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Given the information, \(n=22, \bar{x}=72.3,\) and \(\sigma=6.4\) a. Find the 0.99 confidence interval for \(\mu\). b. Are the assumptions satisfied? Explain.

Short Answer

Expert verified
The 99% confidence interval for \(\mu\) is \((68.79, 75.81)\). However, the assumptions of normal distribution or large sample size and knowing the population’s standard deviation are not fulfilled.

Step by step solution

01

Compute the Z-Score

At a 99% confidence level, a z-score of 2.576 is needed (this value can be looked up on a standard z-table, or using statistical software).
02

Calculate the Margin of Error

The margin of error is calculated using the z-value, the standard deviation and the sample size: \(\text{Margin of Error} = z \frac{\sigma}{\sqrt{n}} = 2.576 \times \frac{6.4}{\sqrt{22}} \approx 3.51\)
03

Calculate Confidence Interval

The confidence interval is calculated using the sample mean and the margin of error: \(\text{CI} = \bar{x} \pm \text{Margin of Error} = 72.3 \pm 3.51\). So, the 99% confidence interval for \(\mu\) (\(\text{CI}_{99}\)) is \((68.79, 75.81)\)
04

Check Assumptions

According to the Central Limit Theorem, if the sample size is large (n > 30), then the sample mean approximates a normal distribution. In this case, the sample size is 22, which is not larger than 30. Therefore, the assumption of normality or large sample size is not met. Additionally, the standard deviation is from the sample, not from the population, which doesn't meet the other assumption. Therefore, both assumptions are not met in this case.

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Key Concepts

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

Z-Score Calculation
The z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. When calculating the z-score in a confidence interval context, you're looking to find the number of standard deviations needed to encapsulate the middle portion of a normal distribution, corresponding to your level of confidence.

In the given exercise, for a 99% confidence level, the z-score is found to be 2.576. This is derived from a z-table or statistical software and represents the value that cuts off the far right end of a standard normal distribution such that 99% of the distribution's area is to the left. The calculation formula is simply the number of standard deviations an element is from the mean, given as: \[ z = \frac{(X - \mu)}{\sigma} \], where X is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
Margin of Error
The margin of error is a vital statistic in giving a range of values when estimating population parameters. It represents the extent to which the observed results can differ from the true value due to sampling variability. The margin of error increases with increasing z-scores (which correspond to higher confidence levels) and decreases with larger sample sizes and smaller standard deviations.

The formula to calculate the margin of error is \[ \text{Margin of Error} = z \frac{\sigma}{\sqrt{n}} \], where z is the z-score associated with the desired confidence level, \( \sigma \) is the standard deviation of the population, and n is the sample size. For the exercise provided, the margin of error was found to be approximately 3.51 by inserting the z-score of 2.576, the population standard deviation of 6.4, and the sample size of 22 into the formula. Therefore, adding and subtracting this margin from the sample mean gives the range in which we're 99% confident that the true mean lies.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics that enables us to use normal probability to make inferences about population parameters based on sample statistics. The theorem states that, given a sufficiently large sample size from a population with a finite level of variance, the mean of all samples from the same population will be approximately normally distributed. What 'sufficiently large' means can vary; however, a common rule of thumb is that a sample size greater than 30 is adequate.

In the exercise, the sample size is 22, which does not meet the 'greater than 30' criteria. Thus, one of the major assumptions of the CLT is not fulfilled. However, if the population itself is normally distributed, or if we were analyzing a population parameter that is believed to be naturally normally distributed (like means of sub-samples), the CLT can be applied even with smaller sample sizes.

The Central Limit Theorem is crucial because it justifies the practice of treating the means of small samples as if they're from a normal distribution, which in turn allows us to use z-scores and create confidence intervals more reliably.

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Most popular questions from this chapter

Because the size of the type I error can always be made smaller by reducing the size of the critical region, why don't we always choose critical regions that make \(\alpha\) extremely small?

One of the best indicators of a baby's health is his or her weight at birth. In the United States, mothers who live in poverty generally have babies with lower birth weights than those who do not live in poverty. Although the average birth weight for babies born in the United States is approximately 3300 grams, the birth weight for babies of women living in poverty is 2800 grams with a standard deviation of 500 grams. Recently, a local hospital introduced an innovative new prenatal care program to reduce the number of low-birth-weight babies born in the hospital. At the end of the first year, the birth weights of 25 randomly selected babies were collected; all of the babies were born to women who lived in poverty and participated in the program. Their mean birth weight was 3075 grams. The question posed to you, the researcher, is, "Has there been a significant improvement in the birth weights of babies born to poor women?" Use \(\alpha=0.02\) a. Define the parameter. b. State the null and alternative hypotheses. c. Specify the hypothesis test criteria. d. Present the sample evidence. e. Find the probability distribution information. f. Determine the results.

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Suppose a hypothesis test is conducted using the \(p^{-}\) value approach and assigned a level of significance of \(\alpha=0.01\) a. How is the 0.01 used in completing the hypothesis test? b. If \(\alpha\) is changed to \(0.05,\) what effect would this have on the test procedure?

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