Confidence Intervals, Proportions, Mean, Standard Deviation and Variance
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1. Three hundred of 450 drivers surveyed indicated that they would consider bying retread tires. Find a 95% confidence interval for the true proportion of all who might buy retread tires.
2. Insurance industry analysts report that the average amount of life insurance in effect in Virginia is $92,500 with a sample standard deviation of $8,400. A newspaper conducted a random survey of 64 Virginia families. What is the probability that the average amount of insurance in effect for these families is more than $95,000?
3. A national health foods manufacturer wishes to determine the average amount people are willing to spend annually on health foods. If previous data indicate that o=variance=7.50, how large should the sample be to ensure with 95% confidence that estimate is within $1.00 of the true value?
4. A traffic survey revealed that the speeds of vehicles along a dangerous stretch of interstate follow a normal distribution with mean 50 mph and standard deviation 8 mph. What speed limit should be posted so that only 8% of all drivers monitored are guilty of speeding?
5. Bank officials claim that the average amount of money in savings accounts at all branches of the Commerce Bank is $7,500, with a standard deviation of $650. Many random samples of size 100 are taken.
a. the mean of these samples
b. the standard error of the mean
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Probability and Stats
1. Three hundred of 450 drivers surveyed indicated that they would consider bying retread tires. Find a 95% confidence interval for the true proportion of all who might buy retread tires.
Solution:
The confidence interval for proportions is given by,
p = 300/450 = 0.67 q = 1-p = 1 - 0.67 = 0.33
Z = 1.96 n = 450
By plug in above values in the formula we get,
So the confidence interval for P lies between 0.634 and 0.706.
2.. Insurance industry analysts report that the ...
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