Machines are used to pack sugar into packets supposedly containing 1.20 kg each. On testing a large number of packets over a long period of time, it was found that the mean weight of the packets was 1.24 kg and the standard deviation was 0.04 Kg. A particular machine is selected to check the total weight of each of the 25 packets filled consecutively by the machine. Calculate the limits within which the weight of the packets should lie assuming that the machine is not been classified as faulty.
Assuming that you are using P = .05, you would want 1.96 SEm (standard error of the mean) in either direction.
Mean ± 1.96SEm
SEm = SD/√(n-1)
To calculate the limits within which the weight of the packets should lie, we need to use the concept of confidence intervals.
1. Determine the sample mean: In this case, the sample mean is given as 1.24 kg.
2. Determine the standard deviation of the population: In this case, the standard deviation of the population is given as 0.04 kg.
3. Determine the sample size: The sample size is 25 packets.
4. Determine the desired level of confidence: Let's assume a 95% confidence level, which is commonly used.
5. Find the standard error: The standard error is calculated by dividing the standard deviation of the population by the square root of the sample size. In this case, the standard error is 0.04 kg / sqrt(25) = 0.008 kg.
6. Determine the critical value: The critical value is based on the desired level of confidence and the sample size. For a 95% confidence level and a sample size of 25, the critical value is 2.064 (obtained from a t-distribution table).
7. Calculate the margin of error: The margin of error is determined by multiplying the standard error by the critical value. In this case, the margin of error is 0.008 kg * 2.064 = 0.0165 kg.
8. Calculate the lower and upper limits: The lower limit is obtained by subtracting the margin of error from the sample mean, and the upper limit is obtained by adding the margin of error to the sample mean. In this case, the lower limit is 1.24 kg - 0.0165 kg = 1.2235 kg, and the upper limit is 1.24 kg + 0.0165 kg = 1.2565 kg.
Therefore, assuming that the machine is not faulty, the weight of the packets should lie within the limits of 1.2235 kg and 1.2565 kg.