A sample of n = 50 bread loaves is taken from the sizable production that left o

Question

A sample of n = 50 bread loaves is taken from the sizable production that left our bakery this morning. We find that the average weight of the 50 loaves is 1.05 pounds; the standard deviation is s = 0.06 pound.

(a) Obtain a 95% confidence interval for the mean weight of this morning's production.

(b) One of the employees claims that the current process produces loaves that are heavier than one pound, on average. Is there enough information in our sample to reject the null hypothesis that , = 1.00 lb, in favor of the alternative that , > 1.00 lb?

(c) Assume that the distribution of weights is normal; furthermore assume that the sample average and standard deviation are good estimates of the corresponding population characteristics , and u . Calculate the proportion of loaves that are underweight (i.e., weigh less than 1.0 pounds).

(d) Predict the weight of a single loaf from this morning's production. Obtain an approximate 95% prediction interval.

Explanation / Answer

a) here std error =std deviation/(n)1/2 =0.0085

for 95% CI and (n-1=49) degree of freedom ; t =2.0096

therefore 95% confidence interval =sample mean -/+ t*Std error =1.0329 ; 1.0671

b)here as our confidence interval contains all values above 1 lb; therefore we have sufficient evidence in favour of , >1.

c) P(X<1) =P(Z<(1-1.05)/0.0085) =P(Z<-5.8926)=0.0000

d)for prediction interval ; std error =std deviation(1+1/n)1/2 =0.0606

for 95% PI and (n-1=49) degree of freedom ; t =2.0096

therefore 95% prediction interval =sample mean -/+ t*Std error =0.9282 ; 1.1718

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