Question

A new kind of typhoid shot is being developed by a medical research team. The old typhoid shot was known to protect the population for a mean time of 36 months, with a standard deviation of 3 months. To test the time variability of the new shot, a random sample of 23 people were given the new shot. Regular blood tests showed that the sample standard deviation of protection times was 1.7 months. Using a 0.05 level of significance, test the claim that the new typhoid shot has a smaller variance of protection times.

(f) Find a 90% confidence interval for the population standard deviation. (Round your answers to two decimal places.)

lower limit

upper limit

Answer 1

(1)

Data:    

n = 23   

σ = 3   

s = 1.7   

Hypotheses:    

Ho: σ ≥ 3   

Ha: σ < 3   

Decision Rule:    

Degrees of freedom = n - 1 =    23 - 1 = 22

α = 0.05   

Critical value =   12.33801468

Reject Ho if the test χ2 value <    12.33801468

Test Statistic:    

χ2 = (n - 1) s^2 / σ^2 =   (23 - 1) * 1.7^2 / 3^2 = 7.064444444

p- value = 0.001095609   

Decision (in terms of the hypotheses):    

Since 7.064444444 < 12.33801468 we reject Ho

Conclusion (in terms of the problem):    

There is sufficient evidence that the population standard deviation is now less than 3 months. The new typhoid shot has a smaller variance of protection times.

(2)

n = 23          

s^2 = 2.8900          

% = 90          

Left χ2 value = 12.3380          

Right χ2 value = 33.9244          

Lower limit of the confidence interval (for variance) = (n - 1) * s^2 /Right χ2 value = (23 -1) * 2.89^2/33.9244385165976

= 1.8742

Upper limit of the confidence interval (for variance) = (n - 1) * s^2 /Left χ2 value = (23 -1) * 2.89^2/12.3380146806528

= 5.1532

The confidence interval is for variance is (1.87, 5.15)     

Lower limit of the confidence interval (for standard deviation) = 1.3690      

Upper limit of the confidence interval (for standard deviation) = 2.2701      

The confidence interval for standard deviation is (1.37, 2.27)