# Statistical Inference and Confidence Intervals: A Comprehensive Guide to Hypothesis Testing and Estimation

Embark on a journey through the realm of statistical inference and confidence intervals with our comprehensive guide for your statistical inference assignment. Explore the intricacies of hypothesis testing, from evaluating pharmaceutical claims to estimating the life of wind turbines and investigating water quality. Gain a solid understanding of these statistical techniques through real-world examples and insightful solutions.

## Problem 1: Cancer Drug Efficacy

### Problem Description:

A pharmaceutical company claims that its cancer drug, X, extends patients' lives by at least 8 years. A sample of 25 patients yielded a sample mean of 8.5 years with a standard deviation of 1.5 years. We test this claim at a significance level of 0.01.

### Solution:

1. Hypotheses:

• Null Hypothesis (H0): µ = 8 years
• Alternative Hypothesis (H1): µ > 8 years
• Claim: Cancer patients using drug X will live at least 8 more years.

2. Critical Value:

• Degrees of freedom (d.f.): 24 (n - 1)
• One-tailed test at α = 0.01
• Critical value (t): 2.492

3. Test Value (t-statistic):

t_24=(8.5-8)/(1.5/√25)=0.5/0.3=1.667

4. Decision:

• Since 1.667<2.4921.667<2.492, we do not reject the null hypothesis.
• Conclusion: There is not sufficient evidence to support the claim that cancer patients using drug X will live at least 8 more years.

5. Confidence Interval (95%):

• Critical value (t): 2.064

Give the CIE at 95%

t0.05;24 = 2.064

8.5±2.064(1.5/√25)=8.5±0.6192

• 7.88 to 9.127.88to9.12 years

We are 95% confident that the mean lies between 7.88 and 9.12 years.

## Problem 2: Wind Turbine Life Estimate

### Problem Description:

A company produces wind turbines, and a sample of 25 turbines has a mean life of 20.00 years and a standard deviation of 2.50 years. A 95% confidence interval for the mean life is to be constructed.

### Solution:

d.f. = n -1 = 25 – 1 = 24

t0.05;24 = 2.064

n = 25

s = 2.50 years

X ̅=20 years

The 95% confidence interval for the mean is given by:

20±2.064(2.50/√25)=20±1.032

Upper limit = 20 + 1.032

Upper limit: 21.032 years

The upper limit of the 95% confidence interval for the mean life is 21.032 years.

## Problem 3: Benzene in Water

### Problem Description:

The Fresh Springs Company claims that benzene in their water is less than 1 ppm. A sample of 25 measurements yielded a mean of 1.10 ppm and a standard deviation of 0.20 ppm. The claim is tested at a significance level of 0.10.

### Solution:

1. Hypotheses:

• 0=1 ppmH0:μ=1ppm
• 1<1 ppmH1: μ<1ppm
• Claim: Benzene in the water is less than 1 ppm.

Critical value (t): t-statistic:

t_24=(1.10-1)/(0.20/√25)=0.1/0.04=2.5

2. Test value (t-statistic): 24=2.5t24=2.5

3. Decision: Do not reject 0H0.

• Conclusion: There is not sufficient evidence to support the claim of benzene being less than 1 ppm in The Fresh Springs Company's water.

## Problem 4: Computer Chip Life Estimate

### Problem Description:

The average life of a new computer chip is estimated using a sample of 16 chips with a mean of 10 years and a standard deviation of 0.80 years. A 90% confidence interval is constructed.

### Solution:

Degrees of freedom (d.f.): 15 (n - 1)

Critical value (t): d.f. = n -1 = 16 – 1 = 15

t0.10; 15 = 1.753

n = 16

s = 0.80 years

X ̅=10 years

10±1.753(0.80/√16)=10±0.3506The 90% confidence interval is (9.65 to 10.35)(9.65to10.35) years.

## Problem 5: Battery Life Claim

### Problem Description:

The Battery & Gadgets store claims a battery life of 600 hours. A sample of 20 batteries has a mean life of 580 hours and a standard deviation of 40 hours. A two-tailed 95% confidence interval for the mean is to be constructed.

### Solution:

1. Degrees of freedom (d.f.): 19 (n - 1)

Critical value (t): The 95% confidence interval for the mean is given by:

580±2.093(40/√20)=580±18.72

2. 580±2.093(4020)=580±18.72580±2.093(2040)=580±18.72

3. Upper limit: 580+18.72=598.72580+18.72=598.72 hours

The upper limit of the 95% confidence interval is 598.72 hours.

## Problem 6: Hospital Infections Investigation

### Problem Description:

An investigation claims the average weekly infections in a hospital in Westchester are over 16. A sample of 10 weeks has a mean of 16.2 infections and a standard deviation of 1.5. The claim is tested at a significance level of 0.25.

### Solution:

Hypotheses:

• 0=16 infections per weekH0: μ=16infections per week
• 1>16 infections per weekH1: μ>16infections per week
• Claim: The average number of infections per week is more than 16.

Critical value (t): 9=0.7027t9=0.7027

Test value (t-statistic):

t_9=(16.2-16)/(1.5/√10)=0.2/0.4743=0.422

3. Decision: Do not reject 0H0.

• Conclusion: There is not enough evidence to support the claim of more than 16 infections per week at the hospital.