\) of adults prefer mint chocolate chip ice cream? Use the \(\alpha=0.01\) level of significance.
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