Respuesta :
Using the z-distribution, it is found that mean weights less than 135.1g and more than 144.9 would lead to a rejection of the hypothesis.
What is the critical value of the test?
We have a two-tailed test, with a significance value of α = 0.05, hence, using a z-distribution calculator, the critical value is given by:
|z| = 1.96 -> z < -1.96 or z > 1.96.
What is the test statistic?
The test statistic is given as follows:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which:
- X is the sample mean.
- [tex]\mu[/tex] is the value tested at the null hypothesis.
- [tex]\sigma[/tex] is the standard deviation of the population.
- n is the sample size.
For this problem, the parameters are given as follows:
[tex]\mu = 140, \sigma = 15, n = 36[/tex]
The lower bound of acception is found as X when z = -1.96, hence:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
-1.96 = (X - 140)/(15/6)
-1.96 = (X - 140)/(2.5)
X - 140 = -1.96 x 2.5
X = 135.1.
The upper bound of acception is found as X when z = 1.96, hence:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
1.96 = (X - 140)/(15/6)
1.96 = (X - 140)/(2.5)
X - 140 = 1.96 x 2.5
X = 144.9.
Hence:
Mean weights less than 135.1g and more than 144.9 would lead to a rejection of the hypothesis.
More can be learned about the z-distribution at brainly.com/question/16313918
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