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stats question please help?
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
2 Answers
- BleakskyLv 61 decade agoFavorite 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.
Source(s): Computations done through excel - GuillermoLv 71 decade ago
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) < μ < 19.01 - tst(8; 0.975)*4.289888889/sqrt(10)) = 0.95
Pf(19.01 - 2.306*4.289888889/sqrt(10) < μ < 19.01 + 2.306*4.289888889/sqrt(10)) = 0.95
Pf( 15.88 < μ < 22.14 ) = 0.95
H0: μx ⥠0.2
H1: μx < 0.2, α = 0.05
Decision Rule:
If (Xmean - μx)/(Sx/sqrt(n)) < - tst(n - 2; α) 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
Source(s): Probabilities t - student Distribution