Please i need help solving these problems. I’ll rate the answer. 1. An irate stu
ID: 3062009 • Letter: P
Question
Please i need help solving these problems. I’ll rate the answer.1. An irate student complained that the cost of textbooks at PPU was too high. He was so angry, he randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was 396.21 dollars. If the standard deviation of the population was 18.33 dollars, find the 90 percent confidence interval of the true mean (population mean). 2. For a certain urban area, it was found that in a sample of 4 months, an average of 4 burglaries occurred each month. The standard deviation was 4. Find the 99 percent confidence interval of the true mean number of burglaries each month. 3. In a study of 150 accidents in the emergency room, thirty-six percent involved children 6 and under. Find the 90 percent confidence interval of the true proportion of accidents involving children 6 and under. 4. Randomly select 30 stocks classified as the Dow Jones industrials as a sample. Note the amount each stock has gained or lost in the last year. Compute the mean and standard deviation of the data set. Compute the 95 percent confidence interval for the mean and standard deviation. Compute the percentage of stocks that had a gain since last year. Find a 95 percent confidence interval for the percentage of stocks with a gain.
Explanation / Answer
1)
n = 36 , mean = 396.21 , s = 18.33
z value at 90% = 1.645
CI = mean + /- z * ( s/sqrt(n))
= 396.21 +/- 1.645 * * 18.33/sqrt(36))
= (391.1845 , 401.2355)
Lower Bound = 391.1845
Upper bound = 401.2355
2)
n = 4 , s = 4 , mean = 4
t value at 99% = 5.8408
CI = mean + /- t * ( s/sqrt(n))
= 4 + /- * 5.8408*( 4/sqrt(4))
= (-7.6816 , 15.6816)
Lower Bound = -7.6816
Upper bound = 15.6816
3)
n =150 , p = 6/150 = 0.04
z value at 90% = 1.645
CI = p + /- z * sqrt(p * ( 1 - p) / n)
= 0.04 =/- 1.645 * sqrt ( 0.04 * 0.96 / 150)
= (0.0137 , 0.0663)
Lower Bound = 0.0137
Upper bound = 0.0663