Mathematics, 14.02.2020 04:51 kaylahill14211

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 medium-sized tree. Actual times from a sample of 10 plantings during the past month follow (times in hours).
1.7 1.5 2.6 2.2 2.4 2.3 2.6 3.0 1.4 2.3
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?

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Answer from: Blaise2653

We conclude that mean tree-planting time is two hours.

Step-by-step explanation:

We are given the following data about plantings in hours in the question:

1.7, 1.5, 2.6, 2.2, 2.4, 2.3, 2.6, 3, 1.4, 2.3

b) Sample mean

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{22}{10} = 2.2$

c) Formula for sample standard deviation

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where x_i are data points, $\bar{x}$ is the mean and n is the number of observations.

Sum of squares of differences = 2.4

$s= \sqrt{\dfrac{2.4}{9}} = 0.516$

Population mean, μ = 2 hours

Sample mean, $\bar{x}$ = 2.2 hours

Sample size, n = 10

Alpha, α = 0.05

Sample standard deviation, s = 0.516 hours

a) the null and the alternate hypothesis are

$H_{0}: \mu = 2\text{ hours}\\H_A: \mu \neq 2\text{ hours}$

We use two-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{2.2 - 2}{\frac{0.516}{\sqrt{10}} } = 1.2256$

d) We calculate the p-value from the table.

P-value = 0.251453

e) Conclusion:

Since, p value is greater than thee significance level, we fail to reject the null hypothesis and accept it.

Thus, we conclude that mean tree-planting time does not differs from two hours.

Answer from: crodriguez87

The mean tree-planting time is two hours.

Given that ,

Actual times from a sample of 10 plantings during the past month n = 10

The observation are = 1.7, 1.5, 2.6, 2.2, 2.4, 2.3, 2.6, 3.0, 1.4, 2.3.

Sum of all the observations = 1.7+1.5+2.6+ 2.2+ 2.4 +2.3 +2.6 +3.0 +1.4 +2.3 = 22.

The sample mean = $\frac{sum of all the observation}{total no. of the observation}$

Mean = $\frac{22}{10}$ = 2.2

The sample mean is 2.2 .

Standard deviation = $\frac{\sqrt {\Sigma(x_1 - x )^{2} } }{n- 1} \\\\$

$= \frac{\sqrt{(2.4 ) } }{10 - 1} \\\\= 0.54$

Where x_1 are data points and x is the mean and the n is the no. of samples

Population mean, μ = 2 hours

Sample mean, = 2.2 hours

,Sample size n = 10

Alpha, α = 0.05

Sample standard deviation, s = 0.516 hours.

The null and alternative hypotheses.

$H_0 =\mu = 2 hours \\\\H_a = \mu \neq 2 hours$ $\mu = 2 hours \\\\H_a = \mu \neq 2 hours$

Using two tailed test to perform this hypothesis

$t_stat = \frac{x-\mu}{\frac{s}{\sqrt{n} } } \\ \\\\= \frac{2.2 - 2}{\frac{0.516}{\sqrt{10} } } \\\\= \frac{0.2}{3.16} \\\\= 1.25$

The value of the p calculate from the table is 0.2514.Conclusion: Since, p value is greater than the significance level, we fail to reject the null hypothesis and accept it.

Thus, we conclude that mean tree-planting time does not differs from two hours.

For more information about probability click the link given below.

link

Answer from: Quest

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong

Answer from: Quest

Then just do it (lol)

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