/*! 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 597 A college dean wants to determin... [FREE SOLUTION] | 91Ó°ÊÓ

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A college dean wants to determine if the students entering the college in a given year have higher IQ's than the students entering the same college in the previous year. The IQ's of the college entrants in the two years are known to be normally distributed and to have equal variances. A random sample of four students from this year's entering class gives IQ scores of \(110,113,116\), and 117. Another random sample of four students from last year's entering class gives IQ scores of \(109,111,112\), and 112 . At the \(5 \%\) level of significance, can we conclude that this year's students have a higher IQ than last years?

Short Answer

Expert verified
At the \(5 \%\) level of significance, we can conclude that this year's students have a higher average IQ than last year's students. We arrived at this conclusion through an independent two-sample t-test, which resulted in a test statistic of \(t = 2.30\), greater than the critical value of \(1.943\). As a result, the null hypothesis was rejected, supporting the claim of a higher average IQ for this year's students.

Step by step solution

01

Identify the null hypothesis and alternative hypothesis

The null hypothesis assumes that there is no significant difference between the average IQs of this year's students and last year's students. The alternative hypothesis states that this year's students have a higher average IQ. \(H_0: \mu_1 = \mu_2\) \(H_a: \mu_1 > \mu_2\) where \(\mu_1\) is the average IQ of this year's students and \(\mu_2\) is the average IQ of last year's students.
02

Calculate the summary statistics

For each sample, we need to calculate the mean and standard deviation. This year's students: - Mean: \(\bar{x}_1 = \frac{110+113+116+117}{4} = 114\) - Standard deviation: \(s_1=\sqrt{\frac{(110-114)^2+(113-114)^2+(116-114)^2+(117-114)^2}{4-1}}=3.51\) Last year's students: - Mean: \(\bar{x}_2 = \frac{109+111+112+112}{4} = 111\) - Standard deviation: \(s_2=\sqrt{\frac{(109-111)^2+(111-111)^2+(112-111)^2+(112-111)^2}{4-1}}=1.54\)
03

Calculate the test statistic

The independent two-sample t-test statistic can be calculated using the following formula: \[t = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\] where \(\bar{x}_1\) and \(\bar{x}_2\) are the means of the two samples, \(s_1\) and \(s_2\) are the standard deviations, and \(n_1\) and \(n_2\) are the sample sizes. In our case, we have: \[t = \frac{(114 - 111) - 0}{\sqrt{\frac{3.51^2}{4} + \frac{1.54^2}{4}}}= 2.30\]
04

Determine the critical value

Since we are performing a one-tailed test (the alternative hypothesis is that \(\mu_1 > \mu_2\)), we will only look at the right tail of the t-distribution. We need to find the critical value for the t-distribution with a \(5 \%\) level of significance. The degrees of freedom can be calculated as \(df = n_1 + n_2 - 2 = 4+4-2=6\). Using a t-table or calculator, we find that \(t_{\alpha=0.05, df=6} = 1.943\).
05

Compare the test statistic to the critical value and make a decision

Our test statistic, \(t = 2.30\), is greater than the critical value, \(1.943\). So, we reject the null hypothesis (\(H_0\)). Since the test statistic is in the rejection region, we conclude that there is enough evidence at the \(5 \%\) level of significance to support the claim that this year's students have a higher average IQ than last year's students.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a method used in statistics to make inferences about a population based on sample data. In this exercise, we are testing whether this year's college entrants have higher IQs than last year's entrants. This involves setting up two opposing statements:
  • The null hypothesis (\(H_0\)) assumes no difference in the average IQs between the two groups. Mathematically, it's written as \(\mu_1 = \mu_2\).
  • The alternative hypothesis (\(H_a\)) suggests that this year's students have a higher average IQ, which is expressed as \(\mu_1 > \mu_2\).
Establishing these hypotheses helps determine the direction and nature of our test. By analyzing sample statistics, we can decide whether there's enough evidence to support the alternative hypothesis.
Normal Distribution
The normal distribution is a probability distribution that is symmetric and bell-shaped, describing how the values of a variable are distributed. Many real-world phenomena, including IQ scores, approximate a normal distribution. When performing hypothesis tests for means, assuming normal distribution is key because:
  • It simplifies calculations since test statistics often rely on the properties of normal or nearly normal distributions.
  • It allows us to apply the Central Limit Theorem, which states that the distribution of sample means tends to be normal, especially with larger sample sizes.
In this exercise, we assume the IQ scores in both years are normally distributed and have equal variances, making it a perfect scenario for using a t-test.
Level of Significance
The level of significance, denoted by \(\alpha\), is the threshold we set to determine whether a statistical result is considered significant. It reflects the probability of rejecting the null hypothesis when it is true, known as a Type I error. In this exercise, we use a \(5\%\) level of significance (\(\alpha = 0.05\)). This means we are willing to accept a \(5\%\) risk of stating that this year's students have higher IQs when there actually is no difference.During hypothesis testing, the test statistic is compared to a critical value, which depends on \(\alpha\). If the test statistic falls beyond this critical value in the direction of the alternative hypothesis, we reject \(H_0\). Here, the calculated \(t\)-value is \(2.30\), which exceeds the critical value of \(1.943\) at \(df = 6\), leading to rejection of the null hypothesis.
Sample Statistics
Sample statistics, such as the mean and standard deviation, provide concrete numerical summaries of the sample data. In hypothesis testing, these statistics are vital in calculating the test statistic.From this year's student sample, the mean IQ is \(114\) and the standard deviation is \(3.51\). Whereas, for last year's sample, the mean IQ is \(111\), with a standard deviation of \(1.54\). These summary statistics feed into the formula for calculating the t-test statistic:\[ t = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]This formula provides a standardized way to compare the means from two samples, considering sample size and variability. Understanding these statistics helps in determining the validity of the hypothesis test.

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Most popular questions from this chapter

Suppose we have a type of battery for which we take a sample of 10 batteries. The mean operating life of these batteries is \(18.0\) hours with a standard deviation of \(3.0\) hours. Suppose also that we have a new type of battery for which we take a sample of 17 batteries. The mean of this sample is \(22.0\) hours with a standard deviation of \(6.0\) hours. Determine for a \(1 \%\) level of significance whether there is a significant difference between the means of the two samples. Also determine at a \(1 \%\) level of significance whether we can conclude that the new batteries are superior to the old ones.

Refer to the data in the previous problem concerning the Selective Service data to test the hypothesis that the proportions are equal against the alternative that \(\pi_{1}>\pi_{2}\). Use \(\alpha=.05\).

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