Answer :
Answer:
Sample mean = 1970.4
Standard Error = 40.025
There is enough evidence to support the claim that the true average viscosity be 2000.
Step-by-step explanation:
We are given the following in the question:
2062, 1906, 1993, 1827, 2064
Formula:
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{9852}{5} = 1970.4[/tex]
Population standard deviation, σ = 89.5
Standard error =
[tex]\dfrac{\sigma}{\sqrt{n}}= \displaystyle\frac{89.5}{\sqrt{5}} = 40.025[/tex]
Now,
Population mean, μ = 2000
Sample mean, [tex]\bar{x}[/tex] = 1970.4
Sample size, n = 5
Alpha, α = 0.05
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 2000\\H_A: \mu \neq 2000[/tex]
We use Two-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{1970.4 - 2000}{\frac{89.5}{\sqrt{5}} } = -0.7395[/tex]
Now, [tex]z_{critical} \text{ at 0.05 level of significance } = \pm 1.96[/tex]
Since, the calculated z statistic lies in the acceptance region, we fail to reject the null hypothesis and accept it.
Thus. there is enough evidence to support the claim that the true average viscosity be 2000.