The five parts are:
i. Null Hypothesis: H0 : µ =5.2
ii. Alternative Hypothesis: HA : µ < 5.2
iii. Rejection Region: Reject H0 if t statistic <−t49,.05 =−1.677
iv. Test Statistics: t = Y−µ0 S/pn = 5−5.2 0.7/p50 =−2.0203 <−t49,.05 =−1.677
v. Conclusion. Reject H0 at α = 5%. The data support that the mean dissolved oxygen count of the water is less than the reading at this location over the past year.
What is the p-value?
Solution:-
Given data:-
H0 : µ =5.2
HA : µ < 5.2
Significance level =0.05
Degree of freedom=n-1=50-1=49
So, df=49.
Test Statistics,t = −1.677
using P value calculator, at 0.05 level of significance,
P-value = 0.099912.
The five parts are: i. Null Hypothesis: H0 : µ =5.2 ii. Alternative Hypothesis: HA :...
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