Exercise 2: A candy company sells 20 ounce bags of candy. A researcher is interested in estimating the mean number of pieces of candy per bag. A random sample of 120 such bags obtained which results in a sample mean of 18.72 pieces of candy. Assuming the population standard deviation is .88, construct a 99% confidence interval to estimate the population mean number of pieces of candy in each 20 ounce bag.Since n 30, we use the Z-distributionFrom the text of the problem we have: = 18.72, = .88, and n = 120.From the table on slide 7: = = 2.5758± = 18.72 ± 2.5758 = 18.72 ± .2069P(18.5131 18.9269) = 99%
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Module 11: Confidence Intervals Part 1
Solutions to Module 11 Exercises
Exercise 3: An owner of a Christmas tree farm wants to estimate the average height of the trees at his farm. he randomly samples 40 of them and obtains a sample mean height of 5.8 feet. Historically the standard deviation of Christmas tree heights at his farm is .7 feet. Construct a 95% confidence for the mean tree height at tree farm. Assume their are a total of 1200 trees at this particular farm.
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Module 11: Confidence Intervals Part 1
Solutions to Module 11 Exercises
Exercise 4: Suppose the following data are selected randomly from a normally distributed population. Construct a 98% confidence interval for the population mean. From Excel:Since n < 30 we use the t-distribution= 45.6154, s = 5.6941, n = 13, n - 1 = 12, and α/2 = .01 (98% Confidence)t/2, n-1 = t.01, 12 = 2.6810± t/2, n-1= 45.6154 ± 2.6810 = 45.6154 ± 4.2340P(41.3814 49.8494) = 98%4051434844575439424845394345.6154=AVERAGE(A1:A13)5.6941=STDEV(A1:A13)tα/2, n-1t.01, 122.6810=TINV(2*0.01,12)
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Module 11: Confidence Intervals Part 1
This concludes Module 11.
The next step is to do the homework for Module 11 (Module 11 HW).
Module 11 HW is due by 11:59 PM on the date indicated in the course calendar.

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- Spring '14
- DebraACasto
- Statistics, Normal Distribution, Standard Deviation