So the question asks: It has been generally believed that average daily caffeine consumption, $μ$, is at most $200$ mg with $σ^2 = 225$. To examine whether at present time $μ$ is greater than $200$ mg, it is agreed that daily caffeine consumption of a random sample of $100$ adults be examined. If the sample mean is greater than $203.50$, the decision will be to reject the null hypothesis.

The following information may be useful: pnorm(0.67)$=0.75$, pnorm(1)$=0.84,$ pnorm(1.56)$=0.94$, pnorm(2.33)$=0.99$

(a) What are the null and alternative hypotheses?

(b) What is the test statistic and what is it distribution?

(c) What is the probability of a Type 1 error?

(d) What is the power of the test if the true mean is 205mg?

(e) Is the assumption that the population distribution of daily caffeine consumption is normal necessary for the hypothesis test? Explain your reasoning.

My answers:

(a) $H_0: μ \le 200$ (average daily caffeine consumption is at most $200mg$)

$H_1: μ > 200$ (average daily caffeine consumption is more than $200mg$)

(b) For large sample, and known population standard deviation, use 1-sample $Z$ test. Distribution: Normal.

$$Z=\frac{\bar X-μ}{\sigma/\sqrt{n}} =\frac{203.50-200}{15/10} =2.3333.$$

(c) probability of Type I error, $\alpha=0.05$

(d) Since it is a right-tailed test, therefore, one commits a Type II error (fail to reject $H_0$), when one gets a test statistic less than 1.645. Compute $X_\text{critical}$ by substituting the value in following Z score formula

$Z =1.64=\frac{X_\text{critical}-\mu}{\sigma/\sqrt{n}}=\frac{X_\text{critical}-200}{1.5}$


$$P(\bar X<202.46)=P\left(Z<\frac{202.46-205}{1.5}\right)=P(Z<-1.69333333)=0.0455.$$

Power of test: $1-0.0455 =0.9545.$

(e) Not necessary because the sample size is sufficiently large.

So I am not sure about my answers in d and e. Should I calculate the new x, then use z-test to get the power of the test? And in e, does my answer sound reasonable enough?

  • $\begingroup$ I have edited your question slightly to improve formatting; please check to make sure I have not changed your meaning. $\endgroup$ – BruceET Dec 2 '16 at 23:51

(c) To check your power computation, here is a printout from Minitab's 'Power and Sample Size' procedure. [The slight discrepanacy might be because you have used 1.64 instead of 1.644854 (or 1.645 as in many tables).]

 1-Sample Z Test

 Testing mean = null (versus > null)
 Calculating power for mean = null + difference
 α = 0.05  Assumed standard deviation = 15

 Difference    Size     Power
          5     100  0.954340

The power curve shows the power against alternatives with differences other than $\Delta = \mu_a - \mu_0 = 205 - 200 = 5.$ [The only point on this curve of 'P(Reject) vs. $\Delta$' that is not a 'power' value is above $\Delta = 0$ where the height of the curve is the significance level $\alpha = .05.$]

enter image description here

(d) My answer would be only slightly more cautious: Because of the large sample size, probably not necessary--unless the actual population distribution is extremely skewed or otherwise very far from normal.

Even for $n=100,$ results would not be accurate for an exponential population, and would be totally wrong for a Cauchy population. (But I think the question clearly expects you to take the Central Limit Theeorem into consideration.)


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