S
sxyvxn28
Guest
...level of significance is 0.05? 3.A local tire store suspects that the mean life of a new discount tire is less that 39,000 miles. To check the claim, the store selects randomly 18 of these new discount tires. When they are tested, it is found that the mean life is 38,250 miles with a sample standard deviation s = 1200 miles.
use the critical value z0 method from the normal distribution to test for the population mean. Test the company's claim at the level of significance 0.05
H0=
Ha=
level of sig =
test statistics =
P-value or critical z0 or t0
rejection region =
decision =
interpritation =
so far I have:
xbar=38250
mean=39000
s=1200
n=18
d.f. = 17 looked on tabel 5 and got 1.740, since its left-tailed it becomes a negative -1.740
(38250-39000)/(1200/sgrt18)
=-750/282.84
=-2.65
p-value = P(z<-2.65)=.0040
after this I'm lost (and I'm not sure I am right so far). Can anyone please help and explain this to me?
use the critical value z0 method from the normal distribution to test for the population mean. Test the company's claim at the level of significance 0.05
H0=
Ha=
level of sig =
test statistics =
P-value or critical z0 or t0
rejection region =
decision =
interpritation =
so far I have:
xbar=38250
mean=39000
s=1200
n=18
d.f. = 17 looked on tabel 5 and got 1.740, since its left-tailed it becomes a negative -1.740
(38250-39000)/(1200/sgrt18)
=-750/282.84
=-2.65
p-value = P(z<-2.65)=.0040
after this I'm lost (and I'm not sure I am right so far). Can anyone please help and explain this to me?