## 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.