A certain trial was done 200 times and the X event occurred 88 times. Find a 95% confidence interval for the true probability p of the event X.
Sol:
p^=x/n=88/200= 0.44
zcrit for 95%=1.96
95% confidence interval for the true probability p of the event is
p^-Z*sqrt(p^*(1-p^)/n),p^+Z*sqrt(p^*(1-p^)/n)
0.44-1.96*sqrt(0.44*(1-0.44)/200),0.44+1.96*sqrt(0.44*(1-0.44)/200)
0.3712043, 0.5087957
95% confidence interval for p
(0.3712043, 0.5087957)
A certain trial was done 200 times and the X event occurred 88 times. Find a...
3. In a study of 200 accidents that required treatment in an emergency room, 80 occurred at work. Find the 90% confidence interval of the true proportion of accidents that occurred at work. a) Write the confidence interval formula: b) Za/2 = c) Substitute the numbers in the confidence interval formula: d) Provide the confidence interval (numerical) lower and upper limits:
In a study of 200 accidents that required treatment in an emergency room, 80 occurred at work. Find the 90% confidence interval of the true proportion of accidents that occurred at work?
R CODE PROGRAM 1. Suppose we want to simulate an experiment that can take outcomes 1; : : : ; n with probabilies p1; : : : ; pn. To be specic, suppose the R-vector p=c(.1,.2,.3,.35, .02, .03) gives the desired probabilities. Write R code that produces a number from 1 to 6 with the given probabilities, without using if statements. I recommend using the R command cumsum to do this, though there many possible approaches. 2. Suppose we are...
A random sample of 88 observations produced a mean x = 25.8 and a standard deviation s = 2.4. a. Find a 95% confidence interval for . b. Find a 90% confidence interval for . c. Find a 99% confidence interval for pl. a. The 95% confidence interval is ( D. (Use integers or decimals for any numbers in the expression. Round to two decimal places as needed.) b. The 90% confidence interval is C . (Use integers or decimals...
A statistician selects a random sample of 200 seeds from a large shipment of a certain variety of tomato seeds and tests the sample for percentage germination. If 155 of the 200 seeds germinate, then a 95% confidence interval for p, the population proportion of seeds that germinate is:
5. A certain production process produces 20 defective parts out of 200. An engineer designs a change in the process which results in 8 defective parts out of 200. Take a two-sided approach to answer the following questions. (a) 5 Do these data indicate that there is a difference in the support for decreasing pro- portion of defective parts due to changing the process? Find the P-value and report all relevant steps of the testing procedure. (b) Construct a 95%...
Assume that a procedure yields a binomial distribution with a trial repeated n = 8 times. Use either the binomial probability formula (or technology) to find the probability of k = 5 successes given the probability p = 0.31 of success on a single trial. (Report answer accurate to 4 decimal places.) P(X = k) = Submit Question Question 8 Assume that a procedure yields a binomial distribution with a trial repeated n = 15 times. Use either the binomial...
9. Assume that a procedure yields a binomial distribution with a trial repeated n times. Use (1 point) the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. Round to three decimal places. n=64, x=3, p=0.04 O 0.091 O 0.139 O 0.221 O 0.375 6. Find the indicated probability. (1 point) An archer is able to hit the bull's-eye 53% of the time. If the archer shoots 10 arrows,...
Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. n= 24, x= 21, p= 0.85 1. P(21)=?
Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. n=5, x=2, p=0.15