/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 150 The same achievement test is giv... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 150

The same achievement test is given to soldiers selected at random from two units. The scores they attained are summarized as follows: $$\begin{aligned}&\text { Unit } 1: n_{1}=70, \bar{x}_{1}=73.2, s_{1}=6.1\\\&\text { Unit } 2: n_{2}=60, \bar{x}_{2}=70.5, s_{2}=5.5 \end{aligned}$$ Construct a \(90 \%\) confidence interval for the difference in the mean level of the two units.

Short Answer

Expert verified
The 90% confidence interval for the difference of the mean scores of the two units lies between the resultant boundaries obtained from step 4.

Step by step solution

01

Identify the given data

From the problem, the given data is as follows: For unit 1, \(n_1 = 70\), \(\overline{x}_1 = 73.2\), \(s_1 = 6.1\), For unit 2, \(n_2 = 60\), \(\overline{x}_2 = 70.5\), \(s_2 = 5.5\)
02

Compute the standard error of the difference

The standard error of the difference between two means \(\overline{x}_1\) and \(\overline{x}_2\) can be expressed as: \(\sigma_{\overline{x}_1 - \overline{x}_2} = \sqrt{\frac{{s_1}^2}{n_1} + \frac{{s_2}^2}{n_2}}\), Substituting the given values in, \(\sigma_{\overline{x}_1 - \overline{x}_2} = \sqrt{\frac{{6.1}^2}{70} + \frac{{5.5}^2}{60}}\)
03

Determine the Z-value

As the problem is asking for a 90% confidence interval, there is 5% of the distribution in each tail (given by 100% - 90% = 10% / 2 = 5%). According to the Z-table, a column labeled 0.05 in the tail corresponds to a Z-score of approximately 1.645 (negative or positive depending on which tail).
04

Compute the confidence interval

The formula to find the confidence interval is: \( (\overline{x}_1 - \overline{x}_2) \pm Z_\alpha \times \sigma_{\overline{x}_1 - \overline{x}_2} \), So, substituting the calculated and given values, we obtain: \( (73.2 - 70.5) \pm 1.645 \times \sigma_{\overline{x}_1 - \overline{x}_2} \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Statistical Inference
Statistical inference is the process of drawing conclusions about a population's characteristics based on a sample taken from it. In educational settings, for instance, it allows us to make assertions about a student's proficiency or an educational intervention's effectiveness without needing to test every single individual.

A key tool in statistical inference is the confidence interval, which provides a range of values within which we can be certain, to a specified probability, the true population parameter lies. So, in our problem involving the soldiers' test scores, we're not just interested in the difference in mean scores between the two units; we're using that difference to infer about the performance differences in the larger population these units represent. This approach uses probability to account for the randomness inherent in sampling and provides a measure of certainty鈥攐r confidence鈥攁round our estimate.
Standard Error
The standard error (SE) is a measure that tells us how far we can expect a sample's statistic (like the mean) to be from the actual population parameter. It's critical in gauging the precision of our estimations.

The standard error's role becomes clear when we look at the formula for calculating the confidence interval. It is used in the calculation as a multiplier to the Z-score, allowing us to expand the interval to the appropriate width for our desired level of confidence. Think of it as adjusting our net when fishing; a larger standard error means we cast a wider net (broader confidence interval), reflecting more uncertainty in our sample's representation of the population.
Z-score
A Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations. When constructing confidence intervals, the Z-score helps us determine how 'far out' we need to go on the normal distribution to encapsulate the desired percentage of the data.

In the exercise, we used a Z-score of 1.645 for a 90% confidence interval. Why? Because for a normal distribution, approximately 90% of the data falls within 1.645 standard deviations from the mean. This value comes in handy when we can't reliably estimate the population standard deviation and need to infer it from the sample.
Mean Difference
The mean difference is a straightforward statistic鈥攊t's the difference between the means of two groups being compared. It's essential in experiments and observational studies where we're interested in whether there is a statistically significant difference between two groups.

In our units' test scores problem, we calculated the difference between the average scores of units 1 and 2. This gave us the point estimate around which our confidence interval was built. This interval then provides a range of likely values for the true difference in mean scores, were we to consider the larger population of all such units. By computing the mean difference, we pave the way for inferring this key comparison across the broader population not captured within our sample.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A study in the New England Journal of Medicine reported that based on 987 deaths in southern California, right-handers died at an average age of 75 and left-handers died at an average age of \(66 .\) In addition, it was found that \(7.9 \%\) of the lefties died from accident-related injuries, excluding vehicles, versus \(1.5 \%\) for the right-handers; and \(5.3 \%\) of the left- handers died while driving vehicles versus \(1.4 \%\) of the right-handers. Suppose you examine 1000 randomly selected death certificates, of which 100 were left-handers and 900 were right-handers. If you found that 5 of the left- handers and 18 of the right-handers died while driving a vehicle, would you have evidence to show that the proportion of left-handers who die at the wheel is significantly higher than the proportion of right-handers who die while driving? Calculate the \(p\) -value and interpret its meaning.

Lauren, a brunette, was tired of hearing, "Blondes have more fun." She set out to "prove" that "brunettes are more intelligent." Lauren randomly (as best she could) selected 40 blondes and 40 brunettes at her high school.$$\begin{array}{llll}\hline \text { Blondes } & n_{\beta}=40 & \bar{x}_{\beta i}=88.375 & s_{\beta i}=6.134\\\\\text { Brunettes } & n_{\mathrm{S}}=40 & \bar{x}_{A r}=87.600 & s_{B r}=6.640\\\\\hline\end{array}$$ Upon seeing the sample results, does Lauren have support for her claim that "brunettes are more intelligent than blondes"? Explain. What could Lauren say about blondes' and brunettes' intelligence?

Twenty people were selected to participate in a psychology experiment. They answered a short multiplechoice quiz about their attitudes on a particular subject and then viewed a 45 -minute film. The following day the same 20 people were asked to answer a follow-up questionnaire about their attitudes. At the completion of the experiment, the experimenter will have had two sets of scores. Do these two samples represent dependent or independent samples? Explain.

State the null and alternative hypotheses that would be used to test the following claims: a. There is a difference between the mean age of employees at two different large companies. b. The mean of population 1 is greater than the mean of population 2 c. The mean yield of sunflower seeds per county in North Dakota is less than the mean yield per county in South Dakota. d. There is no difference in the mean number of hours spent studying per week between male and female college students.

The students at a local high school were assigned to do a project for their statistics class. The project involved having sophomores take a timed test on geometric concepts. The statistics students then used these data to determine whether there was a difference between male and female performances. Would the resulting sets of data represent dependent or independent samples? Explain.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.