The pvalue of 0.0113 < 0.05 means that there is sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance

Step-by-step explanation:

The null hypothesis is:

$H_{0} = 28$

The alternate hypotesis is:

$H_{1} < 28$

The test statistic is:

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

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

A recent survey of 50 executives who were laid off during a recent recession revealed it took a mean of 26 weeks for them to find another position.

This means that $n = 50, X = 26$

Assume the population standard deviation is 6.2 weeks.

This means that $\sigma = 6.2$

Does the data provide sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance

We have to find the pvalue of Z, looking at the z-table, when $\mu = 28$ . It if is lower than 0.05, it provides evidence.

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

$z = \frac{26 - 28}{\frac{6.2}{\sqrt{50}}}$

$z = -2.28$

$z = -2.28$ has a pvalue of 0.0113 < 0.05.

The pvalue of 0.0113 < 0.05 means that there is sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance

5 0

2 years ago

Check my answers? 15 points to whomever answers.

vova2212 [387]

38, 20, 12, 2, 1 and 4 only.

6 0

2 years ago

60.3 60.44 witch one is greater

Sindrei [870]

60.44 is greater than 60.3
-you can see this when you subtract 60.3 from 60.44 this would leave a 0.11 which is a positive number.
-thus meaning 60.44 is larger/greater.

4 0

2 years ago

Help me pls.........!

erica [24]

If each square represents a square foot half of it would be .5 so then add all of them. The formula for Area is A=l•w and the Area formula for a triangle is A=hbb/2. Hope this helps :)

7 0

2 years ago