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In: Statistics and Probability

A new kind of typhoid shot is being developed by a medical research team. The old...

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 27 people were given the new shot. Regular blood tests showed that the sample standard deviation of protection times was 1.9 months. Using a 0.05 level of significance, test the claim that the new typhoid shot has a smaller variance of protection times.

Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)  

What are the degrees of freedom?

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

lower limit
upper limit    

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Expert Solution

SOLUTION:

From given data,

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 27 people were given the new shot. Regular blood tests showed that the sample standard deviation of protection times was 1.9 months. Using a 0.05 level of significance, test the claim that the new typhoid shot has a smaller variance of protection times.

level of significance = 0.05

s = sample standard deviation =1.9

= hypothesized standard deviation = 3

n = sample size = 27

Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)  

chi-square = =(n-1)*(s/σ)2

=(27-1)*(1.9/3)2

=10.4289

=10.43 (Round your answer to two decimal places.)  

What are the degrees of freedom

degrees of freedom = df = n-1

=27-1

  =26

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

90% confidence interval

90/100 = 0.90

= 1-0.90 =0.10

/2 = 0.10/2 = 0.05

= = 38.8

= = 15.4

90% confidence interval for

( n-1) S2 / <   <  ( n-1) S2 /

Lower limit =   ( n-1) S2 /   

= ( 27-1) 1.9 2 / 38.8

= 93.86 /38.8

= 2.419

= 2.42(Round your answers to two decimal places.)

Upper limit =   ( n-1) S2 /   

= ( 27-1) 1.9 2 / 15.4

= 93.86 /15.4

=6.0948

= 6.09(Round your answers to two decimal places.)

orchestra answered 2 years ago
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