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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

$\sigma$ = 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 = $\chi ^{2}$ =(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

$\alpha$ = 1-0.90 =0.10

$\alpha$ /2 = 0.10/2 = 0.05

$\chi _{\alpha/2,df }^{2}$ = $\chi _{0.05,26 }^{2}$ = 38.8

$\chi _{1-\alpha/2,df }^{2}$ = $\chi _{0.95,26 }^{2}$ = 15.4

90% confidence interval for $\sigma ^{2}$

( n-1) S² / $\chi _{\alpha/2 }^{2}$ < $\sigma ^{2}$ < ( n-1) S² / $\chi _{1-\alpha/2}^{2}$

Lower limit = ( n-1) S² / $\chi _{\alpha/2 }^{2}$

= ( 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 / $\chi _{1-\alpha/2}^{2}$

= ( 27-1) 1.9 ² / 15.4

= 93.86 /15.4

=6.0948

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

orchestra answered 1 year ago

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