A student sets up two reactions. Reaction 1 uses 0.180 mol/L of reactant, and Reaction 2 uses 0.510 mol/L of reactant. How many
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Answer:
Since the computed value of t= 0.833 does not fall in the critical region we therefore do not reject H0 and may conclude that population mean is greater than 160. Or the sample comes from population with mean of 165.
Step-by-step explanation:
- State the null and alternative hypothesis as
H0: μ= 160 against the claim Ha :μ ≠160
Sample mean = x`= 165
Sample standard deviation= Sd= 12
2. The test statistic to use is
t= x`-μ/sd/√n
which if H0 is true , has t distribution with n-1 = 36-1= 35 degrees of freedom
3. The critical region is t< t (0.025(35)= 2.0306
t= x`-μ/sd/√n
4. t = (165-160)/[12/√(36)] = 5/[6] = 0.833
5. Since the computed value of t= 0.833 does not fall in the critical region we therefore do not reject H0 and may conclude that population mean is greater than 160. Or the sample comes from population with mean of 165.
Now
6. The p-value is 0 .410326 for t= 0.8333 with 35 degrees of freedom.
