Solved on Nov 03, 2023

Test if $3\%$ of patients treated with a drug develop nausea as an adverse reaction, using a $0.10$ significance level. Identify the null and alternative hypotheses.

STEP 1

Assumptions1. The total number of patients treated with the drug is6090. . The number of patients who developed nausea is149.
3. The claim is that3% of users develop nausea.
4. The significance level for the test is0.10.

STEP 2

First, we need to identify the null and alternative hypotheses. The null hypothesis (H0) is the claim we are testing. The alternative hypothesis (H1) is what we believe might be true if the null hypothesis is rejected.
In this case, the null hypothesis is that the proportion of users who develop nausea is% (0.03). The alternative hypothesis depends on what we are trying to prove. If we are testing whether the proportion is not%, then the alternative hypothesis is that the proportion is not%. If we are testing whether the proportion is less than%, then the alternative hypothesis is that the proportion is less than%. If we are testing whether the proportion is more than%, then the alternative hypothesis is that the proportion is more than%.

STEP 3

Now, we need to calculate the sample proportion (p̂) of patients who developed nausea. This is done by dividing the number of patients who developed nausea by the total number of patients.
$p̂ = \frac{Number\, of\, patients\, who\, developed\, nausea}{Total\, number\, of\, patients}$

STEP 4

Plug in the given values for the number of patients who developed nausea and the total number of patients to calculate the sample proportion.
$p̂ = \frac{149}{6090}$

STEP 5

Calculate the sample proportion.
$p̂ = \frac{149}{6090} \approx0.0245$

STEP 6

Now, we compare the sample proportion (p̂) with the claimed proportion (p). If p̂ > p, then we have evidence to suggest that the proportion of users who develop nausea is more than3%, and the alternative hypothesis is that p >0.03. If p̂ < p, then we have evidence to suggest that the proportion of users who develop nausea is less than3%, and the alternative hypothesis is that p <0.03. If we are only interested in whether the proportion is different from3% (not specifically whether it is more or less), then the alternative hypothesis is that p ≠0.03.
In this case, p̂ < p, so the alternative hypothesis is that p <0.03.

STEP 7

Therefore, the null and alternative hypotheses are $\begin{array}{l} H_{0}: p=0.03 \\ H_{1}: p<0.03\end{array}$ So, the correct answer is B.

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Test if $3\%$ of patients treated with a drug develop nausea as an adverse reaction, using a $0.10$ significance level. Identify the null and alternative hypotheses.

STEP 1

Assumptions1. The total number of patients treated with the drug is6090. . The number of patients who developed nausea is149.
3. The claim is that3% of users develop nausea.
4. The significance level for the test is0.10.

STEP 2

STEP 3

Now, we need to calculate the sample proportion (p̂) of patients who developed nausea. This is done by dividing the number of patients who developed nausea by the total number of patients.
$p̂ = \frac{Number\, of\, patients\, who\, developed\, nausea}{Total\, number\, of\, patients}$

STEP 4

Plug in the given values for the number of patients who developed nausea and the total number of patients to calculate the sample proportion.
$p̂ = \frac{149}{6090}$

STEP 5

Calculate the sample proportion.
$p̂ = \frac{149}{6090} \approx0.0245$

STEP 6

STEP 7

Therefore, the null and alternative hypotheses are $\begin{array}{l} H_{0}: p=0.03 \\ H_{1}: p<0.03\end{array}$ So, the correct answer is B.

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Test if 3%3\%3% of patients treated with a drug develop nausea as an adverse reaction, using a 0.100.100.10 significance level. Identify the null and alternative hypotheses.

STEP 1

Assumptions1. The total number of patients treated with the drug is6090. . The number of patients who developed nausea is149.3. The claim is that3% of users develop nausea.4. The significance level for the test is0.10.

STEP 2

STEP 3

STEP 4

Plug in the given values for the number of patients who developed nausea and the total number of patients to calculate the sample proportion.p^=1496090p̂ = \frac{149}{6090}p^​=6090149​

STEP 5

Calculate the sample proportion.p^=1496090≈0.0245p̂ = \frac{149}{6090} \approx0.0245p^​=6090149​≈0.0245

STEP 6

STEP 7

Therefore, the null and alternative hypotheses areH0:p=0.03H1:p<0.03\begin{array}{l} H_{0}: p=0.03 \\ H_{1}: p<0.03\end{array} H0​:p=0.03H1​:p<0.03​So, the correct answer is B.

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Test if $3\%$ of patients treated with a drug develop nausea as an adverse reaction, using a $0.10$ significance level. Identify the null and alternative hypotheses.

Assumptions1. The total number of patients treated with the drug is6090. . The number of patients who developed nausea is149.
3. The claim is that3% of users develop nausea.
4. The significance level for the test is0.10.

Plug in the given values for the number of patients who developed nausea and the total number of patients to calculate the sample proportion.
$p̂ = \frac{149}{6090}$

Calculate the sample proportion.
$p̂ = \frac{149}{6090} \approx0.0245$

Therefore, the null and alternative hypotheses are $\begin{array}{l} H_{0}: p=0.03 \\ H_{1}: p<0.03\end{array}$ So, the correct answer is B.