91Ó°ÊÓ

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. The average cost for teeth straightening with metal braces is approximately \(\$ 5400\). A nationwide franchise thinks that its cost is below that figure. A random sample of 28 patients across the country had an average cost of \(\$ 5250\) with a standard deviation of \(\$ 629 .\) At \(\alpha=0.025,\) can it be concluded that the mean is less than \(\$ 5400 ?\)

Step by step solution

01

State the hypotheses and identify the claim

First, we need to set up the null and alternative hypotheses. The null hypothesis \(H_0\) assumes that there is no significant difference from the claimed value, whereas the alternative hypothesis \(H_a\) reflects the claim that needs to be tested. In this case, the null hypothesis is \(H_0: \mu = 5400\) and the alternative hypothesis (the claim) is \(H_a: \mu < 5400\).

02

Find the critical value(s)

Since the significance level \(\alpha\) is 0.025 and we are testing whether the mean is less than a specific value, this is a one-tailed test. We need to find the critical value for a t-distribution with \(n-1\) degrees of freedom, where \(n = 28\). This implies \(df = 27\). Using a t-table or calculator, the critical value \(t_c\) for \(df = 27\) and \(\alpha = 0.025\) is approximately \(-2.052\).

03

Find the test value

We calculate the test statistic using the formula for a one-sample t-test: \[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]Substituting the values, we have \(\bar{x} = 5250\), \(\mu = 5400\), \(s = 629\), and \(n = 28\): \[ t = \frac{5250 - 5400}{\frac{629}{\sqrt{28}}} \approx \frac{-150}{119.0} \approx -1.261 \]

04

Make the decision

We compare the calculated test value of \(t = -1.261\) with the critical value \(t_c = -2.052\). Since \(-1.261 > -2.052\), we do not reject the null hypothesis \(H_0\). This means there is not enough evidence to support the claim that the cost is below \(\$5400\).

05

Summarize the results

The statistical analysis indicates that, at the \(\alpha = 0.025\) level of significance, the average cost of teeth straightening at the nationwide franchise cannot be concluded to be less than \(\$5400\) based on the sample data of 28 patients.

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Key Concepts

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

t-distribution

The t-distribution is an essential concept in statistics, especially when dealing with small sample sizes. It is a probability distribution that is symmetrical and bell-shaped like the normal distribution but has thicker tails. This makes it more suitable for samples where the population standard deviation is unknown, and the sample size is small (typically, less than 30). With increased sample sizes, the t-distribution closely resembles the normal distribution.

Key factors influencing the t-distribution are:

• Degrees of Freedom (df): The shape of the t-distribution depends on the degrees of freedom, which is calculated as the sample size minus one ( -1"). In the context of hypothesis testing for means, it is used to locate the correct t-value in a t-table or calculator.
• Shape: As the degrees of freedom increase, the t-distribution approaches a normal distribution. This is because with larger samples, our estimate of the population standard deviation becomes more precise.

In hypothesis testing, t-distribution is crucial for determining the critical value when assessing the t-test statistic.

significance level

The significance level, denoted as \(\alpha\), is a threshold that determines how strong the evidence must be to reject the null hypothesis. It is a predefined probability level that acts as a cut-off point in hypothesis testing.

Key aspects of significance level include:

• Common Values: Typical significance levels are 0.05, 0.01, and 0.10, although it can vary depending on the specific context or field of study. In the given problem, \(\alpha = 0.025\).
• Decision Making: If your test statistic falls beyond the critical value determined by \(\alpha\), you reject the null hypothesis. For example, a significance level of 0.025 indicates that there is a 2.5% risk of rejecting the null hypothesis when it is actually true.
• Type I Error: Choosing a significance level is a trade-off; a lower value decreases the likelihood of a Type I error (false positive), where you incorrectly reject the null hypothesis.

Hence, the significance level helps determine how much evidence is needed to consider a result statistically significant.

one-sample t-test

A one-sample t-test is utilized to compare the mean of a sample from a single group to a known mean, called the population mean. It is appropriate when data is approximately normally distributed and the population standard deviation is unknown.

The procedure in a one-sample t-test includes:

• Setting Hypotheses: Formulate the null hypothesis (), which assumes no effect or difference, and the alternative hypothesis (), which states the effect or difference.
• Calculating the Test Statistic: Use the formula \[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \], where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.
• Decision Making: Compare the calculated t-value with the critical value from the t-distribution. If the test statistic exceeds the critical value in the direction of the alternative hypothesis, reject the null hypothesis.

Understanding this method is crucial for determining whether a sample mean significantly differs from a known population mean.

critical value

In hypothesis testing, the critical value is a point on the scale of the test statistic beyond which we reject the null hypothesis. It separates the critical region, where the null hypothesis can be rejected, from the non-critical region.

To determine the critical value:

• Use of Significance Level: The given significance level ( ) helps identify how extreme the test statistic must be to fall into the critical region.
• T-distribution Lookup: In this context of t-test, you look up the critical value in a t-distribution table using the degrees of freedom ( -1"). In our example with 27 degrees of freedom and =0.025, the critical value was found to be approximately -2.052.

Once the critical value is obtained, you compare it with the test statistic. If the calculated t-value is more extreme than the critical value, the null hypothesis is rejected.

Understanding critical values is essential for interpreting the results of hypothesis tests and making sound conclusions based on data.