Respuesta :
Answer:
[tex] z = \frac{\bar X -\mu}{\sigma_{\bar X}}= \frac{70.9-71.5}{0.77}=-0.779[/tex]
From this result we can conclude that the value of 70.9 is 0.78 deviation below the true mean of 71.5 and that can be considered as unusual. If we conduct a hypothesis test or a confidence interval we will see that we have enough evidence to conclude that the true mean is not significantly different from 71.5.
Step-by-step explanation:
For this case from all the population we know that the population mean and deviation are:
[tex] \mu = 71.5,\sigma = 4.87[/tex]
And we take a random sample of size n =40 and we got a sample mean calculated with the following formula:
[tex] \bar X =\frac{\sum_{i=1}^n X_i}{n}= \frac{\sum_{i=1}^{40} X_i}{40}=70.9[/tex]
And we want to test if this value is unusually low.
Since the sample size is large n>30 we can use the central limit theorem who says that the distribution for the sample mean is given by:
[tex] \bar X \sim N (\mu , \frac{\sigma}{\sqrt{n}})[/tex]
And on this case if we replace the values that we have we got:
[tex] \bar X \sim N (\mu_{\bar X}=71.5,\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}=\frac{4.87}{\sqrt{40}}=0.77)[/tex]
For this case we can calculate how many deviations above or below is our calculated value from the sample of size 40, using the z score given by:
[tex] z = \frac{\bar X -\mu}{\sigma_{\bar X}}= \frac{70.9-71.5}{0.77}=-0.779[/tex]
From this result we can conclude that the value of 70.9 is 0.78 deviation below the true mean of 71.5 and that can be considered as unusual. If we conduct a hypothesis test or a confidence interval we will see that we have enough evidence to conclude that the true mean is not significantly different from 71.5.
