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Chapter 9: Problem 7

Construct the appropriate confidence interval. A simple random sample of size \(n=12\) is drawn from a population that is normally distributed. The sample mean is found to be \(\bar{x}=45,\) and the sample standard deviation is found to be \(s=14\). Construct a \(90 \%\) confidence interval for the population mean.

Short Answer

Expert verified
The 90% confidence interval for the population mean is approximately (37.746, 52.254).

Step by step solution

01

Identify the given values

The sample size is given as \(n = 12\). The sample mean is \(\bar{x} = 45\), and the sample standard deviation is \(s = 14\). The confidence level is \(90\%\).
02

Determine the degrees of freedom

The degrees of freedom is calculated as \(df = n - 1\). So, \(df = 12 - 1 = 11\).
03

Find the critical value

For a \(90\%\) confidence interval and \(df = 11\), the critical t-value (from the t-distribution table) is approximately \(t_{0.05,11} = 1.796\).
04

Calculate the margin of error

The margin of error (E) is calculated using the formula: \[ E = t_{\alpha/2, df} \left( \frac{s}{\sqrt{n}} \right) \] Plugging in the values: \[ E = 1.796 \left( \frac{14}{\sqrt{12}} \right) \approx 7.254 \]
05

Construct the confidence interval

The confidence interval for the population mean is given by: \[ \bar{x} \pm E \] Plugging in the values: Upper limit: \(45 + 7.254 = 52.254\) Lower limit: \(45 - 7.254 = 37.746\) So, the 90\% confidence interval is approximately \((37.746, 52.254)\).

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Key Concepts

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

Sample Mean
When we talk about the sample mean, we are referring to the average of a set of observations from a sample. This offers a good estimate of the population mean when the sample is representative. In our exercise, the sample mean was found to be \( \bar{x} = 45 \). The formula to calculate the sample mean is simpler than it might seem. You just add all the sample values together and then divide by the number of values (n). This will give you the average, representing the central tendency of your data.
  • If your sample is from a normal distribution, the sample mean (\
T-Distribution
The t-distribution is essential in statistics, especially with small sample sizes. It is similar to the normal distribution but has heavier tails. This means it gives more probability to observations farther from the mean. In our exercise, since our sample size is 12, we use t-distribution instead of the normal distribution. The degrees of freedom (df) influence the shape of the t-distribution, which brings us to our next concept.
  • The t-distribution is used when the sample size is small (typically n<30), and the population standard deviation is unknown.
  • As the sample size approaches infinity, the t-distribution approaches the normal distribution.
Margin of Error
The margin of error tells us how much we can expect the sample mean to vary from the population mean. It helps us understand the potential error in our estimate. This is why it is crucial for constructing a confidence interval.
The formula to calculate the margin of error is \[ E = t_{\alpha/2, df} \left( \frac{s}{\sqrt{n}} \right) \]. Here, \( t_{\alpha/2, df} \) is the critical value from the t-distribution. For our exercise, it was 1.796 because we were constructing a 90% confidence interval with 11 degrees of freedom.
  • The margin of error increases if your sample standard deviation or t-value increases.
  • It is an essential component in determining how confident we can be about the sample representing the population.
Degrees of Freedom
Degrees of freedom (df) refer to the number of values in a calculation that are free to vary. In the context of our exercise, the degrees of freedom were calculated as \( df = n - 1 = 11 \). Degrees of freedom generally depend on the sample size. They are critical when using the t-distribution to find the margin of error and subsequently construct the confidence interval.
  • The degrees of freedom influence the shape of the t-distribution; fewer degrees of freedom result in a distribution with thicker tails.
  • The calculation of df changes based on the context, but for a single sample mean, it follows the formula \( df = n - 1 \).

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Most popular questions from this chapter

IQ scores are known to be approximately normally distributed with mean 100 and standard deviation \(15 .\) (a) Simulate obtaining a random sample of 12 IQ scores from this population. (b) Use the data from part (a) to construct a \(95 \%\) confidence interval for the mean IQ using Student's \(t\) -distribution. (c) Use the data from part (a) to obtain 1000 bootstrap samples. For each sample, find the mean. (d) Determine an estimate of the standard error of the mean from the 1000 bootstrap means found in part (c). Compare this result to the theoretical standard error of the mean, \(\frac{\sigma}{\sqrt{n}}\). Compare this result to the estimate of the standard error of the mean based on the sample data, \(\frac{s}{\sqrt{n}}\). (e) Construct a \(95 \%\) confidence interval for the mean IQ using the bootstrap sample from part (c).

Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 20 students, she finds 2 who eat cauliflower. Obtain and interpret a \(95 \%\) confidence interval for the proportion of students who eat cauliflower on Jane's campus using Agresti and Coull's method.

Katrina wants to estimate the proportion of adult Americans who read at least 10 books last year. To do so, she obtains a simple random sample of 100 adult Americans and constructs a \(95 \%\) confidence interval. Matthew also wants to estimate the proportion of adult Americans who read at least 10 books last year. He obtains a simple random sample of 400 adult Americans and constructs a \(99 \%\) confidence interval. Assuming both Katrina and Matthew obtained the same point estimate, whose estimate will have the smaller margin of error? Justify your answer.

Indicate whether a confidence interval for a proportion or mean should be constructed to estimate the variable of interest. Justify your response. Researchers at the Gallup Organization asked a random sample 1016 adult Americans aged 21 years or older, "Right now, do you think the state of moral values in the country as a whole is getting better, or getting worse?"

In a survey conducted by the marketing agency 11 mark, 241 of 1000 adults 19 years of age or older confessed to bringing and using their cell phone every trip to the bathroom (confessions included texting and answering phone calls). (a) What is the sample in this study? What is the population of interest? (b) What is the variable of interest in this study? Is it qualitative or quantitative? (c) Based on the results of this survey, obtain a point estimate for the proportion of adults 19 years of age or older who bring their cell phone every trip to the bathroom. (d) Explain why the point estimate found in part (c) is a statistic. Explain why it is a random variable. What is the source of variability in the random variable? (e) Construct and interpret a \(95 \%\) confidence interval for the population proportion of adults 19 years of age or older who bring their cell phone every trip to the bathroom. (f) What ensures that the results of this study are representative of all adults 19 years of age or older?
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