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MA 2612 Test 1 B '97 Solutions


(10 points) Researchers investigating buying preferences administered a questionnaire to 500 randomly-selected consumers. In their statistical analysis, they tested 102 separate hypotheses. In each instance the null hypothesis stated the company's present way of doing business and the alternate hypothesis suggested that a change was needed. Four of the tests were significant at the 0.05 level, having p-values 0.0113, 0.0226, 0.0367 and 0.0481. The researchers called a big meeting with company executives to urge changes based on the results of these four tests. What is your reaction?

This may be much ado about nothing. At the 5% level, there will be, on average, $102\times 0.05\approx 5$ significant tests among the 102, even if H0 is true in each case.

Laboratory experiments on animals had suggested that a new drug treatment for Alzheimer's disease showed promise. The first human trial of the drug enrolled 800 patients afflicted with the early stages of the disease. Half were randomly assigned to the new treatment and half remained on the old treatment. At the end of six months, each patient was evaluated as to whether he or she needed to be institutionalized during the trial period, or not. The results of the trial were that 46 of the control group and 29 of the treatment group required institutionalization during the trial period. You are to conduct an hypothesis test, at the 0.01 level of significance, on the efficacy of the new treatment. To do so, please answer the following.
(10 points) The scientific hypothesis. That the new treatment is better than the old.

(10 points) The statistical model.

The two-sample binomial. Here, sample 1 is the 400 individuals assigned to the new treatment, and Y1 is the number of those who required institutionalization, while sample 2 and Y2 are the corresponding quantities for those assigned to the old treatment. We assume $Y_1\sim b(400,p_1)$ and $Y_2\sim b(400,p_2)$are independent.

(10 points) The statistical hypotheses being tested.

H0:p1=p2, Ha:p1<p2

(10 points) The test statistic being used.

The standardized test statistic is $Z=(\hat{p}_1-\hat{p}_2)/sqrt{\hat{p}(1-\hat{p})[1/n_1+1/n_2]}$, where $\hat{p}_1=Y_1/n_1=Y_1/400$, $\hat{p}_2=Y_2/n_2=Y_2/400$, and $\hat{p}=(Y_1+Y_2)/(n_1+n_2)=(Y_1+Y_2)/800$.

(20 points) The p-value.

Obtain the observed value of the standardized test statistic, z*, by plugging the observed values y1=29 and y2=46 into the above formulas. The result is:


The p-value is then $P(N(0,1)\leq -2.062)=0.0196$.

(10 points) The assumptions made, and why they are, or are not, justified.

It is assumed that Y1, n1-Y1, Y2 and n2-Y2 are all at least 10, which is the case here.

(10 points) Your conclusions.

Reject H0 in favor of Ha, at any significance level $\geq
0.0196$. So at the 0.05 level, reject, at the 0.01 level, do not reject.

(10 points) In a large-sample test of $H_0:\mu=2.5$ versus $H_a:\mu \neq 2.5$, n=100, the sample variance is 0.371, and the p-value is 0.0548. There are two possible values for $\bar{y}$. What are they? (Hint: note that Ha is two-sided.)

The observed test statistic is $z^*=(\overline{y}-\mu_0)/(s/\sqrt{n})=(\overline{y}-2.5)/(\sqrt{0.371}/\sqrt{100})
=(\overline{y}-2.5)/0.0609$. We know that $0.0548=p\pm=2\min(p_-,p^+)$, so that $\min(p_-,p^+)=0.0274$. Our two possible choices are p-=0.0274, or p*=0.0274.

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Joseph D Petruccelli