Mathematics, 15.07.2020 03:01 myalee1419

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 planting trees, shrubs, and so on to be used for the project. For cost-estimating purposes, managers use two hours of labor time for planting of a medium-sized tree. Actual times from a sample of 10 plantintings 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 0.05 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?

Answers: 3

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Answers

Answer from: pleasehelpme666

A) Null and alternative hypothesis

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

B) M = 2.2 hours

C) s = 0.52 hours

D) P-value = 0.255

E) At a significance level of 0.05, there is not enough evidence to support the claim that the mean tree-planting time significantly differs from two hours.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean tree-planting time significantly differs from two hours.

Then, the null and alternative hypothesis are:

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

The significance level is 0.05.

The sample has a size n=10.

The sample mean is M=2.2.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.52.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.52}{\sqrt{10}}=0.1644$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.2-2}{0.1644}=\dfrac{0.2}{0.1644}=1.216$

The degrees of freedom for this sample size are:

df=n-1=10-1=9

This test is a two-tailed test, with 9 degrees of freedom and t=1.216, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t1.216)=0.255$

As the P-value (0.255) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean tree-planting time significantly differs from two hours.

Sample mean and standard deviation:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{10}(1.7+1.5+2.6+. . .+2.3)\\\\\\M=\dfrac{22}{10}\\\\\\M=2.2\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{9}((1.7-2.2)^2+(1.5-2.2)^2+(2.6-2.2)^2+. . . +(2.3-2.2)^2)}\\\\\\s=\sqrt{\dfrac{2.4}{9}}\\\\\\s=\sqrt{0.27}=0.52\\\\\\$

Answer from: Quest

false

step-by-step explanation:

Answer from: Quest

proofs involving isosceles triangles often require special consideration because an isosceles triangle has several distinct properties that do not apply to normal triangles.(more about triangle types) therefore, when you are trying to prove that two triangles are congruent, and one or both triangles, are isosceles you have a few theorems that you can use to make your life easier.

step-by-step explanation:

an isosceles triangle has two congruent sides and two congruent angles. the congruent angles are called the base angles and the other angle is known as the vertex angle. ∠bac and ∠bca are the base angles of the triangle picture on the left. the vertex angle is ∠abc