Solutions to Practice Problems

Note that the solutions to Number 1 can be found here.

(2) Let X represent the number of odd numbers that result in 10 rolls of the die. Then X is a binomial random variable with n=10 and the value of p is under discussion.

a)If the die is fair, then p = .5. The probability we reject the null hypothesis is P(X > 8) = P(X = 9) + P(X = 10) = 0.01074219. So our significance level is about 1%.

b) Now p = .6 (but we, unsuspecting rubes, don't know this). The probability of rejecting is still P(X = 9) = P(X=10), but now X is binomial with n = 10 and p = .6 so P(X > 8) = .0464. Not terribly high. We could easily be fooled by this.

c)Calculate P(X > 8) when n=10 and p = .75. P(X =9) + P(X = 10) = .244. We stand a better chance of catching this more obvious subterfuge.

d) We could increase the significance level (the probability of rejecting the null hypothesis when in fact it is true) by using "easier" rules. If we decide to reject if X > 7, then the significance level is P(X > 7) when p = .5 or 0.0547. (Closer to the traditional 5% level). If in fact p =.6, then the probability we will catch the deception is now P(X > 7) = .167 (using p =.6). If in fact p =.75 then the probability we will detect that the null hypothesis is false is 0.5257.

In practice, we set the significance level at 5% and then calculate the power for a variety of different scenarios (i.e. different values of p). You'll sometimes here reference to a "power curve". This is a graph of the power as a function of p (or whatever parameter you're talking about.) If you are ambitious, you can write a program to calculate this curve for this scenario. As you can see, we are not going to have a very high probability of catching the deception unless p is quite a ways from 0.5. The problem is that we are only rolling the die 10 times. If we rolled it more times, we could keep our significance level at 5% and have a higher power.

3) We're now in a standard normal random variable context. Let X represent the amount of soda in the drink. Nowhere do we state explicitly that the amount of soda is normally distributed, and so we should state this as an assumption of our solution. (If we don't, the central limit theorem might come to our rescue; but our sample size of 10 isn't necessarily "large".) Notice that probably the normal distribution makes sense: the machine will try to put in 16 ounces, but might put in a little more, or a little less.

a) H0: mean = 16, Ha: mean < 16 (It's one sided, because we won't complain if it puts too much in.)

b) We'll reject the null hypothesis if our average of 10 drinks, Xbar, is too small, say less than some number c. We want to set c so that the probability of reject the null hypothesis, when in fact the mean = 16, is .05. So we want P(Xbar < c) = .05

Under the assumptions of the null hypothesis, Xbar is a normal random variable with mean 16 and standard deviation (.2/sqrt(10)).

So if P(Xbar < c) = .05, we must have P(Z < (c-16)/(.2/sqrt(10))) = .05, which means that

(c-16)/(.2/sqrt(10)) = -1.645 which implies c = 15.896 or about 15.9.

c) We want to find P(reject the null given that mean = 15.8). In other words P(Xbar < 15.896 | mean = 15.8).

Standardizing: P(Xbar < 15.896) = P(Z < (15.896 - 15.8)/.2/sqrt(10))) = P(Z < 1.517893) = .935. Pretty good!

In other words, our power when the true mean is 15. 8 is about 93.5%

4. H0: mean = 15 mg/day

Ha: mean < 15 mg/day

The population SD is unknown, and so we use the sample SD of 6.43. Assuming the zinc intake in the population is normally distributed and that this is a random sample of men from the 65-74 group, we can use a t-test. The observed value of the t-test is

T = (11.3-15)/(6.43/sqrt(115)) = -6.17

It's really not necessary to compute the p-value here, but if this were the AP test we would have to at least go through the motions. The degrees of freedom are 114, and with these degrees of freedom, P(T < -6.17) is on the order of 10 to the negative ninth power.

Because the p-value is less than .05, we reject the null hypothesis at the 5% level and conclude there is evidence to conclude that the mean zinc intake in men aged 65-74 is less than the recommended daily allowance.

Please make sure that your students write out all of the steps (null hypothesis, state assumptions, etc.) and state their conclusion in clear English and in the context of the research question.