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Joan's Nursery specializes in custom-designed landscaping for residential areas. The estimated labor cost associated with a particular landscaping proposal is based on the number of plantings of trees, shrubs, and so on to be used for the project. For costestimating purposes, managers use two hours of labor time for the planting of a mediumsized tree. Actual times from a sample of 10 plantings during the past month follow (times in hours). $$\begin{array}{llllllll} 1.7 & 1.5 & 2.6 & 2.2 & 2.4 & 2.3 & 2.6 & 3.0 & 1.4 & 2.3 \end{array}$$ With a .05 level of significance, test to see whether the mean tree-planting time differs from two hours. a. State the null and alternative hypotheses. b. Compute the sample mean. c. Compute the sample standard deviation. d. What is the \(p\) -value? e. What is your conclusion?

Short Answer

Expert verified
Do not reject the null hypothesis; the mean tree-planting time does not significantly differ from 2 hours.

Step by step solution

01

State the Hypotheses

The null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)) for the test are:- \(H_0: \mu = 2\) hours (The mean tree-planting time is equal to 2 hours)- \(H_1: \mu eq 2\) hours (The mean tree-planting time is different from 2 hours)
02

Compute the Sample Mean

We calculate the sample mean (\(\bar{x}\)) using the given data:\[\bar{x} = \frac{1.7 + 1.5 + 2.6 + 2.2 + 2.4 + 2.3 + 2.6 + 3.0 + 1.4 + 2.3}{10} = \frac{22}{10} = 2.2\]The sample mean is 2.2 hours.
03

Compute the Sample Standard Deviation

The sample standard deviation (\(s\)) is calculated as follows:\[s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2\]Substituting the values, we find:\[s^2 = \frac{1}{9} [(1.7-2.2)^2 + (1.5-2.2)^2 + \ldots + (2.3-2.2)^2]\]\[s^2 = \frac{1}{9} [0.25 + 0.49 + 0.16 + 0.04 + 0.04 + 0.01 + 0.16 + 0.64 + 0.64 + 0.01]\]\[s^2 = \frac{1}{9} [2.44] = 0.2711\]\[s = \sqrt{0.2711} \approx 0.52\]The sample standard deviation is approximately 0.52 hours.
04

Compute the Test Statistic

The test statistic for a t-test is computed as:\[t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\]Plugging in the values we have:\[t = \frac{2.2 - 2}{\frac{0.52}{\sqrt{10}}} = \frac{0.2}{0.164}\approx 1.22\]
05

Determine the p-value and Conclusion

Using a t-distribution table or calculator with 9 degrees of freedom (\(n-1=10-1\)), find the p-value for \(t\approx 1.22\). The p-value is greater than 0.05.Since the p-value is greater than the significance level of 0.05, we do not reject the null hypothesis. There is insufficient evidence to conclude that the mean tree-planting time differs from 2 hours.

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Key Concepts

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

Statistical Analysis
Statistical analysis is the process of collecting and interpreting data to uncover patterns and trends. It is a fundamental component of hypothesis testing, where researchers aim to determine whether there is enough evidence to support or refute a specific statement, called the hypothesis.

In this context, the null hypothesis (\(H_0\)) proposes that there is no effect or difference, and any observed variation is due to chance. The alternative hypothesis (\(H_1\)) suggests that there is an effect or a difference. By using a significance level (like 0.05 in our case), researchers define the threshold for rejecting the null hypothesis. A smaller significance level indicates stricter criteria to conclude a significant effect.

The outcome of statistical analysis guides decision-making. In Joan's Nursery example, the analysis helps managers decide whether the average tree-planting time is consistent with the estimated labor costs. If the null hypothesis is not rejected, they may conclude that the mean planting time does not significantly deviate from two hours.
T-Test
The t-test is a statistical test used to determine if there is a significant difference between the means of two groups. It is particularly useful when dealing with small sample sizes and when the population standard deviation is unknown.

In Joan's Nursery problem, the t-test is applied to assess whether the mean tree-planting time is significantly different from two hours. Here, the sample data has 10 observations, which is small, making the t-test appropriate.

The test statistic is calculated as:
  • First, determine the difference between the sample mean, \(\bar{x}\), and the hypothesized population mean, \(\mu\).
  • Normalize this difference by dividing it by the estimated standard error of the mean, \(s / \sqrt{n}\), where \(s\) is the sample standard deviation, and \(n\) is the sample size.
  • This results in a t-value, which is compared against critical values from the t-distribution tables based on the degrees of freedom (in this case, \(9\), since \(n-1 = 10-1\)).

If the p-value derived from this t-value is greater than the significance level, the null hypothesis is not rejected, indicating that the observed mean is not statistically different from the hypothesized mean of two hours.
Sample Standard Deviation
Sample standard deviation is a measure of the amount of variation or dispersion in a set of sample observations. It provides insight into how spread out the data points are around the mean.

Calculating the sample standard deviation involves several steps:
  • Subtract the sample mean from each data point to find the deviation of each observation.
  • Square each deviation to eliminate negative values and sum all squared deviations.
  • Divide this sum by \(n-1\) (where \(n\) is the sample size) to find the variance. This step acknowledges the degrees of freedom in the sample.
  • Take the square root of the variance to arrive at the standard deviation, \(s\).

In Joan's Nursery example, the calculation resulted in a sample standard deviation of approximately 0.52 hours. This value helps determine the standard error in the t-test, which is crucial for assessing the reliability of the conclusions drawn from the statistical analysis.

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Most popular questions from this chapter

Carpetland salespersons average \(\$ 8000\) per week in sales. Steve Contois, the firm's vice president, proposes a compensation plan with new selling incentives. Steve hopes that the results of a trial selling period will enable him to conclude that the compensation plan increases the average sales per salesperson. a. Develop the appropriate null and alternative hypotheses. b. What is the Type I error in this situation? What are the consequences of making this error? c. What is the Type II error in this situation? What are the consequences of making this error?

A production line operates with a mean filling weight of 16 ounces per container. Overfilling or underfilling presents a serious problem and when detected requires the operator to shut down the production line to readjust the filling mechanism. From past data, a population standard deviation \(\sigma=.8\) ounces is assumed. A quality control inspector selects a sample of 30 items every hour and at that time makes the decision of whether to shut down the line for readjustment. The level of significance is \(\alpha=.05\) a. State the hypothesis test for this quality control application. b. If a sample mean of \(\bar{x}=16.32\) ounces were found, what is the \(p\) -value? What action would you recommend? c. If a sample mean of \(\bar{x}=15.82\) ounces were found, what is the \(p\) -value? What action would you recommend? d. Use the critical value approach. What is the rejection rule for the preceding hypothesis testing procedure? Repeat parts (b) and (c). Do you reach the same conclusion?

Wall Street securities firms paid out record year-end bonuses of \(\$ 125,500\) per employee for 2005 (Fortune, February 6,2006 ). Suppose we would like to take a sample of employees at the Jones \& Ryan securities firm to see whether the mean year-end bonus is different from the reported mean of \(\$ 125,500\) for the population. a. State the null and alternative hypotheses you would use to test whether the year-end bonuses paid by Jones \& Ryan were different from the population mean. b. Suppose a sample of 40 Jones \& Ryan employees showed a sample mean year- end bonus of \(\$ 118,000 .\) Assume a population standard deviation of \(\sigma=\$ 30,000\) and \(\mathrm{com}-\) pute the \(p\) -value. c. With \(\alpha=.05\) as the level of significance, what is your conclusion? d. Repeat the preceding hypothesis test using the critical value approach.

Virtual call centers are staffed by individuals working out of their homes. Most home agents earn \(\$ 10\) to \(\$ 15\) per hour without benefits versus \(\$ 7\) to \(\$ 9\) per hour with benefits at a traditional call center (BusinessWeek, January 23, 2006). Regional Airways is considering employing home agents, but only if a level of customer satisfaction greater than \(80 \%\) can be maintained. A test was conducted with home service agents. In a sample of 300 customers 252 reported that they were satisfied with service. a. Develop hypotheses for a test to determine whether the sample data support the conclusion that customer service with home agents meets the Regional Airways criterion. b. What is your point estimate of the percentage of satisfied customers? c. What is the \(p\) -value provided by the sample data? d. What is your hypothesis testing conclusion? Use \(\alpha=.05\) as the level of significance.

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu=22 \\ H_{\mathrm{a}}: \mu \neq 22 \end{array} \\] A sample of 75 is used and the population standard deviation is \(10 .\) Compute the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.01\) a. \(\bar{x}=23\) b. \(\bar{x}=25.1\) c. \(\bar{x}=20\)

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