A random sample of 26 nutrition bars from a particular company has standard dvia

Question

A random sample of 26 nutrition bars from a particular company has standard dviation of 1.19 grams of carbohydrates. The Population of carbohydrates in this particular company's nutrition bars is normally distributed.

a. Construct a 99% confidence interval for the true variance of carbohydrates in the nutrition bar.

b. Construct a 99% confidence interval for the true standard deviation of carbohydrates in the nutrition bar.

c. The manufacturer claims that the standard deviation of carbohydrates in the nutrition bars is 1.11 grams. Test the claim of the manufacturer at the = 0.05 level.

Explanation / Answer

a)

As n = 26, s = 1.19,

As              
              
df = n - 1 =    25          
alpha = (1 - confidence level)/2 =    0.005          
              
Then the critical values for chi^2 are              
              
chi^2(alpha/2) =    46.92789016          
chi^2(alpha/2) =    10.51965211          
              
Thus, as              
              
lower bound = (n - 1) s^2 / chi^2(alpha/2) =    0.754402124          
upper bound = (n - 1) s^2 / chi^2(1 - alpha/2) =    3.365367944          
              
Thus, the confidence interval for the variance is              
              
(   0.754402124   ,   3.365367944   ) [ANSWER]

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b)

Also, for the standard deviation, getting the square root of the bounds,              
              
(   0.868563253   ,   1.83449392   ) [ANSWER]

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c)

Formulating the null and alternative hypotheses,              
              
Ho:   sigma   =   1.11  
Ha:    sigma   =/   1.11  
              
As we can see, this is a    two   tailed test.      
              
Thus, getting the critical chi^2, as alpha =    0.05   ,      
alpha/2 =    0.025          
df = N - 1 =    25          
chi^2 (crit) =    13.11972002   and   40.64646912  
              
Getting the test statistic, as              
s = sample standard deviation =    1.19          
sigmao = hypothesized standard deviation =    1.11          
n = sample size =    26          
              
              
Thus, chi^2 = (N - 1)(s/sigmao)^2 =    28.73346319          
              
As chi^2 is between the two critical values, we FAIL TO REJECT THE NULL HYPOTHESIS.              

Hence, we cannot reject the manufacturer's claim that the standard deviation of carbohydrates in the nutrition bars is 1.11 grams at 0.05 level. [CONCLUSION]

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