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if the sample mean and the standard deviation for the fill weights of 81 boxes 1

ID: 3160646 • Letter: I

Question

if the sample mean and the standard deviation for the fill weights of 81 boxes 11.98 oz and 0.16 oz respectively a) Find 95% confidence interval for the mean fill of the boxes
b) Find a 99% confidence interval for the mean fill of the boxes
c) Compared the two confidence intervals and comment on this
d) Interpret the meaning of the confidence interval you found in part (a) If the sample mean and standard deviation for the fill weights of 8I boxes are 11.98 oz and 0.16 oz, respectively 3. (20 Points) a. Find a 95% confidence interval for the mean fill of the boxes b. Find a 99% confidence interval for the mean fill of the boxes c. Compare the two confidence intervals and comment on this. d. Interpret the meaning of the confidence interval you found in part (a). Let X be a random standard deviation 8. Find probability that X lies in the interval [55, 85]. variable with a normal distribution with mean 81and 4. (10 Points)

Explanation / Answer

a)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    11.98          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    0.16          
n = sample size =    81          
              
Thus,              
Margin of Error E =    0.034843804          
Lower bound =    11.9451562          
Upper bound =    12.0148438          
              
Thus, the confidence interval is              
              
(   11.9451562   ,   12.0148438   ) [ANSWER]

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b)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    11.98          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    0.16          
n = sample size =    81          
              
Thus,              
Margin of Error E =    0.045792521          
Lower bound =    11.93420748          
Upper bound =    12.02579252          
              
Thus, the confidence interval is              
              
(   11.93420748   ,   12.02579252   ) [ANSWER]

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c)

The 99% confidence interval is wider. It is because the critical z value is larger for 99%.

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d)

We are 95% confident that the true mean fill weight is between 11.9451562 and 12.0148438 oz. [CONCLUSION]

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