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Chapter 10: Problem 16

Given the following hypothesis: $$\begin{array}{l}H_{0}: \mu=400 \\\H_{1}: \mu \neq 400\end{array}$$ For a random sample of 12 observations, the sample mean was 407 and the sample standard deviation \(6 .\) Using the .01 significance level: a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis?

Short Answer

Expert verified
Reject \(H_0\); the test statistic is significant.

Step by step solution

01

Identify the Type of Test

The hypotheses given are of the form where we are testing whether the population mean \(\mu\) equals 400 or is not equal to 400. This indicates a two-tailed test, as there are two possible alternatives to \(H_0\).
02

State the Decision Rule

For a two-tailed test at the 0.01 significance level with 11 degrees of freedom (since degrees of freedom is \(n - 1 = 12 - 1 = 11\)), find the critical t-values from the t-distribution table. The critical values for \(\alpha = 0.01\) and 11 degrees of freedom are approximately \(-3.106\) and \(3.106\). The decision rule is to reject \(H_0\) if the test statistic \(t\) is less than \(-3.106\) or greater than \(3.106\).
03

Compute the Test Statistic

Use the formula for the t-statistic:\[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{407 - 400}{6/\sqrt{12}} = \frac{7}{1.732} \approx 4.04.\]This calculation involves finding the difference between the sample mean \(\bar{x}\) and the proposed population mean \(\mu_0\), divided by the standard error of the mean.
04

Make a Decision

The calculated test statistic \(t = 4.04\) is greater than the critical value of \(3.106\). According to the decision rule, since \(t\) lies in the rejection region (beyond \(3.106\)), we reject the null hypothesis \(H_0\). This means we conclude there is significant evidence to suggest \(\mu eq 400\).

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Key Concepts

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

Two-Tailed Test
In hypothesis testing, a two-tailed test is applied when we want to determine if there is a statistically significant difference, but we do not specify in advance the direction of the difference. In other words, we test for the possibility of the relationship in both directions. This is illustrated in our problem where the hypotheses are stated as:

\( H_0: \mu = 400 \) and \( H_1: \mu eq 400 \).

A two-tailed test is used here because we are interested in seeing if the mean is different from 400, rather than strictly greater or less.

**Key Features of a Two-Tailed Test:**
  • The null hypothesis (\( H_0 \)) denotes no difference or effect, while the alternative (\( H_1 \)) captures both directions of difference.
  • Two critical regions: one at each end of the distribution, making it possible for the test statistic to fall into the rejection region at either tail.
  • Used when deviations from the mean could occur in both directions.
T-Distribution
The t-distribution is a type of probability distribution that is symmetric and bell-shaped, much like the standard normal distribution, but has thicker tails. It is particularly useful when dealing with small sample sizes (typically less than 30) or when the population standard deviation is not known.

In the context of our exercise, we use the t-distribution because we have a sample size of 12 (less than 30), and the population standard deviation is unknown. Instead, we have a sample standard deviation.

**Properties of T-Distribution:**
  • It is more spread out than the normal distribution, reflecting the additional uncertainty when estimating the standard deviation from a small sample.
  • The shape of the t-distribution changes with sample size: it approaches the normal distribution as the sample size increases.
  • Characterized by degrees of freedom (df), calculated as \( n - 1 \) for a sample size \( n \).
In our problem, we have 11 degrees of freedom (12 observations minus 1).

The t-distribution allows for the calculation of critical values, which in turn help us determine whether our sample mean is significantly different from our tested population mean.
Significance Level
The significance level, often denoted as \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. It essentially reflects the risk the researcher is willing to take for this type of error, also known as a Type I error.

In the problem given, we use a significance level of 0.01. This means we are willing to accept a 1% chance of incorrectly rejecting \( H_0 \) if it is true.

**Understanding Significance Level:**
  • The lower the significance level, the stronger the evidence must be to reject the null hypothesis.
  • Common significance levels are 0.05, 0.01, and 0.10, though these may vary depending on the field or study design.
  • It determines the width of the rejection region in your test, influencing the critical values.
Applying a 0.01 significance level in a two-tailed test leads to two critical regions each with a size of \( \alpha/2 \) or 0.005. This strict criterion ensures that only very strong evidence can lead to the rejection of \( H_0 \).

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Most popular questions from this chapter

A recent article in the Wall Street Journal reported that the 30-year mortgage rate is now less than 6 percent. A sample of eight small banks in the Midwest revealed the following 30 year rates (in percent): \(\begin{array}{llllllll}4.8 & 5.3 & 6.5 & 4.8 & 6.1 & 5.8 & 6.2 & 5.6\end{array}\) At the .01 significance level, can we conclude that the 30 -year mortgage rate for small banks is less than 6 percent? Estimate the \(p\) -value.

Most air travelers now use e-tickets. Electronic ticketing allows passengers to not worry about a paper ticket, and it costs the airline companies less to handle than a paper ticketing. However, in recent times the airlines have received complaints from passengers regarding their e-tickets, particularly when connecting flights and a change of airlines were involved. To investigate the problem an independent watchdog agency contacted a random sample of 20 airports and collected information on the number of complaints the airport had with e-tickets for the month of March. The information is reported below. $$\begin{array}{|rrrrrrrrrr}\hline 14 & 14 & 16 & 12 & 12 & 14 & 13 & 16 & 15 & 14 \\\12 & 15 & 15 & 14 & 13 & 13 & 12 & 13 & 10 & 13 \\\\\hline\end{array}$$ At the . 05 significance level can the watchdog agency conclude the mean number of complaints per airport is less than 15 per month? a. What assumption is necessary before conducting a test of hypothesis? b. Plot the number of complaints per airport in a frequency distribution or a dot plot. Is it reasonable to conclude that the population follows a normal distribution? c. Conduct a test of hypothesis and interpret the results.

The postanesthesia care area (recovery room) at St. Luke's Hospital in Maumee, Ohio, was recently enlarged. The hope was that with the enlargement the mean number of patients per day would be more than \(25 .\) A random sample of 15 days revealed the following numbers of patients. $$\begin{array}{lllllllllllllll}25 & 27 & 25 & 26 & 25 & 28 & 28 & 27 & 24 & 26 & 25 & 29 & 25 & 27 & 24 \\\\\hline\end{array}$$ At the .01 significance level, can we conclude that the mean number of patients per day is more than 25 ? Estimate the \(p\) -value and interpret it.

Hugger Polls contends that an agent conducts a mean of 53 in-depth home surveys every week. A streamlined survey form has been introduced, and Hugger wants to evaluate its effectiveness. The number of in-depth surveys conducted during a week by a random sample of agents are: $$\begin{array}{|lllllllllllllll|}\hline 53 & 57 & 50 & 55 & 58 & 54 & 60 & 52 & 59 & 62 & 60 & 60 & 51 & 59 & 56 \\\\\hline\end{array}$$ At the .05 level of significance, can we conclude that the mean number of interviews conducted by the agents is more than 53 per week? Estimate the \(p\) -value.

The following information is available. $$\begin{array}{l}H_{0}: \mu=50 \\\H_{1}: \mu \neq 50\end{array}$$ The sample mean is \(49,\) and the sample size is \(36 .\) The population follows the normal distribution and the standard deviation is \(5 .\) Use the .05 significance level.
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