Respuesta :
Answer:
Calculated value Z=0.976 < 1.64 at 0.10 significance level
hence we accepted null hypothesis
The claim that the population mean of all such employees weights is less than 200 lb.
Step-by-step explanation:
Given sample size 'n' = 54
Given mean x⁻ = 183.9 lb
Assuming that σ is known to be 121.2 lb
Population standard deviation (σ) =121.2 lb
given the population mean μ = 200 lb
Null hypothesis:-H₀:μ< 200
Alternative hypothesis:-H₁:μ >200
Level of significance: ∝=0.10
Test statistic
[tex]Z = \frac{x^{-} - mean}{\frac{S.D}{\sqrt{n} } }[/tex] ...(i)
substitute all values x⁻ = 183.9 , (σ) =121.2 , μ = 200 and n = 54 in (i)
[tex]Z = \frac{183.9 - 200}{\frac{121.2}{\sqrt{54} } }[/tex]
on calculation we get Z- value
Z=-0.976
|Z| =|-0.976| = 0.976
The tabulated value z = 1.64 at 0.10 significance level
Calculated value Z=0.976 < 1.64 at 0.10 significance level
hence we accepted null hypothesis
Conclusion:-
The claim that the population mean of all such employees weights is less than 200 lb.
Using the z-distribution, it is found that since the test statistic is greater than the critical value for the left-tailed test, there is not enough evidence to conclude that the population mean of all such employees weights is less than 200 lb.
At the null hypothesis, it is tested if the mean weigh is not less than 200 lb, that is:
[tex]H_0: \mu \geq 200[/tex]
At the alternative hypothesis, it is tested if it is less, that is:
[tex]H_1: \mu < 200[/tex]
We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
The parameters are:
- [tex]\overline{x}[/tex] 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 values of the parameters are: [tex]\overline{x} = 183.9, \mu = 200, \sigma = 121.2, n = 54[/tex].
The value of the test statistic is:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{183.9 - 200}{\frac{121.2}{\sqrt{54}}}[/tex]
[tex]z = -0.98[/tex]
The critical value for a left-tailed test, as we are testing if the mean is less than a value, with a 0.1 significance level is [tex]z^{\ast} = -1.28[/tex]
Since the test statistic is greater than the critical value for the left-tailed test, there is not enough evidence to conclude that the population mean of all such employees weights is less than 200 lb.
A similar problem is given at https://brainly.com/question/16225455
