Question

1. Richard is asked to perform a hypothesis test to determine whether the mean resistance of pyrite from a particular site is 2 Ohms (Ω). Richard collects 50 samples, and obtains a sample mean of 1.8 Ohms and a standard deviation of 0.6 Ohms. He then constructs the following hypotheses:

   H0 : x ̄ = 2.0Ω H1 : x ̄ ̸= 2.0Ω

   and concludes that, since x ̄ = 1.8 is not equal to 2.0, we should reject the null hypothesis. Explain why Richard’s reasoning is incorrect, and explain what we actually need to do to correctly perform a hypothesis test for this experiment.

Answer 1

Richard’s reasoning is incorrect, because the hypotheses should involve population parameters and not sample statistics. Further, the difference should be tested for significance statistically using t test.

Explanation of what we actually need to do to correctly perform a hypothesis test for this experiment.:

H0: Null Hypothesis: = 2.00 ( the mean resistance of pyrite from a particular site is 2 Ohms (Ω). ) (Claim)

HA: Alternative Hypothesis: 2.00 ( the mean resistance of pyrite from a particular site is not 2 Ohms (Ω). )

SE = s/

= 0.6/

= 0.0849

Test Statistic is given by:

t = (1.8 - 2.00)/0.0849

= - 2.3570

Take = 0.05

ndf = 50 - 1 = 49

From Table, critical values of t = 2.0096

Since the calculated value of t = - 2.3570 is less than critical value of t = - 2.0096, the difference is significant. Reject null hypothesis.

Conclusion:
The data do not support the claim that the mean resistance of pyrite from a particular site is 2 Ohms (Ω).