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Going to the prom Tonya wants to estimate what proportion of her school’s seniors plan to attend the prom. She interviews an SRS of 50 of the 750 seniors in her school and finds that 36 plan to go to the prom. (a) Identify the population and parameter of interest. (b) Check conditions for constructing a confidence interval for the parameter. (c) Construct a 90% confidence interval for p. Show your method. (d) Interpret the interval in context.

Short Answer

Expert verified
90% confidence interval for prom attendance is [61.24%, 82.76%].

Step by step solution

01

Identify the Population and Parameter of Interest

The population of interest is all the seniors in Tonya's school, which totals to 750 students. The parameter of interest is the proportion of seniors who plan to attend the prom, denoted by \( p \).
02

Check Conditions for Constructing a Confidence Interval

To construct a confidence interval for the sample proportion, we need to check the following conditions: 1. **Random Sample**: Tonya interviews a simple random sample (SRS) of 50 seniors.2. **Normality**: We check if both \( np \) and \( n(1-p) \) are greater than 10. Here, \( n = 50 \) and \( p = \frac{36}{50} = 0.72 \). Calculating gives \( 50 \times 0.72 = 36 \) and \( 50 \times 0.28 = 14 \), both of which are greater than 10.3. **10% Condition**: Since 50 seniors represent a sample of less than 10% of the total senior class (750), this condition is met.
03

Construct the 90% Confidence Interval

First, calculate the sample proportion \( \hat{p} = \frac{36}{50} = 0.72 \). Find the standard error: \( SE = \sqrt{\frac{0.72 \times (1-0.72)}{50}} = \sqrt{\frac{0.72 \times 0.28}{50}} = 0.0655 \).For a 90% confidence interval, the z* value (critical value) is approximately 1.645.Calculate the margin of error (MOE) as \( MOE = 1.645 \times 0.0655 = 0.1076 \).Determine the confidence interval: \( 0.72 \pm 0.1076 \), resulting in an interval of \( [0.6124, 0.8276] \).
04

Interpret the Interval in Context

We are 90% confident that the true proportion of seniors at Tonya's school who plan to attend the prom is between approximately 61.24% and 82.76%.

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Key Concepts

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

Population and Parameter of Interest
Tonya's goal is to determine how many seniors from her school intend to attend the prom.
In statistical terms, she is trying to establish information about a certain population.
Here, the **population of interest** consists of all the senior students at Tonya's school. In total, there are 750 such students. Her research aims to find out about the whole group using a statistical technique.
The **parameter of interest**, on the other hand, is the specific aspect Tonya wants to measure or infer about this population. In this case, she wants to estimate the proportion of seniors planning to go to the prom.
This proportion is denoted by the symbol \( p \). Essentially, understanding \( p \) would tell us what portion of the entire senior class is fun and party-ready.
Random Sample Condition
One crucial requirement for making predictions about the whole population is that the sample must be random.
Tonya achieved this by selecting a simple random sample (SRS) of 50 students from the 750 seniors.
By ensuring that every senior had an equal chance of being chosen, her sample accurately reflects the diversity and behavior of the entire population.
A random sample helps in reducing biases and represents different sub-groups within the population.
With a fair sample selection process, there's increased confidence that the insights derived from this sample hold true for the entire group of seniors.
  • Ensures fair and unbiased selection
  • Enhances the accuracy of the study
  • Reflects the overall population's variation
Thus, Tonya's strategy of random sampling plays a pivotal role in making her findings reliable.
Normality Condition
When estimating proportions, it's vital to ensure that the sample size and expected frequencies uphold the **Normality Condition**.
This condition states that both \( np \) and \( n(1-p) \) should be greater than 10. This ensures the sample data is sufficiently large enough to approximate a normal distribution, which is essential for constructing confidence intervals.In Tonya's case, the sample size \( n = 50 \) and the sample proportion \( \hat{p} = 0.72 \).
Therefore, \( np = 50 \times 0.72 = 36 \) and \( n(1-p) = 50 \times 0.28 = 14 \). Both values are indeed greater than 10.
This implies that her sample satisfies the Normality Condition, providing a solid foundation for Tonya to estimate the proportion of seniors heading to the prom using a confidence interval.
  • Ensures the sample size is adequate
  • Allows approximation to a normal distribution
  • Builds a foundation for reliable interval estimation
10% Condition
The **10% Condition** is another crucial factor to check when creating a confidence interval. This ensures that the sample size should be less than 10% of the population size, keeping the sampling process independent.In Tonya's situation, she sampled 50 seniors out of the total 750.
This gives us a percentage of \( \frac{50}{750} \times 100 \% = 6.67\% \), which is clearly under the 10% threshold.
By meeting the 10% Condition, Tonya can maintain the assumption of independence between each selected student in her sample.
  • Keeps the sample size within a reasonable range
  • Ensures independence among sample measurements
  • Ensures the reliability of the statistical inferences made
This vital check assures that her subsequent analyses, like the construction of a confidence interval, are legitimate and sound.

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Most popular questions from this chapter

In Exercises 1 to 4, determine the point estimator you would use and calculate the value of the point estimate. Going to the prom Tonya wants to estimate what proportion of the seniors in her school plan to attend the prom. She interviews an SRS of 50 of the 750 seniors in her school and finds that 36 plan to go to the prom.

93% confidence Find z* for a 93% confidence interval using Table A or your calculator. Show your method.

In Exercises 1 to 4, determine the point estimator you would use and calculate the value of the point estimate. Got shoes? How many pairs of shoes, on average, do female teens have? To find out, an AP Statistics class conducted a survey. They selected an SRS of 20 female students from their school. Then they recorded the number of pairs of shoes that each student reported having. Here are the data: \(\begin{array}{llllllllll}{50} & {26} & {26} & {31} & {57} & {19} & {24} & {22} & {23} & {38} \\ {13} & {50} & {13} & {34} & {23} & {30} & {49} & {13} & {15} & {51} \\ \hline\end{array}\)

NAEP scores Young people have a better chance of full-time employment and good wages if they are good with numbers. How strong are the quantitative skills of young Americans of working age? One source of data is the National Assessment of Educational Progress (NAEP) Young Adult Literacy Assessment Survey, which is based on a nationwide probability sample of households. The NAEP survey includes a short test of quantitative skills, covering mainly basic arithmetic and the ability to apply it to realistic problems. Scores on the test range from 0 to 500. For example, a person who scores 233 can add the amounts of two checks appearing on a bank deposit slip; someone scoring 325 can determine the price of a meal from a menu; a person scoring 375 can transform a price in cents per ounce into dollars per pound. \(^{4}\) Suppose that you give the NAEP test to an SRS of 840 people from a large population in which the scores have mean 280 and standard deviation S 60. The mean \(\overline{x}\) of the 840 scores will vary if you take repeated samples. (a) Describe the shape, center, and spread of the sampling distribution of \(\overline{x} .\) (b) Sketch the sampling distribution of \(\overline{x}\) . Mark its mean and the values one, two, and three standard deviations on either side of the mean. (c) According to the \(68-95-99.7\) rule, about 95\(\%\) of all values of \(\overline{x}\) lie within a distance \(m\) of the mean of the sampling distribution. What is \(m ?\) Shade the region on the axis of your sketch that is within \(m\) of the mean. (d) Whenever \(\overline{x}\) falls in the region you shaded, the population mean \(\mu\) lies in the confidence interval \(\overline{x} \pm m .\) For what percent of all possible samples does the interval capture \(\mu\) ?

Multiple choice: Select the best answer for Exercises 49 to 52. I collect an SRS of size n from a population and compute a 95% confidence interval for the population proportion. Which of the following would produce a new confidence interval with larger width (larger margin of error) based on these same data? (a) Use a larger confidence level. (b) Use a smaller confidence level. (c) Increase the sample size. (d) Use the same confidence level, but compute the interval n times. Approximately 5% of these intervals will be larger. (e) Nothing can guarantee absolutely that you will get a larger interval. One can only say that the chance of obtaining a larger interval is 0.05.

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