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A consumer watchdog organization estimates the mean weight of 1-ounce "Fun-Size"
A consumer watchdog organization estimates the mean weight of 1-ounce "Fun-Size"candy bars to see if customers are getting full value for their money. A random sample
of 25 bars is selected and weighted, and the organization reports that a 95% confidence
interval for the true mean weight of the candy bars is 0.982 to 0.988 ounces.
a) What is the point estimate (=sample mean) from this sample?
b) What is the margin of error?
(Hint: find the distance between the sample mean and the upper limit).
c) Interpret the confidence level of 90% in the context of the problem?
Pointestimatefrom the sample is 0.985, margin error is 0.003.What is Confidence Interval?Confidence interval is defined as theintervalwhich is the estimate for the parameter of the sample or population to be contained.(a) To calculate point estimate orsamplemean :Point estimate is the mid point of the confidence interval.Given that truemeanweight of candy bars is 0.982 ounces to 0.988 ounces.Point estimate = (0.982 + 0.988) / 2 = 1.97 / 2 = 0.985(b)Marginerror is the one half of the total width of the interval.Marginerror= (0.988 - 0.982) / 2 = 0.003(c) The confidence level of 90% in this problem can be interpreted as , if we do the interval construction for many times, about 90% of the total constructed intervals has the truepopulationmean of weight of fun size candy bars.Hence the point estimate and margin error are 0.985 and 0.003 respectively.Learn more aboutConfidence Intervalhere :brainly.com/question/18914334#SPJ1...