Question

Honeydew bottles honey jars and sells them through retail channels. The weight on the sticker says 20 oz and Honeydew claims its bottles have a weight that is normally distributed with a mean of 20 oz and std dev of 2 oz. The retailer has been receiving several complaints lately and decides to measure a sample of 4 jars and finds weights of 18, 20, 17 and 19 oz respectively.

a. The retailer complains to Honeydew if he is 95% confident that the mean is lower than advertised. Does he complain?

b. If the retailer’s customers return any jar under 18 oz in weight, what fraction of the retailer’s sales result in a return?

c. Assuming Honeydew’s claim is true, what is their current sigma capability? What can they do to reduce the fraction of returns to 5%? What then would the new sigma capability be?

Answer 1

Answer:-

Honeydew bottles honey jars and sells them through retail channels. The weight on the sticker says 20 oz and Honeydew claims its bottles have a weight that is normally distributed with a mean of 20 oz and std dev of 2 oz. The retailer has been receiving several complaints lately and decides to measure a sample of 4 jars and finds weights of 18, 20, 17 and 19 oz respectively.

a)

For 95% confidence, value of z = NORMSINV(0.95) = 1.65

Lower control limit

In the samples measured by him, there are two samples having weights of 18 and 17 oz, which are below the lower control limit. Therefore, he will complain.

b)

For 18oz weight, value of z = (18-20)/2 =-2/2= -1

P(z) = NORMSDIST(-1) = 0.1586

So, 15.86 % of the retailer's sales would result in a return.

c)

For z = -1, the spread of process variation is from -sigma to +sigma. So, current sigma capability = 2

For 5% return, value of z = NORMSINV(0.05) = 1.65.

Therefore, required standard deviation = (18-20)/1.65 = 1.21

In otder to reduce the fraction of returns to 5%, they need to reduce the standard deviation to 1.21 as determined above.

New sigma capability = 1.65*2 = 3.3