STATISTICS: Acceptance sampling, please help?

Hi,


I'm studying stats independently so don't have a teacher to ask how to finish this question. Any help would be much appreciated.





Before being delivered, bags of lawn seed are chosen at random from large batches and the contents weighed. The weight is known to be normally distributed with a standard deviation of 0.5kg.





A batch is accepted if the mean of 4 bags exceeds 24.8kg.





The manufacturer requires that the probability of accepting a batch with mean contents of 25kg should be at least 0.95. The garden centre requires that the probability of accepting a batch with mean contents of 24.6kg should be less than 0.01.





Question: Determine the minimum sample size and the weight which must be exceeded by the sample mean for a batch to be accepted.





I got n=17 and n=34 for the first part, but the answer book says n=25. I couldn't work out where to start for the second part. The answer is 24.84 but I don't know how to get there.

STATISTICS: Acceptance sampling, please help?
We know the true variance so we use Z-tables instead of T-tables. There is an interactive one here:





http://davidmlane.com/hyperstat/z_table....





The standard deviation is 0.5kg, so the variance is 0.25 kg^2.





%26gt;%26gt; A batch is accepted if the mean of 4 bags exceeds 24.8kg.





Okay, if the true mean is u, then the mean of a random sample of 4 is also normally distributed, also with mean u, but with standard deviation 0.5 kg / sqrt(4) = 0.25 kg.





%26gt;%26gt; The manufacturer requires that the probability of accepting a batch with mean contents of 25kg should be at least 0.95





Go to the web site mentioned above, use the upper form, put in 25 for the mean, 0.25 for the standard deviation, and above 24.8 for the test. The probability of acceptance is 0.788145, which is not at least 0.95.





%26gt;%26gt; The garden centre requires that the probability of accepting a batch with mean contents of 24.6kg should be less than 0.01.





Mean 24.6, standard deviation 0.25, test below 24.8: the probability of rejecting is 0.788145, which is not less than 0.01.





Conclusion: "A batch is accepted if the mean of 4 bags exceeds 24.8kg." isn't working for either party here.





%26gt;%26gt; Question: Determine the minimum sample size and the weight which must be exceeded by the sample mean for a batch to be accepted.





We expect the cutoff to be above 24.6 and below 25.0, and n to be large enough so that 0.5 / sqrt(n) is small enough so that both:





1) P(cutoff %26lt; Z) %26gt;= 0.95 for a normal with


mean 25, standard deviation 0.5 / sqrt(n)





2) P(Z %26lt; cutoff) %26lt; 0.01 for a normal with


mean 24.6, standard deviation 0.5 / sqrt(n)





Here I would normally switch to something programmable instead of trying various values by hand. But they say the answer has n = 25, so we make an inspired guess and check both n = 24 and n = 25.





For n = 24, then 0.5 / sqrt(n) = (approx) 0.102062


For n = 25, then 0.5 / sqrt(n) = (exactly) 0.1





Now we use the bottom chart:





mean 25.0, sd 0.102062 shaded area 0.95 test Above: cutoff %26lt; 24.8321


mean 24.6, sd 0.102062 shaded area 0.99 test Below: cutoff %26gt; 24.8374





So n = 24 does not work, the appriate cutoff regions do not overlap.





mean 25.0, sd 0.1 shaded area 0.95 test Above: cutoff %26lt; 24.8355


mean 24.6, sd 0.1 shaded area 0.99 test Below: cutoff %26gt; 24.8326





So n = 25 does work, assuming the cutoff weight is taken so that 24.8326 %26lt; cutoff %26lt; 24.8355.





Dan