Respuesta :
Answer:
Step-by-step explanation:
From the given information:
The null hypothesis is:
[tex]H_o: \mu = 8[/tex]
The alternative hypothesis is:
[tex]H_1 : \mu \ne 8[/tex]
The sample size n = 35
Sample mean = 7.91
variance = 0.03
Standard deviation = [tex]\sqrt{0.03} = 0.173[/tex]
The t-test statistics is computed as:
[tex]t = \dfrac{\bar x - \mu}{\dfrac{s}{\sqrt{n}}}[/tex]
[tex]t = \dfrac{7.91 -8}{\dfrac{0.1732}{\sqrt{35}}}[/tex]
[tex]t = \dfrac{-0.09}{\dfrac{0.1732}{5.916}}[/tex]
t = - 3.0741
Degree of freedom:
df = n - 1
df = 35 - 1
df = 34
Level of significance = 1 - C.I
= 1 - 0.99
= 0.01
The t_{critical} value for the two-tailed test at 0.01 is within the rejection region:
[tex]t_{critical} = -2.728 \ and \ 2.728[/tex]
Decision rule: To reject [tex]H_o[/tex] if [tex]t< -2.728 \ or \ t > 2.728[/tex]
As [tex]t_{statistics}[/tex] fails the rejection region, we reject [tex]H_o[/tex]
Conclusion: There is sufficient evidence to believe that the machine will not dispense 8 ounces & and it must be stopped for repairs.
You can use hypothesis testing with t test to know if there is enough evidence that he machine should be stopped and production wait for repairs.
There is enough evidence that the machine should be stopped and production wait for repairs.
Given information about taken sample are:
- Sample mean [tex] \overline{x} = 7.91[/tex]
- Sample size n = 35
- Sample standard deviation [tex]s = \sqrt{variance} = \sqrt{0.03} = 0.1732[/tex]
- Confidence should be of 99%.
Let the population mean be denoted by [tex]\mu[/tex]
Forming hypotheses:
Null hypothesis: [tex]\mu = 8[/tex] (that is, machine needs no repair and working fine)
Alternate hypothesis: [tex]\mu \neq 8[/tex] (that is, machine is not dispensing right amount and need repairs)
Using t test:
[tex]t = \dfrac{\overline{x} - \mu}{\dfrac{s}{\sqrt{n}}}[/tex]
[tex]t = \dfrac{7.91- 8}{\dfrac{0.1732}{\sqrt{35}}} = -0.30821[/tex]
Degree of freedom = sample size - 1 = 34
Level of significance = [tex]\alpha = 1 -0.99 = 0.01[/tex]
From t tables we have:
[tex]t_{critical} = t_{0.01} = \pm 0.278[/tex] (two tailed test) (at 34 degree of freedom)
Since obtained value of t -0.30821 is less than the minimum value of range [tex](-0.278, 0.278)[/tex], thus lying outside the range and thus, we cannot accept the null hypothesis and end up accepting the alternate hypothesis.
Thus, there is enough evidence that the machine should be stopped and production wait for repairs.
Learn more about testing of hypothesis for single mean here:
https://brainly.com/question/25316952
