Question 1198184: Consider a study conducted to evaluate the effectiveness of a new treatment to stop smoking. Before the study, it was known that the average time to quit was 2.3 months, with a standard deviation of 1.1. Using a sample of 50 adults on the new treatment, the sample average time was 1.9 months.
What should the p-value for the test be?
It should be one of the following.
A) 0.0038
B) 0.3594
C) 0.9962
D) 0.6406
Which one of the above is the low-value?
Answer by onyulee(41)   (Show Source):
You can put this solution on YOUR website! **1. Calculate the Standard Error**
* Standard Error (SE) = σ / √n
* σ = population standard deviation = 1.1 months
* n = sample size = 50
* SE = 1.1 / √50 ≈ 0.1556
**2. Calculate the Test Statistic (z-score)**
* z = (sample mean - population mean) / standard error
* z = (1.9 - 2.3) / 0.1556
* z = -4 / 0.1556 ≈ -2.57
**3. Calculate the p-value**
* Since we're interested in whether the new treatment is *more* effective (meaning it takes *less* time to quit), we're conducting a one-tailed test.
* We want to find the probability of getting a z-score as extreme or more extreme than -2.57 in the left tail of the standard normal distribution.
* Using a z-table or statistical software, we find:
* p-value = P(Z ≤ -2.57) ≈ 0.0050
**Therefore, the closest p-value to 0.0050 from the given options is A) 0.0038.**
**Key Points:**
* We use a z-test because the population standard deviation is known and the sample size is large enough (n ≥ 30).
* The p-value is small (0.0050), suggesting strong evidence against the null hypothesis (that the new treatment has no effect on the average time to quit smoking).
**In summary, the correct answer is A) 0.0038.**
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