Question

In: Statistics and Probability

John measured the total playing time of 35 randomly chosen CDs from his very large collection...

John measured the total playing time of 35 randomly chosen CDs from his very large collection and found a mean of 79.7 minutes and a standard deviation of 9.8 minutes. Give the 93% confidence interval for the population mean, assuming that the population is approximately normally distributed. Round the endpoints to two decimal places.

Solutions

Expert Solution

Solution :

Given that,

= 79.7

= 9.8

n = 35

At 93% confidence level the z is ,

  = 1 - 93% = 1 - 0.93 = 0.07

/ 2 = 0.07 / 2 = 0.035

Z/2 = Z0.035 = 1.81

Margin of error = E = Z/2* ( /n)

= 1.81 * (9.8 / 35)

= 3.00

At 93% confidence interval estimate of the population mean is,

- E < < + E

79.7 - 3.00 < < 79.7 + 3.00

76.70 < < 82.70

(76.70 , 82.70)

orchestra answered 3 years ago
Next > < Previous

Related Solutions

The table below shows the lengths of some randomly chosen CDs in John’s very large collection....
The table below shows the lengths of some randomly chosen CDs in John’s very large collection. Give the 92% confidence interval for the population mean, assuming that the population is approximately normally distributed. Round the endpoints to two decimal places. 69.33 74.64 65.28 61.01 38.52 69.07 40.01 42.14 41.52 56.76 58.14 70.80 39.21 71.36 46.94 52.90 56.16 41.01 56.57 60.97 64.62 69.12 66.58 58.26 62.79 Confidence interval: ( , ) John measured the total playing time of 30 randomly chosen...
The table below shows the lengths of some randomly chosen CDs in John’s very large collection....
The table below shows the lengths of some randomly chosen CDs in John’s very large collection. Give the 92% confidence interval for the population mean, assuming that the population is approximately normally distributed. Round the endpoints to two decimal places. 69.33 74.64 65.28 61.01 38.52 69.07 40.01 42.14 41.52 56.76 58.14 70.80 39.21 71.36 46.94 52.90 56.16 41.01 56.57 60.97 64.62 69.12 66.58 58.26 62.79 Confidence interval:
The average playing time of compact discs in a large collection is 35 minutes, and the...
The average playing time of compact discs in a large collection is 35 minutes, and the standard deviation is 3 minutes. (a) What value is 1 standard deviation above the mean? 1 standard deviation below the mean? What values are 2 standard deviations away from the mean? 1 standard deviation above the mean 1 standard deviation below the mean 2 standard deviations above the mean 2 standard deviations below the mean     b) Without assuming anything about the distribution of times,...
The average playing time of compact discs in a large collection is 37 min, and the...
The average playing time of compact discs in a large collection is 37 min, and the standard deviation is 2 min. (a) What value is 1 standard deviation above the mean? 1 standard deviation below the mean? What values are 2 standard deviations away from the mean? 1 standard deviation above the mean 1 standard deviation below the mean 2 standard deviation above the mean 2 standard deviation below the mean (b) Without assuming anything about the distribution of times,...
The average playing time of compact discs in a large collection is 37 minutes, and the...
The average playing time of compact discs in a large collection is 37 minutes, and the standard deviation is 7 minutes. (a) What value is 1 standard deviation above the mean? 1 standard deviation below the mean? What values are 2 standard deviations away from the mean? 1 standard deviation above the mean 44 1 standard deviation below the mean 30 2 standard deviations above the mean 51 2 standard deviations below the mean     23 b) Without assuming anything about...
The average playing time of compact discs in a large collection is 37 min, and the...
The average playing time of compact discs in a large collection is 37 min, and the standard deviation is 5 min. (a) What value is 1 standard deviation above the mean? 1 standard deviation below the mean? What values are 2 standard deviations away from the mean? 1 standard deviation above the mean ___ 1 standard deviation below the mean ____ 2 standard deviation above the mean ____ 2 standard deviation below the mean _____ (b) Without assuming anything about...
The mean playing time for a large collection of compact discs is 37 minutes, and the...
The mean playing time for a large collection of compact discs is 37 minutes, and the standard deviation is 4 minutes. (a) What value (in minutes) is 1 standard deviation above the mean? One standard deviation below the mean? What values are 2 standard deviations away from the mean? 1 standard deviation above the mean   1 standard deviation below the mean   2 standard deviations above the mean   2 standard deviations below the mean (b) Assuming that the distribution of times...
The quality control department of John Deere measured the length of 100 bolts randomly selected from...
The quality control department of John Deere measured the length of 100 bolts randomly selected from a specified order. The mean length was found to be 9.75 cm, and the standard deviation was 0.01 cm. if the bolt lengths are normally distributed, find: a) The percentage of bolts shorter than 9.74 cm b) The percentage of bolts longer than 9.78 cm c) The percentage of bolts that meet the length specification of 9.75 +/- 0.02 cm d) The percentage of...
A. Recall that Benford's Law claims that numbers chosen from very large data files tend to...
A. Recall that Benford's Law claims that numbers chosen from very large data files tend to have "1" as the first nonzero digit disproportionately often. In fact, research has shown that if you randomly draw a number from a very large data file, the probability of getting a number with "1" as the leading digit is about 0.301. Now suppose you are an auditor for a very large corporation. The revenue report involves millions of numbers in a large computer...
A farmer weighs 10 randomly chosen watermelons from his farm and he obtains the following values...
A farmer weighs 10 randomly chosen watermelons from his farm and he obtains the following values (in lbs): 7.72,9.58,12.38,7.77,11.27,8.80,11.10,7.80,10.17,6.00 Assuming that the weight is normally distributed with mean μ and and variance σ2, find a 95% confidence interval for μ.
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT