Math / Data & Statistics

QuestionConsider a drug that is used to help prevent blood clots in certain patients. In clinical trials, among 6178 patients treated with this drug, 155 developed the adverse reaction of nausea. Use a 0.01 significance level to test the claim that $3 \%$ of users develop nausea. Does nausea appear to be a problematic adverse reaction?
Identify the null and alternative hypotheses for this test. Choose the correct answer below. A. $\mathrm{H}_{0}: p=0.03$ $H_{1}: p \neq 0.03$ B. $H_{0}: p=0.03$ $H_{1}: p<0.03$ C. $H_{0}: p=0.03$ $H_{1}: p>0.03$ D. $H_{0}: p \neq 0.03$ $H_{1}: p=0.03$ Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is $\square$ . (Round to two decimal places as needed.)

Studdy Solution

STEP 1

What is this asking? Out of nearly 6200 patients, 155 got nauseous after taking a new drug.
We need to figure out if this nausea rate is significantly different from 3% and decide if it's a big problem. Watch out! Don't mix up the actual nausea rate with the claimed 3% rate.
Also, pay close attention to whether we're looking for a rate greater than, less than, or different from the claim!

STEP 2

1. Set up the Hypothesis Test
2. Calculate the Sample Proportion
3. Calculate the Test Statistic

STEP 3

We're testing if the true proportion of patients experiencing nausea ( $p$ ) is different from 3%, so our null hypothesis ( $H_0$ ) is that $p$ equals 0.03.
Mathematically, that's $H_0: p = 0.03$ .

STEP 4

The alternative hypothesis ( $H_1$ ) is that the true proportion is not equal to 3%.
We write this as $H_1: p \neq 0.03$ .
This means we're looking for a difference in either direction – more or less nausea than 3%.
So the correct answer for the multiple-choice question is A.

STEP 5

The sample proportion ( $\hat{p}$ ) is the number of patients with nausea divided by the total number of patients.
We have 155 nauseous patients out of a total of 6178 patients.

STEP 6

So, $\hat{p} = \frac{155}{6178} \approx 0.0251$ .
This means approximately 2.51% of the patients in the sample experienced nausea.

STEP 7

The test statistic ( $z$ ) tells us how far our sample proportion is from the claimed proportion (0.03) in terms of standard deviations.
The formula is: $z = \frac{\hat{p} - p}{\sqrt{\frac{p \cdot (1 - p)}{n}}}$ Where $p$ is the claimed proportion (0.03), $\hat{p}$ is our calculated sample proportion (0.0251), and $n$ is the total number of patients (6178).

STEP 8

Let's plug in the numbers: $z = \frac{0.0251 - 0.03}{\sqrt{\frac{0.03 \cdot (1 - 0.03)}{6178}}}$ $z = \frac{-0.0049}{\sqrt{\frac{0.03 \cdot 0.97}{6178}}}$ $z = \frac{-0.0049}{\sqrt{\frac{0.0291}{6178}}}$ $z = \frac{-0.0049}{\sqrt{0.00000471}}$ $z = \frac{-0.0049}{0.00217}$ $z \approx -2.26$

STEP 9

So, our test statistic is approximately -2.26.
This means our sample proportion is about 2.26 standard deviations below the claimed proportion.

STEP 10

The correct hypothesis test is A. $H_0: p = 0.03$ , $H_1: p \neq 0.03$ .
The test statistic is approximately -2.26.
Since this value is quite far from zero, it suggests that the nausea rate might indeed be different from 3%.
Further calculations (p-value and comparison with the significance level) are needed to make a definitive conclusion about whether nausea is a "problematic" adverse reaction.

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Studdy Solution

STEP 1

STEP 2

1. Set up the Hypothesis Test
2. Calculate the Sample Proportion
3. Calculate the Test Statistic

STEP 3

We're testing if the true proportion of patients experiencing nausea ( $p$ ) is different from 3%, so our null hypothesis ( $H_0$ ) is that $p$ equals 0.03.
Mathematically, that's $H_0: p = 0.03$ .

STEP 4

The alternative hypothesis ( $H_1$ ) is that the true proportion is not equal to 3%.
We write this as $H_1: p \neq 0.03$ .
This means we're looking for a difference in either direction – more or less nausea than 3%.
So the correct answer for the multiple-choice question is A.

STEP 5

The sample proportion ( $\hat{p}$ ) is the number of patients with nausea divided by the total number of patients.
We have 155 nauseous patients out of a total of 6178 patients.

STEP 6

So, $\hat{p} = \frac{155}{6178} \approx 0.0251$ .
This means approximately 2.51% of the patients in the sample experienced nausea.

STEP 7

STEP 8

Let's plug in the numbers: $z = \frac{0.0251 - 0.03}{\sqrt{\frac{0.03 \cdot (1 - 0.03)}{6178}}}$ $z = \frac{-0.0049}{\sqrt{\frac{0.03 \cdot 0.97}{6178}}}$ $z = \frac{-0.0049}{\sqrt{\frac{0.0291}{6178}}}$ $z = \frac{-0.0049}{\sqrt{0.00000471}}$ $z = \frac{-0.0049}{0.00217}$ $z \approx -2.26$

STEP 9

So, our test statistic is approximately -2.26.
This means our sample proportion is about 2.26 standard deviations below the claimed proportion.

STEP 10

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Studdy Solution

STEP 1

STEP 2

1. Set up the Hypothesis Test2. Calculate the Sample Proportion3. Calculate the Test Statistic

STEP 3

We're testing if the *true* proportion of patients experiencing nausea (ppp) is different from **3%**, so our **null hypothesis** (H0H_0H0​) is that ppp *equals* **0.03**.Mathematically, that's H0:p=0.03H_0: p = 0.03H0​:p=0.03.

STEP 4

STEP 5

The **sample proportion** (p^\hat{p}p^​) is the number of patients with nausea divided by the total number of patients.We have **155** nauseous patients out of a total of **6178** patients.

STEP 6

So, p^=1556178≈0.0251\hat{p} = \frac{155}{6178} \approx 0.0251p^​=6178155​≈0.0251.This means approximately **2.51%** of the patients in the sample experienced nausea.

STEP 7

STEP 8

STEP 9

So, our **test statistic** is approximately **-2.26**.This means our sample proportion is about **2.26 standard deviations** *below* the claimed proportion.

STEP 10

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1. Set up the Hypothesis Test
2. Calculate the Sample Proportion
3. Calculate the Test Statistic

We're testing if the true proportion of patients experiencing nausea ( $p$ ) is different from 3%, so our null hypothesis ( $H_0$ ) is that $p$ equals 0.03.
Mathematically, that's $H_0: p = 0.03$ .

The sample proportion ( $\hat{p}$ ) is the number of patients with nausea divided by the total number of patients.
We have 155 nauseous patients out of a total of 6178 patients.

So, $\hat{p} = \frac{155}{6178} \approx 0.0251$ .
This means approximately 2.51% of the patients in the sample experienced nausea.

So, our test statistic is approximately -2.26.
This means our sample proportion is about 2.26 standard deviations below the claimed proportion.