In a recent year the fuel economy of all passenger vehicles was 19.8 mpg. A trade organization sampled 50 passenger vehicles for fuel economy and obtained a sample mean of 20.1 mpg with standard deviation 2.45 mpg. The sample mean 20.1 exceeds 19.8, but perhaps the increase is only a result of sampling error. Perform the relevant test of hypotheses at the 20% level of significance using the critical value approach. Compute the observed significance of the test. Perform the test at the 20% level of significance using the p-value approach. You need not repeat the first three steps, already done in part (a).

Answers and Step-by-step explanation:

Hypotheses are:

H o=19.8

H 1 ≠19.8

Here we have following information:

s=2.45, X bar=20.1, n=50

So test statistics will be

t=20.1-19.8/(2.45/ =0.87

Degree of freedom : d f = n - 1 = 50 - 1 =49

Test is two tailed so critical value of the test is -1.30 and 1.30.

Rejection region:

if t < -1.30 and t > 1.30, reject H o

Since test statistics lie in the rejection region so we fail to reject the null hypothesis.

(b)

The observed significance of the test is

P value =0.3885

(c)

Since p value is greater than 0.20 so we fail to reject the null hypothesis.

13cm

step-by-step explanation:

(apex)

$2416.6

step-by-step explanation:

per hour = $ 25

total work time = 35

commission = 15%

total she has to make = $ 4,500

let she sold x dollar products

so her commission will be 15 % of x

now

total = weekly basic pay + commission

4500= (25*35) + 15 % of x

4500=875 + * x

4500-875 =   * x

3625= * x

multiplying wwith 100 on both sides gives

36250 = 15 x

now

x =  

x= $2416.6