Math  /  Data & Statistics

QuestionPart 1 of 4 Points: 0 of 1
In randomized, double-blind clinical trials of a new vaccine, monkeys were randomly divided into two groups. Subjects in group 1 received the new vaccine while subjects in group 2 received a control vaccine. After the second dose, 134 of 460 subjects in the experimental group (group 1) experienced drowsiness as a side effect. After the second dose, 26 of 84 of the subjects in the control group (group 2) experienced drowsiness as a side effect. Does the evidence suggest that a different proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the α=0.01\alpha=0.01 level of significance?
Determine the null and alternative hypotheses. Choose the correct answer below. A. H0:p1=p2H_{0}: p_{1}=p_{2} versus H1:p1<p2H_{1}: p_{1}<p_{2} B. H0:p1=0H_{0}: p_{1}=0 versus H0:p10H_{0}: p_{1} \neq 0 C. H0:p1=p2H_{0}: p_{1}=p_{2} versus H1:p1p2H_{1}: p_{1} \neq p_{2} D. H0:p1=p2H_{0}: p_{1}=p_{2} versus H1:p1>p2H_{1}: p_{1}>p_{2}

Studdy Solution

STEP 1

What is this asking? We want to see if there's a real difference in drowsiness rates between monkeys who got the new vaccine and those who got the control vaccine. Watch out! Don't mix up the groups!
We're comparing the *proportions* of sleepy monkeys, not just the raw counts.

STEP 2

1. Set up the hypotheses
2. Calculate the pooled proportion
3. Calculate the test statistic
4. Find the p-value
5. Make a decision

STEP 3

We're looking for a *difference* in drowsiness rates, but we don't know beforehand if one group will be sleepier than the other.
This means we're doing a **two-tailed test**.

STEP 4

Our **null hypothesis** \(H_0\) is that the proportions are the same: p1=p2p_1 = p_2.

STEP 5

Our **alternative hypothesis** \(H_1\) is that the proportions are *different*: p1p2p_1 \neq p_2.
This matches answer choice **C**.

STEP 6

The **pooled proportion** is like a weighted average of the drowsiness rates in both groups.
It assumes the null hypothesis is true (that the proportions are the same) and gives us an overall estimate of that shared proportion.

STEP 7

The formula is: p^=x1+x2n1+n2 \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} where x1x_1 and x2x_2 are the number of drowsy monkeys in each group, and n1n_1 and n2n_2 are the total number of monkeys in each group.

STEP 8

Plugging in our **values**: x1=134x_1 = \textbf{134}, n1=460n_1 = \textbf{460}, x2=26x_2 = \textbf{26}, and n2=84n_2 = \textbf{84}, we get: p^=134+26460+84=1605440.294 \hat{p} = \frac{134 + 26}{460 + 84} = \frac{160}{544} \approx \textbf{0.294}

STEP 9

The **test statistic** tells us how far apart our observed proportions are from what we'd expect if the null hypothesis were true.
We'll use the **z-test** for proportions.

STEP 10

The formula is: z=(p^1p^2)0p^(1p^)(1n1+1n2) z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}} where p^1\hat{p}_1 and p^2\hat{p}_2 are the sample proportions for each group.

STEP 11

First, let's calculate p^1\hat{p}_1 and p^2\hat{p}_2: p^1=x1n1=1344600.291 \hat{p}_1 = \frac{x_1}{n_1} = \frac{134}{460} \approx \textbf{0.291} p^2=x2n2=26840.310 \hat{p}_2 = \frac{x_2}{n_2} = \frac{26}{84} \approx \textbf{0.310}

STEP 12

Now, plug everything into the **z-test formula**: z=(0.2910.310)00.294(10.294)(1460+184)0.0190.2070.0140.0190.054-0.35 z = \frac{(0.291 - 0.310) - 0}{\sqrt{0.294(1 - 0.294)(\frac{1}{460} + \frac{1}{84})}} \approx \frac{-0.019}{\sqrt{0.207 \cdot 0.014}} \approx \frac{-0.019}{0.054} \approx \textbf{-0.35}

STEP 13

The **p-value** is the probability of observing a test statistic as extreme as ours (or more extreme) if the null hypothesis were true.
Since this is a two-tailed test, we need to consider both tails of the distribution.

STEP 14

Using a z-table or calculator, we find the area to the left of z=0.35z = -0.35 is approximately 0.363\textbf{0.363}.
Since it's a two-tailed test, we double this value to get our p-value: 20.3630.7262 \cdot 0.363 \approx \textbf{0.726}.

STEP 15

Our **p-value** (0.726\textbf{0.726}) is much larger than our **significance level** (α=0.01\alpha = 0.01).

STEP 16

This means we **fail to reject the null hypothesis**.
There's not enough evidence to say there's a real difference in drowsiness rates between the two groups of monkeys.

STEP 17

The correct answer is C.
We fail to reject the null hypothesis.
There's not enough evidence to suggest a difference in drowsiness rates between the two groups at the 1% significance level.

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