. A used car dealer says that the mean price of a two-year old sedan (in good condition) is at least $20,500. You suspect this claim is incorrect and find that a random sample of 14 similar vehicles has a mean price of $19,850 and a standard deviation of $1084. Is there enough evidence to reject the dealer's claim at a significance level (alpha) =0.05?

There is sufficient evidence to reject the dealer's claim that the mean price is at least $20,500

Step-by-step explanation:

given that;

n = 14

mean Ж = 19,850

standard deviation S = 1,084

degree of freedom df = n - 1 = ( 14 -1 ) = 13

H₀ : ц ≥ 20,500

H₁ : ц < 20,500

Now the test statistics

t = (Ж - ц) / ( s/√n)

t = ( 19850 - 20500) / ( 1084/√14)

t = -2.244

we know that our degree of freedom df = 13

from the table, the area under the t-distribution of the left of (t=-2.244) and for (df=13) is 0.0215

so P = 0.0215

significance ∝ = 0.05

we can confidently say that since our p value is less than the significance level, we reject the null hypothesis ( H₀ : ц ≥ 20,500 )

There is sufficient evidence to reject the dealer's claim that the mean price is at least $20,500

answer by yourhope:

hi! : )

a farmer that lived in ancient egypt would most likely be barried in the sand, in the desert!

: )

too many. (jk)

31 v bucks

step-by-step explanation:

since you bought 25 on friday and 6 on tuesday, you can just add them up.

25+6=31

check your work.

31-6=25

i hope this !

-yand3r3