Math Problem Statement
A survey found that in 2 0 0 8 , doctors in residency programs at U . S . hospitals worked an average of 5 9 . 3 hours per week. A new residency program guideline went into effect in 2 0 0 9 , with the aim of reducing the number of hours residents work per week. In 2 0 0 9 , a small survey of 2 0 residents found the following weekly hours worked: 5 3 ,
5 7 ,
5 2 ,
6 5 ,
7 0 ,
5 1 ,
5 5 ,
6 1 ,
4 9 ,
6 1 , 5 4 ,
6 4 ,
4 4 ,
5 7 ,
5 5 ,
6 0 ,
5 3 ,
6 1 ,
5 1 ,
5 6 You would like to show that that the new guidelines have reduced the number of hours worked per week. Perform a test using alpha
0 . 0 5 , and answer the following: a . State the Null and Alternative hypotheses b . Does the sample provide enough evidence to suggest that the new guidelines have reduced the number of hours worked per week? c . For the test in ( b ) , what is the p
value?
Solution
To determine whether the new guidelines have reduced the average number of hours worked per week, we will perform a one-sample t-test since the sample size is small () and the population standard deviation is unknown. Let’s proceed step by step.
a. Null and Alternative Hypotheses
- Null Hypothesis (): The average weekly hours worked under the new guidelines is the same as or more than before:
- Alternative Hypothesis (): The average weekly hours worked under the new guidelines has been reduced:
This is a one-tailed test since we are testing for a reduction.
b. Does the sample provide enough evidence to suggest that the new guidelines have reduced the number of hours worked per week?
-
Given Data:
- Sample size () = 20
- Sample data:
- Population mean () = 59.3
- Significance level () = 0.05
-
Calculate the sample mean ():
-
Calculate the sample standard deviation ():
-
Compute the t-statistic:
-
**Compare the t-statistic to the critical value of for a one-tailed test or compute as np from scipy.stats import t
Data
data = [53, 57, 52, 65, 70, 51, 55, 61, 49, 61, 54, 64, 44, 57, 55, 60, 53, 61, 51, 56] n = len(data) mu_0 = 59.3 # Population mean
Sample statistics
sample_mean = np.mean(data) sample_std = np.std(data, ddof=1) # Use ddof=1 for sample standard deviation
Test statistic (t)
t_stat = (sample_mean - mu_0) / (sample_std / np.sqrt(n))
Degrees of freedom
df = n - 1
P-value (one-tailed test)
p_value = t.cdf(t_stat, df)
sample_mean, sample_std, t_stat,1. Sample Mean (): 2. Sample Standard Deviation (): (rounded to 2 decimal places) 3. t-statistic: 4. p-value: (rounded to 3 decimal places)
Interpretation
-
Compare p-value to :
- Since the p-value () is less than the significance level (), we reject the null hypothesis ().
-
Conclusion:
- There is enough evidence to suggest that the new guidelines have reduced the average number of weekly hours worked by residents.
c. p-value
The p-value for the test is .
Summary
The test indicates that the new guidelines were effective in reducing the average weekly hours worked by residents, as the p-value is statistically significant.
Do you have questions or need further clarification? Here are five questions to consider:
- How does the t-test differ from other statistical tests?
- What assumptions are required for this t-test to be valid?
- How does the sample size impact the reliability of this test?
- Could we perform a similar test if we had a larger dataset?
- Why do we use a one-tailed test here instead of a two-tailed test?
Tip: Always verify the assumptions of normality when performing t-tests,
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Math Problem Analysis
Mathematical Concepts
Hypothesis Testing
One-Sample t-Test
Statistical Significance
Formulas
Sample Mean: \(\bar{x} = \frac{\text{Sum of all sample values}}{n}\)
Sample Standard Deviation: \(s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\)
t-Statistic: \(t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}\)
Theorems
Central Limit Theorem
Student's t-Distribution
Suitable Grade Level
Grades 11-12, Undergraduate Statistics
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