Answer :
The provided sample mean is x̅ =7.1pounds/square inch and the sample standard deviation is s=1.0, and the sample size is n = 16.
(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
H0 : μ ≤ 6.9
H1 : μ > 6.9
This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.
(2) Rejection Region
Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is at 0.05 significance level with (16 - 1)= 15 degrees of freedom is tc =1.753.
The rejection region for this right-tailed test is R={t:t>1.753}
That is,
Reject H0 if t > 1.753
(3) Test Statistics
The t-statistic is computed as follows:
t= (x̅ −μ0)/(s/√n)
=(7.1−6.9)/(1.0/√16)
=0.2/0.25
=0.8
(4) Decision about the null hypothesis
Since it is observed that t=0.8 ≤ tc=1.753, it is then concluded that the null hypothesis is not rejected.
Using the P-value approach: The p-value for t score= 0.8 with (16 - 1) = 15 degrees of freedom for right tailed test is p=0.2181, and since p = 0.2181 ≥ 0.05, it is concluded that the null hypothesis is not rejected.
(5) Conclusion
It is concluded that the null hypothesis H0 is not rejected. Therefore, there is not enough evidence to claim that the value perform the above the specification (that is mean pressure is 6.9 pounds/square inch) its means that the population mean μ is greater than 6.9, at the 0.05 significance level.
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