Mathematics, 15.04.2020 00:22 querline87

The health of employees is monitored by periodically weighing them in. A sample of 54 employees has a mean weight of 183.9 lb. Assuming that σ is known to be 121.2 lb, use a 0.10 significance level to test the claim that the population mean of all such employees weights is less than 200 lb.

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Answer from: jay9253

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:

$H_0: \mu \geq 200$

At the alternative hypothesis, it is tested if it is less, that is:

$H_1: \mu < 200$

We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean. $\mu$ is the value tested at the null hypothesis. $\sigma$ is the standard deviation of the population.n is the sample size.

For this problem, the values of the parameters are: $\overline{x} = 183.9, \mu = 200, \sigma = 121.2, n = 54$ .

The value of the test statistic is:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{183.9 - 200}{\frac{121.2}{\sqrt{54}}}$

z = -0.98

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 $z^{\ast} = -1.28$

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 link

Answer from: jeremytheprimerapper

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

$Z = \frac{x^{-} - mean}{\frac{S.D}{\sqrt{n} } }$ ...(i)

substitute all values x⁻ = 183.9 , (σ) =121.2 , μ = 200 and n = 54 in (i)

$Z = \frac{183.9 - 200}{\frac{121.2}{\sqrt{54} } }$

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.

Answer from: Quest

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Answer from: Quest

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