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.