Respuesta :
Answer:
1) A. H0: μ ≥ 7.2; H1: μ < 7.2
2) D. Reject H0 if z < –1.645
3) [tex]t=\frac{6.2-7.2}{\frac{0.9}{\sqrt{35}}}=-6.573[/tex]
4) [tex]p_v =P(z<-6.573)=2,47x10^{-11}[/tex]
Step-by-step explanation:
Information provided
[tex]\bar X=6.2[/tex] represent the sample mean for the number of movies watched last month
[tex]\sigma=0.9[/tex] represent the population deviation
[tex]n=35[/tex] sample size selected
[tex]\mu_o =7.2[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
t would represent the statistic
[tex]p_v[/tex] represent the p value for the test
1) System of hypothesis
We want to check if college students watch fewer movies a month than high school students, and the best system of hypothesis are:
Null hypothesis:[tex]\mu \geq 7.2[/tex]
Alternative hypothesis:[tex]\mu < 7.2[/tex]
A. H0: μ ≥ 7.2; H1: μ < 7.2
2) Decision rule
For this case we are ocnduting a left tailed test so then we need to find a critical value in the normal standard distribution who accumulates 0.05 of the area in the left and we got:
[tex]z_{crit}= -1.645[/tex]
And the rejection zone would be:
D. Reject H0 if z < –1.645
3) Statistic
Since we know the population deviation the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing we got:
[tex]t=\frac{6.2-7.2}{\frac{0.9}{\sqrt{35}}}=-6.573[/tex]
4) P value
We have a left tailed test then the p value would be:
[tex]p_v =P(z<-6.573)=2,47x10^{-11}[/tex]
