stats question please help?

Question

a company selling silences of a new software advertised that firms using this software could obtain , on average during the first year, a minimum yield (in cost savings) of 20 %on average on their software investment. to disprove the claim, we checked with 10 different firms which used the software. these firms reported the following savings cost  
17,     19.2     21.5   18.6    22.1    14.9    18.4    20.1    19.4    18.9
 
1.compute a 95% interval for the average  cost- saving yield estimate


do we have significant evidence to show the the softwares companies claims of a minimum 20 % in cost savings was not valid? test using α=.05

Bleaksky · Accepted Answer

Ho: μ ≥ 20% 
Ha: μ < 20%

Sample ......................10
Sample mean..............19.01
Standard deviation........2.071204695
Hypothesized Value......20
Standard Error..............0.654972434
P-value........................0.065328773

Since the P-value > α=.05, then we accept the null hypothesis that the software companies claim of a minimum 20% in cost savings is valid.

Guillermo · Answer

Xmean = (1/10)*(17+19.2+21.5+18.6+22.1+14.9+18.4+20.1+19.4+18.9) = 19.01

S^2(X) = (17^2+19.2^2+21.5^2+18.6^2+22.1^2+14.9^2+18.4^2+20.1^2+19.4^2+18.9^2-10*19.01^2)*(1/9) = 4.289888889

Pf(19.01 - tst(8; 0.975)*4.289888889/sqrt(10) < Î¼ < 19.01 - tst(8; 0.975)*4.289888889/sqrt(10)) = 0.95

Pf(19.01 - 2.306*4.289888889/sqrt(10) < Î¼ < 19.01 + 2.306*4.289888889/sqrt(10)) = 0.95

Pf( 15.88 <  Î¼ < 22.14 ) = 0.95

H0: Î¼x â¥ 0.2
H1: Î¼x < 0.2,  Î± = 0.05

Decision Rule:

If (Xmean - Î¼x)/(Sx/sqrt(n)) < - tst(n - 2; Î±) Reject H0.

If (0.1901- 0.2)/(4.289888889/3.162277660) < - 1.86, Reject H0.

If  - 0.73e-2 â® - 1.86, we don't Reject H0.

We don't have significant evidence to show the the softwares companies claims of a minimum 20 % in cost savings was not valid

stats question please help?

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