Respuesta :
Answer:
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
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{22}{10} = 2.2[/tex]
c) Formula for sample standard deviation
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
Sum of squares of differences = 2.4
[tex]s= \sqrt{\dfrac{2.4}{9}} = 0.516[/tex]
Population mean, μ = 2 hours
Sample mean, [tex]\bar{x}[/tex] = 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
[tex]H_{0}: \mu = 2\text{ hours}\\H_A: \mu \neq 2\text{ hours}[/tex]
We use two-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{2.2 - 2}{\frac{0.516}{\sqrt{10}} } = 1.2256[/tex]
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.
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 = [tex]\frac{sum of all the observation}{total no. of the observation}[/tex]
Mean = [tex]\frac{22}{10}[/tex] = 2.2
The sample mean is 2.2 .
- Standard deviation = [tex]\frac{\sqrt {\Sigma(x_1 - x )^{2} } }{n- 1} \\\\[/tex]
[tex]= \frac{\sqrt{(2.4 ) } }{10 - 1} \\\\= 0.54[/tex]
Where [tex]x_1[/tex] 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.
[tex]H_0 =\mu = 2 hours \\\\H_a = \mu \neq 2 hours[/tex] [tex]\mu = 2 hours \\\\H_a = \mu \neq 2 hours[/tex]
Using two tailed test to perform this hypothesis
[tex]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[/tex]
- 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.
https://brainly.com/question/11234923
