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Chapter 8: Problem 119

A mathematics test is given to a class of 50 students randomly selected from high school 1 and also to a class of 45 students randomly selected from high school 2 . For the class at high school 1 , the sample mean is 75 points, and the sample standard deviation is 10 points. For the class at high school 2, the sample mean is 72 points, and the sample standard deviation is 8 points. Construct a 95\% confidence interval for the difference in the mean scores. What assumptions are necessary?

Short Answer

Expert verified
The 95% confidence interval for the difference in mean scores is (-0.626, 6.626). Assumptions include normality, random sampling, and equal variances.

Step by step solution

01

Understand the Problem

We need to construct a 95% confidence interval for the difference in mean scores between two independent samples from high school 1 and high school 2.
02

Gather the Necessary Information

- High School 1: sample size \( n_1 = 50 \), sample mean \( \bar{x}_1 = 75 \), and sample standard deviation \( s_1 = 10 \). - High School 2: sample size \( n_2 = 45 \), sample mean \( \bar{x}_2 = 72 \), and sample standard deviation \( s_2 = 8 \). - Confidence level required is 95%.
03

Calculate Standard Error of the Mean Difference

The standard error \( SE \) for the difference between two means is calculated using the formula: \[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]Substitute the values: \[ SE = \sqrt{\frac{10^2}{50} + \frac{8^2}{45}} = \sqrt{\frac{100}{50} + \frac{64}{45}} \] Calculate for simplicity in steps: \[ SE = \sqrt{2 + 1.422} = \sqrt{3.422} \approx 1.85 \]
04

Find Critical Value

For a 95% confidence interval for the difference of two independent means, we use the standard normal distribution (Z-distribution) as both samples are large (\( n_1 = 50 \) and \( n_2 = 45 \)). The critical value for 95% confidence is approximately 1.96.
05

Calculate the Confidence Interval

The formula for the confidence interval is: \[ (\bar{x}_1 - \bar{x}_2) \pm Z \times SE \] Substitute the values: \[ (75 - 72) \pm 1.96 \times 1.85 \] Calculate the margin of error: \[ 3 \pm 3.626 \] Thus, the confidence interval is: \[ (-0.626, 6.626) \]
06

List Necessary Assumptions

1. Both samples are random and independent of each other. 2. The scores are approximately normally distributed, as the sample sizes are large. 3. The variances of scores at both schools are approximately equal (an assumption inherent in assuming equal variances).

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Key Concepts

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

Sample Mean
The sample mean is a fundamental concept in statistics, often used to approximate the average value in a population. Think of it as the average score of a group of individuals. In our exercise, we have two sample means: one for high school 1 and another for high school 2.
- For high school 1, the sample mean is 75, which tells us that, on average, the students scored 75 points on the exam. - Similarly, for high school 2, the sample mean is 72.
Calculating the sample mean involves adding all the scores in a sample and dividing by the number of samples. If you think of it like balancing a seesaw, the sample mean is where the seesaw is perfectly balanced.
Using the sample mean, we can draw insights and make educated guesses about the population average, which is particularly useful when it's infeasible to collect data from the entire population.
Standard Deviation
Standard deviation is a measure of how much variability or spread exists in a set of data. It provides an idea of how much individual scores differ from the mean of the data set. For the current problem, the standard deviation at high school 1 is 10, and at high school 2, it is 8.
- A higher standard deviation (like 10) indicates that scores are more spread out around the mean. - Conversely, a lower standard deviation (like 8) signifies that the scores are closer to the mean.
To calculate standard deviation, you find the square root of the variance, which itself is the average of the squared differences from the Sample Mean.
Understanding standard deviation is crucial when determining how confident we can be in our measurements. It helps us understand the certainty or uncertainty about our sample mean, especially when constructing confidence intervals.
Normal Distribution
In statistics, normal distribution is a common continuous probability distribution, often depicted as a bell-shaped curve. This distribution is symmetric around its mean, which means most of your data points are clustered around the central peak.
The exercise assumes that the test scores are approximately normally distributed. When we say - 'assumes normally distributed', we're implying that the scores should follow a bell-shaped curve with more scores occurring near the mean. - This assumption holds well so long as the sample sizes are large enough, as in our example with 50 and 45 students.
Understanding normal distribution and its properties allows statisticians to make more predictions and conclusions about the population. For example, it supports calculations like confidence intervals, where knowing the properties of the normal distribution helps determine margins of error and critical values.

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Most popular questions from this chapter

Suppose that \(Y_{1}, Y_{2}, \ldots, Y_{n}\) constitute a random sample from a population with probability density function $$f(y)=\left\\{\begin{array}{ll} \left(\frac{1}{\theta+1}\right) e^{-y /(\theta+1)}, & y>0, \theta>-1 \\ 0, & \text { elsewhere } \end{array}\right.$$ Suggest a suitable statistic to use as an unbiased estimator for \(\theta\)

When it comes to advertising, "tweens" are not ready for the hard-line messages that advertisers often use to reach teenagers. The Geppeto Group study \(^{\star}\) found that \(78 \%\) of 'tweens understand and enjoy ads that are silly in nature. Suppose that the study involved \(n=1030\) 'tweens. a. Construct a \(90 \%\) confidence interval for the proportion of 'tweens who understand and enjoy ads that are silly in nature. b. Do you think that "more than \(75 \%\) " of all 'tweens enjoy ads that are silly in nature? Why?

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We can place a 2-standard-deviation bound on the error of estimation with any estimator for which we can find a reasonable estimate of the standard error. Suppose that \(Y_{1}, Y_{2}, \ldots, Y_{n}\) represent a random sample from a Poisson distribution with mean \(\lambda\). We know that \(V\left(Y_{i}\right)=\lambda\), and hence \(E(\bar{Y})=\lambda\) and \(V(\bar{Y})=\lambda / n .\) How would you employ \(Y_{1}, Y_{2}, \ldots, Y_{n}\) to estimate \(\lambda ?\) How would you estimate the standard error of your estimator?

A pharmaceutical manufacturer purchases raw material from two different suppliers. The mean level of impurities is approximately the same for both suppliers, but the manufacturer is concerned about the variability in the amount of impurities from shipment to shipment. If the level of impurities tends to vary excessively for one source of supply, this could affect the quality of the final product. To compare the variation in percentage impurities for the two suppliers, the manufacturer selects ten shipments from each supplier and measures the percentage of impurities in each shipment. The sample variances were \(s_{1}^{2}=.273\) and \(s_{2}^{2}=.094\), respectively. Form a \(95 \%\) confidence interval for the ratio of the true population variances.
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