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Chapter 8: Problem 18

One water quality standard for water that is discharged into a particular type of stream or pond is that the average daily water temperature be at most \(18^{\circ} \mathrm{C}\). Six samples taken throughout the day gave the data: $$ 16.821 .519 .112 .818 .020 .7 $$ The sample mean \(x^{\wedge-=18.15}\) exceeds \(18,\) but perhaps this is only sampling error. Determine whether the data provide sufficient evidence, at the \(10 \%\) level of significance, to conclude that the mean temperature for the entire day exceeds \(18^{\circ} \mathrm{C}\).

Short Answer

Expert verified
There is sufficient evidence to conclude that the mean temperature exceeds \(18^{\circ} \mathrm{C}\) at the 10% significance level.

Step by step solution

01

State the Hypotheses

We need to determine whether there is sufficient evidence that the mean daily water temperature exceeds \(18^{\circ} \mathrm{C}\). The null hypothesis \(H_0\) and the alternative hypothesis \(H_a\) can be stated as follows: \[H_0: \mu \leq 18\]\[H_a: \mu > 18\]
02

Identify the Significance Level

The level of significance is given as \(10\%\). This means we are using \(\alpha = 0.10\). This alpha level is our threshold for determining the cutoff for rejecting the null hypothesis.
03

Calculate the Sample Mean

The sample mean is already given as \(\bar{x} = 18.15\). We use this to test our hypothesis.
04

Calculate the Sample Standard Deviation

Calculate the sample standard deviation \(s\) using the following formula, where the sample data are \(n = 6\).\[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\]However, the standard deviation is not provided here and also can be determined through calculation: \[s \approx 0.38\].
05

Calculate the Test Statistic

Use the sample mean, the hypothesized population mean, the standard deviation, and the sample size to calculate the t-statistic.\[t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{18.15 - 18}{0.38/\sqrt{6}}\]The calculated t-statistic value is approximately \(1.65\).
06

Determine the Critical Value

Since \(n = 6\), degrees of freedom are \(n-1 = 5\). Using a t-distribution table for \(\alpha = 0.10\) and 5 degrees of freedom, find the critical t-value for a one-tailed test. The critical value is approximately \(t_{critical} = 1.476\).
07

Make a Decision

Compare the calculated t-statistic from Step 5 with the critical value from Step 6. Since the t-statistic \(1.65\) is greater than the critical value \(1.476\), we reject the null hypothesis.

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Key Concepts

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

Critical Value
The term "critical value" is crucial in hypothesis testing. It is a point on the test's distribution that is compared to the calculated test statistic to decide whether to reject the null hypothesis.
In our exercise, we use the critical value to determine if the sample mean suggests a significant deviation from the hypothesized population mean.
  • The critical value helps define the boundary for the critical region, along which rejecting the null hypothesis becomes rational.
  • We look up the critical value using a t-distribution table, given an alpha level, here \( \alpha = 0.10 \), and degrees of freedom \( df = n-1 \), where \( n \) is the sample size.
  • For our example, at a \( 10\% \) level of significance with \( df = 5 \), the critical value is approximately \( t_{\text{critical}} = 1.476 \).
Null Hypothesis
In statistical hypothesis testing, we begin by stating the null hypothesis. This hypothesis represents a statement of no effect or no difference, often symbolized as \( H_0 \).
It serves as a baseline or a starting assumption.
  • The null hypothesis for our exercise is \( H_0: \mu \leq 18 \), suggesting that the average daily temperature does not exceed \( 18^\circ \text{C} \).
  • This is put against an alternative hypothesis, here \( H_a: \mu > 18 \), indicating that the mean temperature might exceed \( 18^\circ \text{C} \).
  • If the null hypothesis is rejected, it implies that the sample data provides sufficient evidence to favor the alternative hypothesis.
Rejection or acceptance of the null hypothesis is what guides the conclusion of the test.
T-Statistic
The t-statistic is a crucial part of hypothesis testing, quantifying the distance between a sample mean and the population mean, measured in terms of the sample's variability.
It helps determine if there is enough evidence to reject the null hypothesis.
  • Calculating the t-statistic involves the formula \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \), where \( \bar{x} \) is the sample mean, \( \mu \) is the hypothesized population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size.
  • In our example, the sample mean is \( 18.15 \), \( \mu = 18 \), and \( s \approx 0.38 \), resulting in a t-statistic of approximately \( 1.65 \).
  • The t-statistic is then compared to the critical value to decide on the null hypothesis.
A higher t-statistic implies greater evidence against \( H_0 \).
Sample Mean
The sample mean, denoted as \( \bar{x} \), is a measure of central tendency, helping summarize data by providing the average point.
It acts as an estimate of the population mean from which the sample is drawn.
  • In the context of hypothesis testing, the sample mean \( \bar{x} = 18.15 \) is compared against the benchmark or hypothesized population mean \( \mu \).
  • It is computed straightforwardly by summing all sample observations and dividing by the total number of samples, here \( n = 6 \).
  • A sample mean greater than the hypothesized value could suggest a significant difference, though this needs to be tested.
The sample mean's difference from the hypothesized mean forms the basis of hypothesis tests.
Level of Significance
The level of significance, denoted by \( \alpha \), is the threshold used in hypothesis testing that defines the probability of rejecting the null hypothesis when it is actually true.
It reflects the risk we're willing to take in making a mistake.
  • The chosen \( \alpha \) influences the critical value set for the test, with a smaller \( \alpha \) presenting a stricter criterion for rejection.
  • In our example, \( \alpha = 0.10 \), meaning we're accepting a \( 10\% \) chance of wrongly rejecting \( H_0 \) if it's correct.
  • This value is a subjective choice and often depends on the context and consequences of potential errors.
This balance of risk informs the precision and rigor of the statistical conclusions.

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Most popular questions from this chapter

State the null and alternative hypotheses for each of the following situations. (That is, identify the correct number \(\mu_{0}\) and write \(H_{0} \cdot \mu=\mu_{0}\) and the appropriate analogous expression for \(H_{a}\).) a. The average July temperature in a region historically has been \(74.5^{\circ} \mathrm{F}\). Perhaps it is higher now. b. The average weight of a female airline passenger with luggage was 145 pounds ten years ago. The FAA believes it to be higher now. c. The average stipend for doctoral students in a particular discipline at a state university is \(\$ 14,756\). The department chairman believes that the national average is higher. d. The average room rate in hotels in a certain region is \(\$ 82.53 .\) A travel agent believes that the average in a particular resort area is different. e. The average farm size in a predominately rural state was 69.4 acres. The secretary of agriculture of that state asserts that it is less today.

Perform the indicated test of hypotheses, based on the information given. a. Test \(H_{0: \mu}=105\) vs. Ha: \(\mu>105 @ \alpha=0.05, \sigma\) unknown, \(n=30, x-=108, s=7.2\) b. Test \(H 0: \mu=21.6\) vs. Ha: \(\mu<21.6 @ \alpha=0.01, \sigma\) unknown, \(n=78, x-=20.5, s=3.9\) c. Test \(H 0: \mu=-375\) VS. Ha: \(\mu \neq-375 @ \alpha=0.01, \sigma=18.5, n=31, x-=-388, s=18.0\)

The mean household income in a region served by a chain of clothing stores is $$\$ 48,750 .$$ In a sample of 40 customers taken at various stores the mean income of the customers was $$\$ 51,505$$ with standard deviation $$\$ 6,852 .$$ a. Test at the \(10 \%\) level of significance the null hypothesis that the mean household income of customers of the chain is $$\$ 48,750$$ against that alternative that it is different from $$\$ 48,750.$$ b. The sample mean is greater than $$\$ 48,750,$$ suggesting that the actual mean of people who patronize this store is greater than $$\$ 48,750 .$$ Perform this test, also at the \(10 \%\) level of significance. (The computation of the test statistic done in part (a) still applies here.)

In the past the average length of an outgoing telephone call from a business office has been 143 seconds. A manager wishes to check whether that average has decreased after the introduction of policy changes. A sample of 100 telephone calls produced a mean of 133 seconds, with a standard deviation of 35 seconds. Perform the relevant test at the \(1 \%\) level of significance.

Compute the value of the test statistic for the indicated test, based on the information given. a. Testing \(H_{0}: \mu=72.2\) VS. Ha: \(\mu>72.2, \sigma\) unknown, \(n=55, x-=75.1, s=9.25\) b. Testing \(H 0: \mu=58\) vs. Ha: \(\mu>58, \sigma=1.22, n=40, x-58.5, s=1.29\) c. Testing \(H 0: \mu=-19.5\) VS. Ha: \(\mu<-19.5, \sigma\) unknown, \(n=30, x-=-23.2, s=9.55\) d. Testing \(H_{0}: \mu=805\) Vs. \(H a: \mu \neq 805, \sigma=37.5, n=75, x-=818, s=36.2\)
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