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Answer the questions: (a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. A sample of 36 observations is selected from a normal population. The sample mean is 21 , and the population standard deviation is 5 . Conduct the following test of hypothesis using the .05 significance level. $$ \begin{array}{l} H_{0}: \mu \leq 20 \\ H_{1}: \mu>20 \end{array} $$

Short Answer

Expert verified
One-tailed test; do not reject \\(H_0\\); p-value is 0.1151, indicating insufficient evidence.

Step by step solution

01

Identify Test Type

This hypothesis test involves checking if the population mean is greater than 20. Since the alternative hypothesis is \(H_{1}: \mu > 20\), this is a one-tailed test focusing on values greater than 20.
02

Define Decision Rule

The decision rule involves comparing the test statistic to a critical value from the standard normal distribution (Z-distribution) with a significance level of 0.05. For a one-tailed test at \alpha = 0.05\, reject \(H_0\) if the Z-statistic is greater than 1.645.
03

Calculate Test Statistic

The test statistic is calculated using the formula: \[ Z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \] where \bar{x} = 21\ (sample mean), \mu_0 = 20\ (hypothesized mean), \sigma = 5\ (population standard deviation), and \(n = 36\) (sample size).Plugging in the values:\[ Z = \frac{21 - 20}{\frac{5}{\sqrt{36}}} = \frac{1}{\frac{5}{6}} = \frac{1}{0.8333} \approx 1.20 \]
04

Make Decision Regarding H0

The calculated Z-statistic is approximately 1.20. According to the decision rule, since 1.20 is less than the critical value 1.645, we do not reject \(H_0\).
05

Determine p-value and Interpret

The p-value corresponds to the probability of observing a sample mean as extreme as 21 or more, assuming \(H_0\) is true. Using standard normal distribution tables or software, the p-value for Z = 1.20 is approximately 0.1151. Since the p-value (0.1151) is greater than \alpha = 0.05\, we do not reject \(H_0\). This means there is not enough evidence to conclude that the population mean is greater than 20.

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Key Concepts

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

One-Tailed Test
When conducting a hypothesis test, one of the first steps is determining whether to use a one-tailed or two-tailed test. In this scenario, the alternative hypothesis is that the population mean is greater than 20, represented as \(H_{1}: \mu > 20\). This indicates we are only interested in deviations in one direction (greater than 20). Thus, this is known as a one-tailed test.

A one-tailed test checks if the parameter falls only in one direction from the hypothesized value. It is more powerful for detecting an effect in that specific direction compared to a two-tailed test, which considers deviations on both sides.

If the direction was not specified, or we were interested in deviations on both sides, we would use a two-tailed test. However, here, we have a clear direction defined in the alternative hypothesis, confirming it's a one-tailed test.
Test Statistic
The test statistic is a crucial part of hypothesis testing. It's a standardized value that helps compare your sample results against the null hypothesis. For a population mean, when the population standard deviation is known, we use the formula for the Z-statistic:

\[ Z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \]

Here, the sample mean is \(\bar{x} = 21\), the hypothesized mean \(\mu_0 = 20\), the population standard deviation \(\sigma = 5\), and the sample size \(n = 36\).

Plugging these values into the formula gives us a test statistic \(Z \approx 1.20\). This statistic tells us how many standard deviations the sample mean is above the hypothesized mean. A larger Z would suggest more evidence against the null hypothesis, but here, the calculated Z of 1.20 is not enough on its own to reject \(H_0\).
Decision Rule
The decision rule in hypothesis testing is essential for determining whether to accept or reject the null hypothesis. It uses the test statistic and the significance level (\(\alpha\)) to make this decision.

For a one-tailed test at \(\alpha = 0.05\), the critical Z-value is 1.645. If the calculated test statistic is greater than this critical value, we reject the null hypothesis. In our example, the calculated Z is 1.20, which is less than 1.645.

This decision rule helps maintain the integrity of the test by setting a threshold for error. It minimizes the chances of making a Type I error (incorrectly rejecting a true null hypothesis).

Since 1.20 is not greater than 1.645, we adhere to the decision rule and do not reject the null hypothesis.
p-value Interpretation
The p-value is a probability that provides an alternative way to make decisions in hypothesis testing. It measures how extreme the test statistic is, under the assumption that the null hypothesis is true.

For the calculated Z of 1.20, the p-value is approximately 0.1151. This figure represents the probability of observing a sample mean as extreme as 21 or more, assuming the null hypothesis \(H_0\) is true.

A p-value is compared to the significance level \(\alpha\) to decide on rejecting \(H_0\). If the p-value is less than or equal to \(\alpha\), we reject \(H_0\); otherwise, we do not reject it.

In this case, the p-value (0.1151) is greater than \(\alpha = 0.05\). This means we do not have enough evidence to reject \(H_0\). Simply put, there's not enough statistical proof to say that the population mean is greater than 20.

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Most popular questions from this chapter

(a) State the null hypothesis and the alternate hypothesis. (b) State the decision rule. (c) Compute the value of the test statistic. (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. The waiting time for customers at MacBurger Restaurants follows a normal distribution with a population standard deviation of 1 minute. At the Warren Road MacBurger, the quality assurance department sampled 50 customers and found that the mean waiting time was 2.75 minutes. At the . 05 significance level, can we conclude that the mean waiting time is less than 3 minutes?

(a) State the null hypothesis and the alternate hypothesis. (b) State the decision rule. (c) Compute the value of the test statistic. (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. A recent national survey found that high school students watched an average (mean) of 6.8 movies per month with a population standard deviation of 1.8 . The distribution of number of movies watched per month follows the normal distribution. A random sample of 36 college students revealed that the mean number of movies watched last month was 6.2 . At the .05 significance level, can we conclude that college students watch fewer movies a month than high school students?

For Exercises 5-8: (a) State the null hypothesis and the alternate hypothesis. (b) State the decision rule. (c) Compute the value of the test statistic. (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. The manufacturer of the \(X-15\) steel-belted radial truck tire claims that the mean mileage the tire can be driven before the tread wears out is 60,000 miles. Assume the mileage wear follows the normal distribution and the standard deviation of the distribution is 5,000 miles. Crosset Truck Company bought 48 tires and found that the mean mileage for its trucks is 59,500 miles. Is Crosset's experience different from that claimed by the manufacturer at the .05 significance level?

According to the Census Bureau, 3.13 people reside in the typical American household. A sample of 25 households in Arizona retirement communities showed the mean number of residents per household was 2.86 residents. The standard deviation of this sample was 1.20 residents. At the .05 significance level, is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.13 persons?

For a recent year, the mean fare to fly from Charlotte, North Carolina, to Chicago, Illinois, on a discount ticket was \(\$ 267 .\) A random sample of 13 round-trip discount fares on this route last month shows: \(\begin{array}{llllllllll}\$ 321 & \$ 286 & \$ 290 & \$ 330 & \$ 310 & \$ 250 & \$ 270 & \$ 280 & \$ 299 & \$ 265 & \$ 291 & \$ 275 & \$ 281\end{array}\) At the .01 significance level, can we conclude that the mean fare has increased? What is the \(p\) -value?

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