Math Problem Statement

a popular theory is that presidential candidates have an advantage if they are taller than thier opponents presidents height = 176,186,176,186,193,170 hieght of main opponent 167,186,180,170,188,183 the null hypothesis is Ho=miud=0 the alternative is H1:miud > 0cm identify the T statistic and the p-value construct a confiden interval <miud<__ in this ecxample miud is the mean value of the difference d for the population of all pairs of data the difference have a mean greater than 0 and the significance interval is 0.01

Solution

We are conducting a one-sample t-test for paired data (dependent samples) to determine if the mean difference (μd\mu_d) in height between presidents and their main opponents is significantly greater than 0.

Step 1: Compute the Differences (dd)

The differences between the heights of the presidents and their main opponents are:

d=Presidents’ HeightsOpponents’ Heightsd = \text{Presidents' Heights} - \text{Opponents' Heights}

Presidents' Heights (PP)Opponents' Heights (OO)Difference (d=POd = P - O)
1761679
1861860
176180-4
18617016
1931885
170183-13

The differences (dd) are:
9,0,4,16,5,139, 0, -4, 16, 5, -13

Step 2: Compute Sample Statistics

We compute:

  • Mean of differences (dˉ\bar{d})
  • Standard deviation of differences (sds_d)
  • Sample size (n=6n = 6)

Let's calculate these values. numpy as np

Given differences

d = np.array([9, 0, -4, 16, 5, -13])

Compute mean and standard deviation

d_mean = np.mean(d) d_std = np.std(d, ddof=1) # Sample standard deviation (ddof=1 for unbiased estimate) n = len(d)

d_mean, d_std, nThe computed values are:

  • Mean difference: dˉ=2.167\bar{d} = 2.167
  • Standard deviation: sd=10.187s_d = 10.187
  • Sample size: n=6n = 6

Step 3: Compute the T-Statistic

The t-statistic is given by:

t=dˉ0sd/nt = \frac{\bar{d} - 0}{s_d / \sqrt{n}}

Let's compute it. Compute t-statistic t_stat = d_mean / (d_std / np.sqrt(n)) t_statThe t-statistic is t=0.521t = 0.521.

Step 4: Compute the p-value

Since this is a one-tailed test (H1:μd>0H_1: \mu_d > 0), we compute:

p=P(T>t) for df=n1=5p = P(T > t) \text{ for } df = n - 1 = 5

Let's calculate the p-value. scipy.stats import t

Degrees of freedom

df = n - 1

Compute one-tailed p-value

p_value = 1 - t.cdf(t_stat, df) p_valueThe p-value is p=0.312p = 0.312.

Step 5: Construct the Confidence Interval

The 99% confidence interval for μd\mu_d is given by:

dˉ±t×sdn\bar{d} \pm t^* \times \frac{s_d}{\sqrt{n}}

where tt^* is the critical value from the t-distribution with df=5df = 5 at a 99% confidence level.

Let's compute this interval. Get the critical t-value for 99% confidence level (two-tailed) t_critical = t.ppf(0.995, df) # 0.995 because it's two-tailed

Compute margin of error

margin_error = t_critical * (d_std / np.sqrt(n))

Compute confidence interval

lower_bound = d_mean - margin_error upper_bound = d_mean + margin_error

lower_bound, upper_boundIt seems like I can’t do more advanced data analysis right now. Please try again later.

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Math Problem Analysis

Mathematical Concepts

Inferential Statistics
Hypothesis Testing
Paired Sample t-Test
Confidence Intervals

Formulas

Mean of Differences: ȳd = (Σd) / n
Standard Deviation of Differences: sd = sqrt[(Σ(d - ȳd)²) / (n-1)]
T-Statistic: t = (ȳd - μ0) / (sd / sqrt(n))
Confidence Interval: ȳd ± t* × (sd / sqrt(n))

Theorems

Student's t-distribution
Central Limit Theorem

Suitable Grade Level

Undergraduate Statistics, Advanced High School AP Statistics