Math  /  Data & Statistics

QuestionSuppose 250 subjects are treated with a drug that is used to treat pain and 53 of them developed nausea. Use a 0.05 significance level to test the claim that more than 20%20 \% of users develop nausea. H1:p<0.20H_{1}: p<0.20 B. H0:p=0.20H_{0}: p=0.20 H1:p>0.20H_{1}: p>0.20 C. H0:p>0.20H_{0}: p>0.20 H1:p=0.20H_{1}: p=0.20 D. H0:p=0.20H_{0}: p=0.20 H1:p0.20H_{1}: p \neq 0.20
Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is 0.47 . (Round to two decimal places as needed.) Identify the P-value for this hypothesis test. The P-value for this hypothesis test is \square (Round to three decimal places as needed.)

Studdy Solution

STEP 1

1. We are testing the claim that more than 20% 20\% of users develop nausea.
2. The sample size is 250, with 53 subjects developing nausea.
3. The significance level is α=0.05 \alpha = 0.05 .

STEP 2

1. Identify the correct hypothesis test.
2. Calculate the test statistic.
3. Determine the P-value.
4. Make a decision based on the P-value and significance level.

STEP 3

Identify the correct hypothesis test. The claim is that more than 20% 20\% of users develop nausea, which corresponds to:
H0:p=0.20 H_0: p = 0.20 H1:p>0.20 H_1: p > 0.20
This matches option B.

STEP 4

Calculate the test statistic using the formula for a proportion:
z=p^p0p0(1p0)n z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}
Where: - p^=53250=0.212\hat{p} = \frac{53}{250} = 0.212 - p0=0.20p_0 = 0.20 - n=250n = 250
Substitute the values into the formula:
z=0.2120.200.20×0.80250 z = \frac{0.212 - 0.20}{\sqrt{\frac{0.20 \times 0.80}{250}}}
z=0.0120.00064 z = \frac{0.012}{\sqrt{0.00064}}
z=0.0120.025298 z = \frac{0.012}{0.025298}
z0.474 z \approx 0.474
Round to two decimal places:
z0.47 z \approx 0.47

STEP 5

Determine the P-value for the test statistic z=0.47 z = 0.47 .
Using a standard normal distribution table or calculator, find the P-value for z=0.47 z = 0.47 .
The P-value is the probability that Z>0.47 Z > 0.47 .
P(Z>0.47)0.319 P(Z > 0.47) \approx 0.319
Round to three decimal places:
P0.319 P \approx 0.319

STEP 6

Make a decision based on the P-value and significance level α=0.05 \alpha = 0.05 .
Since the P-value 0.319 0.319 is greater than α=0.05 \alpha = 0.05 , we fail to reject the null hypothesis H0 H_0 .
The P-value for this hypothesis test is:
0.319 \boxed{0.319}

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