Question

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?

Answer 1

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.