Respuesta :
Answer:
a) A. The relevant variable is the population mean work week (in hours) for workers aged 18-25.
b) Null hypothesis:[tex]\mu \leq 40[/tex]
Alternative hypothesis:[tex]\mu > 40[/tex]
c) [tex]t=\frac{45.6-40}{\frac{14.6}{\sqrt{893}}}=11.46[/tex]
The p value for this case would be:
[tex] p_v = P(t_{110} >11.46) \approx =0[/tex]
d) Since the p value is a very low value we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case exceeds 40 hours.
Step-by-step explanation:
Information provided
[tex]\bar X=45.6[/tex] represent the sample mean
[tex]s=14.6[/tex] represent the sample standard deviation
[tex]n=893[/tex] sample size
[tex]\mu_o =40[/tex] represent the value to verify
t would represent the statistic
[tex]p_v[/tex] represent the p value
a. Identify the relevant and parameter variable. Choose the correct relevant variable below.
A. The relevant variable is the population mean work week (in hours) for workers aged 18-25.
b. State the null and alternative hypotheses. State the null hypothesis.
We want to verify if the population mean is higher than 40, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 40[/tex]
Alternative hypothesis:[tex]\mu > 40[/tex]
c. Calculate the statistic
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the info given we got:
[tex]t=\frac{45.6-40}{\frac{14.6}{\sqrt{893}}}=11.46[/tex]
The p value for this case would be:
[tex] p_v = P(t_{110} >11.46) \approx =0[/tex]
d. Conclusion
Since the p value is a very low value we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case exceeds 40 hours.
