Math  /  Data & Statistics

QuestionRhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 36 of the 42 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 83 of the 99 subjects developed rhinovirus infections. Use a 0.01 significance level to test the claim that echinacea has an effect on rhinovirus infections. Complete parts (a) through (b) below. a. Test the claim using a hypothesis test.
Consider the first sample to be the sample of subjects treated with echinacea and the second sample to be the sample of subjects treated with a placebo. What are the null and alternative hypotheses for the hypothesis test? A. H0:p1=p2H_{0}: p_{1}=p_{2} B. H0:p1p2H_{0}: p_{1} \leq p_{2} C. H0:p1=p2H_{0}: p_{1}=p_{2} H1:p1p2H_{1}: p_{1} \neq p_{2} H1:p1p2H_{1}: p_{1} \neq p_{2} D. H0:p1p2H_{0}: p_{1} \geq p_{2} E. H0:p1p2H_{0}: p_{1} \neq p_{2} H1:p1p2H_{1}: p_{1} \neq p_{2} H1:p1=p2H_{1}: p_{1}=p_{2} F. H0:p1=p2H_{0}: p_{1}=p_{2} H1:p1>p2H_{1}: p_{1}>p_{2}
Identify the test statistic. z=z= \square (Round to two decimal places as needed.) Identify the P -value. P -value == \square (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? The PP-value is \square the significance level of α=0.01\alpha=0.01, so \square the null hypothesis. There \square sufficient evidence to support the

Studdy Solution

STEP 1

What is this asking? We want to see if echinacea makes a difference in how many people get rhinovirus infections, compared to a placebo. Watch out! Don't mix up the echinacea and placebo groups!
Also, remember that a small P-value means there's a big difference!

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 conclusion

STEP 3

We're testing if echinacea has *any* effect, meaning it could either increase or decrease infections.
So, our **null hypothesis** H0H_0 is that the proportions are equal: p1=p2p_1 = p_2.
The **alternative hypothesis** H1H_1 is that the proportions are *not* equal: p1p2p_1 \neq p_2.
This corresponds to option **C**.

STEP 4

The **pooled proportion** is the total number of infections divided by the total number of people.
This assumes the null hypothesis is true, so we can combine the groups.
There were **36** infections in the echinacea group and **83** in the placebo group, totaling **36 + 83 = 119** infections.
There were **42** people in the echinacea group and **99** in the placebo group, totaling **42 + 99 = 141** people.
The **pooled proportion** is p^=1191410.844\hat{p} = \frac{119}{141} \approx \textbf{0.844}.

STEP 5

The **test statistic**, *z*, measures how far apart the two sample proportions are, considering the pooled proportion.
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, n1n_1 and n2n_2 are the sample sizes, and p^\hat{p} is the pooled proportion.

STEP 6

Let's plug in the values. p^1=36420.857\hat{p}_1 = \frac{36}{42} \approx \textbf{0.857}, p^2=83990.838\hat{p}_2 = \frac{83}{99} \approx \textbf{0.838}, n1=42n_1 = \textbf{42}, n2=99n_2 = \textbf{99}, and p^0.844\hat{p} \approx \textbf{0.844}. z=(0.8570.838)0.844(10.844)(142+199) z = \frac{(0.857 - 0.838)}{\sqrt{0.844 \cdot (1 - 0.844) \cdot (\frac{1}{42} + \frac{1}{99})}}

STEP 7

Let's simplify: z=0.0190.8440.156(142+199)0.0190.13150.03380.0190.004440.0190.06660.29 z = \frac{0.019}{\sqrt{0.844 \cdot 0.156 \cdot (\frac{1}{42} + \frac{1}{99})}} \approx \frac{0.019}{\sqrt{0.1315 \cdot 0.0338}} \approx \frac{0.019}{\sqrt{0.00444}} \approx \frac{0.019}{0.0666} \approx \textbf{0.29} So, z0.29z \approx \textbf{0.29}.

STEP 8

The **P-value** is the probability of getting a test statistic as extreme as ours (or more extreme) if the null hypothesis were true.
Since our alternative hypothesis is p1p2p_1 \neq p_2, this is a **two-tailed test**.
We look up the area in both tails corresponding to our *z*-score of **0.29**.
Using a *z*-table or calculator, the area in one tail is approximately 0.3859.
Since it's a two-tailed test, we multiply this by 2: 20.38590.7722 \cdot 0.3859 \approx \textbf{0.772}.

STEP 9

Our **P-value** (0.772\approx 0.772) is *much larger* than our **significance level** of α=0.01\alpha = 0.01.
This means we *fail to reject* the null hypothesis.

STEP 10

Test statistic z0.29z \approx 0.29.
P-value 0.772\approx 0.772.
We fail to reject the null hypothesis.
There is not sufficient evidence to support the claim that echinacea has an effect on rhinovirus infections.

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