Respuesta :
Answer:
Option B) Accept [tex]H_0[/tex]
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 115 lbs
Sample mean, [tex]\bar{x}[/tex] = 111 lbs
Sample size, n = 41
Alpha, α = 0.01
Sample standard deviation, s = 12.4 lbs
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 115\text{ lbs}\\H_A: \mu < 115\text{ lbs}[/tex]
We use One-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{111- 115}{\frac{12.4}{\sqrt{41}} } = -2.065[/tex]
Now,
[tex]t_{critical} \text{ at 0.01 level of significance, 40 degree of freedom } = -2.423[/tex]
Since,
[tex]t_{stat} > t_{critical}[/tex]
We fail to reject the null hypothesis and accept the null hypothesis.
