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The target temperature for a hot beverage the moment it is dispensed from a vending machine is \(170^{\circ} \mathrm{F}\). A sample of ten randomly selected servings from a new machine undergoing a pre- shipment inspection gave mean temperature \(173^{\circ} \mathrm{F}\) with sample standard deviation \(6.3^{\circ} \mathrm{F}\). a. Assuming that temperature is normally distributed, perform the test that the mean temperature of dispensed beverages is different from \(170^{\circ} \mathrm{F}\), at the \(10 \%\) level of significance. b. The sample mean is greater than 170 , suggesting that the actual population mean is greater than \(170^{\circ} \mathrm{F}\). Perform this test, also at the \(10 \%\) level of significance. (The computation of the test statistic done in part (a) still applies here.)

Short Answer

Expert verified
a) Fail to reject \( H_0 \); not different. b) Reject \( H_0 \); mean is greater.

Step by step solution

01

Determine Hypotheses for Part (a)

For the two-tailed test, we set up the hypotheses:- Null hypothesis, \( H_0: \mu = 170 \)- Alternative hypothesis, \( H_a: \mu eq 170 \)where \( \mu \) represents the true mean temperature of the beverages.
02

Calculate the Test Statistic for Part (a)

The test statistic for the sample mean is calculated using the formula for a t-test:\[ t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}\]where \( \bar{x} = 173 \), \( \mu_0 = 170 \), \( s = 6.3 \), and \( n = 10 \).\[ t = \frac{173 - 170}{\frac{6.3}{\sqrt{10}}} \approx \frac{3}{1.992} \approx 1.506 \]
03

Determine Critical Value and Decision for Part (a)

Using a t-distribution with \( n - 1 = 9 \) degrees of freedom and a significance level of \( \alpha = 0.10 \), the critical t-value for the two-tailed test is approximately \( \pm 1.833 \).Since \( 1.506 \) is less than \( 1.833 \), we fail to reject the null hypothesis. Thus, there is not enough evidence to say that the mean temperature differs from \( 170^{\circ} \mathrm{F} \).
04

Determine Hypotheses for Part (b)

For the one-tailed test, we set up the hypotheses:- Null hypothesis, \( H_0: \mu = 170 \)- Alternative hypothesis, \( H_a: \mu > 170 \)
05

Determine Critical Value and Decision for Part (b)

Using a t-distribution with \( n - 1 = 9 \) degrees of freedom and a significance level of \( \alpha = 0.10 \), the critical t-value for a right-tailed test is approximately \( 1.383 \).Since \( 1.506 \) is greater than \( 1.383 \), we reject the null hypothesis. This suggests that the mean temperature is greater than \( 170^{\circ} \mathrm{F} \).

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Key Concepts

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

Understanding the Null Hypothesis
The null hypothesis is a fundamental concept in statistics that serves as a starting point for any hypothesis test. In the context of a t-test, it represents the assumption that there is no effect or no difference. For testing the mean temperature of beverages, the null hypothesis (\( H_0 \)) might state that the mean temperature (\( \mu \)) is equal to the target temperature, such as 170°F.

  • The purpose of the null hypothesis is to propose a position of no difference, which can be tested against the data.
  • It is often a default or baseline assumption.
  • Rejecting the null hypothesis suggests that there is enough statistical evidence to support the alternative hypothesis.

When conducting experiments or data analysis, scientists start with the null hypothesis, and then perform statistical tests to see if there is enough evidence to reject it. Rejection depends on the results of the test statistic compared to a critical value.

In hypothesis testing steps, the null hypothesis serves as a pivot point to measure outcomes and helps in maintaining objectivity.
Exploring the Alternative Hypothesis
The alternative hypothesis (\( H_a \)) is the complement of the null hypothesis. It purports the effect or a significant change. In a t-test, this hypothesis suggests that there is a significant statistical difference in the parameter being tested, like the mean temperature.

For example, when considering the beverage's temperature:
  • A two-tailed alternative hypothesis might state that the mean is not equal to the target, \( \mu eq 170 \)°F.
  • A one-tailed alternative might claim that the mean temperature is greater than the target, \( \mu > 170 \)°F.

Choosing between one-tailed and two-tailed tests depends on the direction of interest:
  • Use a two-tailed test when you are interested in deviations in both directions.
  • Use a one-tailed test when you only want to detect deviation in one direction.

In hypothesis tests, the aim is often to gather evidence to support the alternative hypothesis, by reducing the chances of Type I errors (incorrectly rejecting a true null hypothesis).
Significance Level Demystified
The significance level, denoted by \( \alpha \), gauges how confident we are about our test results by setting a threshold for statistical decision-making. It delineates the probability of rejecting the null hypothesis given it is true.

  • The most common significance levels are 0.05, 0.01, or 0.10.
  • An \( \alpha = 0.10 \) allows for a 10% risk of concluding that a difference exists, when in fact it does not.
  • A lower \( \alpha \) level means stricter criteria for rejection, lowering the risk of false positives but making it harder to detect a true difference.

In our beverage temperature example, using \( \alpha = 0.10 \) means that there's a 10% chance of observing more extreme values if the null hypothesis holds. This helps in balancing test sensitivity with decision confidence.

Ultimately, the significance level shapes how we interpret and trust the statistical hypothesis test's outcomes, influencing whether the null hypothesis is rejected or accepted based on the data.
Unveiling the Critical Value
Critical value is a key concept in hypothesis tests, establishing the point beyond which we reject the null hypothesis. It is contingent on the chosen significance level and the distribution of the test statistic.

  • A critical value works like a boundary in a statistical test—lying beyond it suggests a significant result.
  • For a t-test, the critical value depends on degrees of freedom, significance level, and whether it’s a one-tailed or two-tailed test.
  • With a 10% level of significance and 9 degrees of freedom, the critical values are approximately ±1.833 for two-tailed tests and 1.383 for right-tailed tests.

In practice, we compute the test statistic based on sample data and compare it to the critical value:
  • If the statistic exceeds the critical value (in the direction of the alternative hypothesis), the null hypothesis is rejected.
  • If it does not exceed, the evidence is insufficient to reject the null hypothesis.

Thus, critical values provide a standardized decision framework, helping us determine whether our sample data provides enough evidence against the null hypothesis.

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Most popular questions from this chapter

Perform the indicated test of hypotheses, based on the information given. a. Test \(\mathrm{Ho}: \mu=212\) vs. Ha: \(\mu<212 @ \alpha=0.10, \sigma\) unknown, \(n=36, x-211.2, \mathrm{~s}=2.2\) b. Test \(H_{0}: \mu=-18\) vs. Ha: \(\mu>-18\) \(@ \alpha=0.05, \sigma=3.3, n=44, x-=-17.2, s=3.1\) c. Test \(H_{0}: \mu=24\) vs. Ha: \(\mu \neq 24 @ \alpha=0.02, \sigma\) unknown, \(n=50, x-=22.8, s=1.9\)

State the null and alternative hypotheses for each of the following situations. (That is, identify the correct number \(\mu_{0}\) and write \(H_{0} \mu=\mu_{0}\) and the appropriate analogous expression for \(H_{a}\).) a. The average time workers spent commuting to work in Verona five years ago was 38.2 minutes. The Verona Chamber of Commerce asserts that the average is less now. b. The mean salary for all men in a certain profession is \(\$ 58,291\). A special interest group thinks that the mean salary for women in the same profession is different. c. The accepted figure for the caffeine content of an 8 -ounce cup of coffee is \(133 \mathrm{mg}\). A dietitian believes that the average for coffee served in a local restaurants is higher. d. The average yield per acre for all types of corn in a recent year was 161.9 bushels. An economist believes that the average yield per acre is different this year. e. An industry association asserts that the average age of all self-described fly fishermen is 42.8 years. A sociologist suspects that it is higher.

Five years ago \(3.9 \%\) of children in a certain region lived with someone other than a parent. A sociologist wishes to test whether the current proportion is different. Perform the relevant test at the \(5 \%\) level of significance using the following data: in a random sample of 2,759 children, 119 lived with someone other than a parent.

A calculator has a built-in algorithm for generating a random number according to the standard normal distribution. Twenty-five numbers thus generated have mean 0.15 and sample standard deviation 0.94 . Test the null hypothesis that the mean of all numbers so generated is 0 versus the alternative that it is different from 0 , at the \(20 \%\) level of significance. Assume that the numbers do follow a normal distribution.

A report five years ago stated that \(35.5 \%\) of all state-owned bridges in a particular state were "deficient." An advocacy group took a random sample of 100 state-owned bridges in the state and found 33 to be currently rated as being "deficient." Test whether the current proportion of bridges in such condition is \(35.5 \%\) versus the alternative that it is different from \(35.5 \%,\) at the \(10 \%\) level of significance.

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