91Ó°ÊÓ

Suppose you wish to determine if the mean IQ of students on your campus is different from the mean IQ in the general population, \(100 .\) To conduct this study, you obtain a simple random sample of 50 students on your campus, administer an IQ test, and record the results. The mean IQ of the sample of 50 students is found to be 107.3 with a standard deviation of \(13.6 .\) (a) Conduct a hypothesis test (preferably using technology) \(H_{0}: \mu=\mu_{0}\) versus \(H_{1}: \mu \neq \mu_{0}\) for \(\mu_{0}=103,104,105,106,107,108,109,110,111,112\) at the \(\alpha=0.05\) level of significance. For which values of \(\mu_{0}\) do you not reject the null hypothesis? (b) Construct a \(95 \%\) confidence interval for the mean IQ of students on your campus. What might you conclude about how the lower and upper bounds of a confidence interval relate to the values for which the null hypothesis is rejected? (c) Suppose you changed the level of significance in conducting the hypothesis test to \(\alpha=0.01\). What would happen to the range of values of \(\mu_{0}\) for which the null hypothesis is not rejected? Why does this make sense?

Step by step solution

01

Define the Hypotheses

The null hypothesis is defined as: H_0: \( \mu = \mu_0 \) The alternative hypothesis is defined as: H_1: \( \mu eq \mu_0 \)

02

Set the Level of Significance

For this exercise, we use \( \alpha = 0.05 \). This means there is a 5% risk of rejecting the null hypothesis when it is true.

03

Calculate the Test Statistic

The test statistic for a sample mean is calculated using the formula: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] where \( \bar{x} \) is the sample mean, \( \mu_0 \) is the population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size. Plugging in the values: \[ t = \frac{107.3 - \mu_0}{13.6 / \sqrt{50}} \]

04

Determine the Critical Value

Using a t-distribution table and \( \alpha / 2 = 0.025 \) for a two-tailed test with \( df = 49 \) degrees of freedom, the critical t-value is approximately \( \pm 2.009 \).

05

Compare the Test Statistic to the Critical Value

Calculate the test statistic for each \( \mu_0 = 103, 104, 105, 106, 107, 108, 109, 110, 111, 112 \) and determine whether it falls within the range \( -2.009 \) to \( 2.009 \).

06

Identify Values of \( \mu_0 \) for Which \( H_0 \) is Not Rejected

For \( \mu_0 \) values that make the test statistic fall within \( -2.009 \) to \( 2.009 \), the null hypothesis is not rejected.

07

Construct a 95% Confidence Interval

The confidence interval for the mean is calculated using the formula: \[ \bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}} \] where \( \bar{x} \) is the sample mean, \( t_{\alpha/2} \) is the critical value from the t-distribution, \( s \) is the sample standard deviation, and \( n \) is the sample size. Plugging in the values: \[ 107.3 \pm 2.009 \frac{13.6}{\sqrt{50}} \] This results in the interval: \[ 107.3 \pm 3.86 \] or \[ [103.44, 111.16] \]

08

Relate Confidence Interval to Hypothesis Test

The confidence interval \[ [103.44, 111.16] \] indicates that values of \( \mu_0 \) within this range are not rejected at the \( \alpha = 0.05 \) significance level.

09

Analyze Impact of Changing \( \alpha \) to 0.01

A lower significance level \( \alpha = 0.01 \) results in a wider critical value range, from \( -2.677 \) to \( 2.677 \). This would likely narrow down the range of \( \mu_0 \) values for which \( H_0 \) is not rejected.

10

Conclusion

A lower significance level tightens the criteria for rejection, ensuring higher confidence in decisions but reducing the range of accepted \( \mu_0 \) values.

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Key Concepts

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

mean IQ

In this problem, we want to compare the mean IQ of students on a campus to the general population mean IQ, which is 100. The mean (or average) IQ score is a central point around which the IQ scores of individuals within a group vary. We calculated the sample mean IQ from our simple random sample of 50 students, which is found to be 107.3. The sample mean gives us a point estimate of the population mean but it is not enough on its own to determine if there is a significant difference from the known population mean.

t-distribution

The t-distribution plays a crucial role in hypothesis testing, especially when the sample size is small. This distribution is similar to the normal distribution but has thicker tails, which is appropriate when dealing with small sample sizes and unknown population standard deviations. Here, we use the t-distribution because our sample size is 50, and we need to estimate the population standard deviation from the sample. We use the t-distribution to find the critical values for our test statistic, which is calculated using the formula: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] where \( \bar{x} \) is the sample mean, \( \mu_0 \) is the hypothesized population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size.

confidence interval

A confidence interval provides a range of values within which the true population mean is expected to lie with a certain level of confidence. In this problem, we constructed a 95% confidence interval for the mean IQ of students on campus. The confidence interval is calculated using the formula: \[ \bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}} \] where \( \bar{x} \) is the sample mean, \( t_{\alpha/2} \) is the critical value from the t-distribution, \( s \) is the sample standard deviation, and \( n \) is the sample size. By plugging in the values, we get: \[ 107.3 \pm 2.009 \frac{13.6}{\sqrt{50}} \] which results in the confidence interval: \[ 107.3 \pm 3.86 \] or \[ [103.44, 111.16] \]. This means we are 95% confident that the true mean IQ of students on campus lies between 103.44 and 111.16.

level of significance

The level of significance, denoted as \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. In other words, it represents the risk of making a Type I error. For this problem, we used a significance level of \( \alpha = 0.05 \). This means we accept a 5% chance of mistakenly rejecting the null hypothesis. If we change the significance level to \( \alpha = 0.01 \), the criteria to reject the null hypothesis becomes stricter, reducing the risk of Type I error but potentially increasing the range of values for which \( \mu_0 \) is not rejected. This makes our test more conservative.

critical value

The critical value is a threshold that the test statistic must exceed in order for us to reject the null hypothesis. In a two-tailed test, which is used here, there are two critical values: negative and positive. We use a t-distribution table to find these critical values based on our chosen level of significance \( \alpha \) and degrees of freedom \( df \), which is calculated as \( df = n - 1 \). For an \( \alpha = 0.05 \) level of significance and \( df = 49 \), the critical t-value is approximately \( \pm 2.009 \). We then compare our calculated test statistic to these critical values to determine if it falls within the acceptance range, which makes all the values of \( \mu_0 \) where the test statistic falls between -2.009 to 2.009 the values for which the null hypothesis is not rejected.