Question

A researcher wishes to see if the average number of sick days a worker takes per year is less than 5. A random sample of 30 workers at a large department store had a mean of 4.8. The standard deviation of the population is 1.2 days. Is there enough evidence to support the claim at alpha = 0.01?

Answer 1

Answer:

No. At a significance level of 0.01, there is not enough evidence to support the claim that the average number of sick days a worker takes per year is significantly less than 5.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the average number of sick days a worker takes per year is significantly less than 5.

Then, the null and alternative hypothesis are:

$H_0: \mu=5\\\\H_a:\mu< 5$

The significance level is 0.01.

The sample has a size n=30.

The sample mean is M=4.8.

The standard deviation of the population is known and has a value of σ=1.2.

We can calculate the standard error as:

$\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{1.2}{\sqrt{30}}=0.219$

Then, we can calculate the z-statistic as:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{4.8-5}{0.219}=\dfrac{-0.2}{0.219}=-0.913$

This test is a left-tailed test, so the P-value for this test is calculated as:

$\text{P-value}=P(z<-0.913)=0.181$

As the P-value (0.181) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.01, there is not enough evidence to support the claim that the average number of sick days a worker takes per year is significantly less than 5.

Answer 2

Using the z-distribution, it is found that since the test statistic is greater then the critical value for the left-tailed test, this is not enough evidence to support the claim at $\alpja = 0.01$.

At the null hypothesis, it is tested if the number of sick days is of at least 5, that is:

$H_0: \mu \geq 5$

At the alternative hypothesis, we test if it is less than 5, that is:

$H_1: \mu < 5$

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 sample.

• n is the sample size.

For this problem, the values of the parameters are: $\overline{x} = 4.8, \mu = 5, \sigma = 1.2, n = 30$

Hence, the value of the test statistic is:

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

$z = \frac{4.8 - 5}{\frac{1.2}{\sqrt{30}}}$

$z = -0.91$

The critical value for a left-tailed test, as we are testing if the mean is less than a value, with a significance level of 0.01 is of $z^{-\ast} = -2.327$.

Since the test statistic is greater then the critical value for the left-tailed test, this is not enough evidence to support the claim at $\alpja = 0.01$.

A similar problem is given at https://brainly.com/question/16194574