Math Problem Statement
A random sample of students at a college reported what they believed to be their heights in inches. Then the students measured each others' heights in centimeters, without shoes. The data provided are for the men, with their believed heights converted from inches to centimeters. Assume that conditions for t-tests hold. Complete parts a and b below.Centimeters (measured) Centimeters (believed) 165 167.64 174 172.72 182 187.96 164 167.64 191 195.58 172 175.26 176 177.80 190 193.04 180 180.34 179 177.80 181 182.88 175 172.72 173 167.64 172 167.64 184 180.34. a. Find a 95% confidence interval for the mean difference as measured in centimeters. Does it capture 0? What does that show? The 95% confidence interval for measuredminusbelieved is left parenthesis nothing comma nothing right parenthesis . (Round to three decimal places as needed.)
Solution
To calculate a 95% confidence interval for the mean difference between measured and believed heights, we follow these steps:
Step 1: Define the differences
Calculate the differences for each pair.
Step 2: Compute statistics for the differences
- Find the mean difference ().
- Find the standard deviation of the differences ().
- Calculate the standard error of the mean ().
Step 3: Determine the t-value
Find the critical -value for degrees of freedom at a 95% confidence level.
Step 4: Compute the confidence interval
The confidence interval is given by:
Perform the calculations:
I will calculate this for you.import numpy as np from scipy.stats import t
Data
measured = np.array([165, 174, 182, 164, 191, 172, 176, 190, 180, 179, 181, 175, 173, 172, 184]) believed = np.array([167.64, 172.72, 187.96, 167.64, 195.58, 175.26, 177.80, 193.04, 180.34, 177.80, 182.88, 172.72, 167.64, 167.64, 180.34])
Calculate differences
differences = measured - believed
Mean and standard deviation of differences
mean_diff = np.mean(differences) std_diff = np.std(differences, ddof=1) # sample standard deviation n = len(differences)
Standard error of the mean
SE = std_diff / np.sqrt(n)
Degrees of freedom
df = n - 1
t critical value for 95% confidence
t_critical = t.ppf(0.975, df)
Confidence interval
lower_bound = mean_diff - t_critical * SE upper_bound = mean_diff + t_critical * SE
lower_bound, upper_bound(-2.5154300935302305, 1.3154300935302343)(-2.5154300935302305, 1.3154300935302343)successThe 95% confidence interval for the mean difference (measured - believed) is approximately:
Interpretation:
- The interval captures , indicating that there is no statistically significant difference between the measured and believed heights at the 95% confidence level. This suggests that the discrepancies between measured and believed heights could be due to random variation rather than a systematic difference.
Would you like a detailed explanation of any of these steps?
5 Related Questions:
- How do you compute a confidence interval for paired data?
- What is the importance of checking the assumptions for a -test?
- Can a confidence interval that captures still be practically meaningful?
- How does the sample size affect the width of a confidence interval?
- What are alternative methods to analyze paired data?
Tip: Always verify the assumptions of normality and independence before conducting a -test for paired samples.
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Math Problem Analysis
Mathematical Concepts
Confidence Intervals
Paired Data Analysis
t-Tests
Statistical Inference
Formulas
Mean difference: \( \bar{d} = \frac{1}{n} \sum d_i \)
Standard error: \( SE = \frac{s_d}{\sqrt{n}} \)
Confidence interval: \( \bar{d} \pm t \cdot SE \)
Theorems
Central Limit Theorem
t-Distribution Properties
Suitable Grade Level
Undergraduate Statistics or AP Statistics
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