Statistics math question help please?
1. A confidence interval for a parameter gives an interval within which the parameter lies with the level of confidence used. If you find that a 95% confidence interval for the population mean (the parameter) is (a,b) it means that if you find , say, 100 independent 95% confidence intervals, then about 95 of the intervals will contain the population mean. 2. A point estimate gives only one value, a confidence interval gives a range of values. 3.Assuming that the standard deviation ,s,is known, and the ages have a normal distribution then X-1.96s/sqrt(n)=33.0, where X is the sample mean an n is the sample size. X+1.96s/sqrt(n)=39.8 Subtracting gives 3.92s/sqrt(n)=6.8, so s/sqrt(n)=6.8/3,92=1.735; adding gives X=36,4 For a 95% CI z=1.645, X-1.645s/sqrt(n)=36.4-1.645(1.735)=??? X+1.645s/sqrt(n)=36.4+1.645(1.735)=??? Which gives the 90% CI. The bigger the confidence level, the wider the confidence interval 4. You use z when the standard deviation is known or the sample size is large (>30) Use a t-interval when the standard deviation is unknown or sample size is small.
Statistics Interval Estimation: Develop a 95% confidence interval estimate of the proportion of adults,?
aged 22-55 who are risk neutral. You can be 95% confident that the interval estimate ___?___ to __?___ includes the population proportion p, the population of adults aged 22-55 considered risk neutral. Here are some sample results: you receive $10; for tails, you get nothing. **The game is a 10-trial binomial experiment. A toss that lands on heads is defined as a success, and because the probability of a success if 0.5, the expected number of successes is 10(0.5)=5. Since each success pays $10, the expected value of the game is 5($10)=$50. Each person in a random sample of 1,536 adults between the ages of 22 and 55 was invited to play this game. Data were collected on the maximum amount each person was willing to pay. Define someone as "risk neutral" if the maximum amount they are willing to pay to play the game is equal to $50, the game's expected value. Using this definition, 34 people in the 1,536-person sample are risk neutral. Use the sample results to estimate the proportion p of the adult population aged 22 to 55 who are risk neutral. Let p-bar denote the proportion of adults in the sample who are risk neutral.
Statistics estimation problem?
Hello, For 90%. We have a two tailed stats here. So we want (100%-90%)/2 = 0.05 on each side of our tail. So look up the z-score for 0.05 and we have 1.645. Now our formula is [ mean - (z-score of 0.05 * sd)/sqr(n) , mean + (z-score of 0.05 * sd)/sqr(n) ] Hence [ 1570 - (1.645*250)/Sqr(370), 1570 + (1.645*250)/Sqr(370)] [1570-21.38,1570+21.38] [1548.62,1591.38] Now for a 99% confidence level we use (100%-99%)/2 = 0.005 Look up z of 0.005 and we have 2.57. Now put this into you interval and it should be correct! [ 1570 - (2.57*250)/Sqr(370), 1570 + (2.57*250)/Sqr(370)] Hope This Helps!
Statistics help. Estimate the value of the population proportion.?
If you have only the one sample to go by, its proportion would be your best estimate of the population proportion, so .65. You could do a confidence interval to say the likelihood the proportion is within a certain range, if you want. the standard error would be sqrt (pq/n) or sqrt [(.65 X.35)/100]. I get .0477. If you want to be 95 percent sure the population is within your interval, you could use a z score of + or -1.96. Take that times your standard error; adding the negative one to the .65 mean will give the bottom of the interval, adding the positve one to the mean will give the top of the interval.
What are some every day uses for confidence intervals in statistics?
For most things in daily life, you'll never need a formal CI. However, we humans use informal CIs all the time. A formal CI is a mathematical function; an informal CI is a pretty good guess of a range of values. For instance, you may not know how much money is in your pocket down to the penny, but you do know if it's between $100 and $1. That's an informal one, and you'd get pushback from many statisticians about whether it's a CI.But let's say that you want to maintain a certain range of cash, so at the end of every day you count all the cash you're carrying, and after several months you calculate the mean and standard deviation. Then you calculate out a CI at whatever confidence level you like. That's a pretty good indicator for how much money you're likely to carry on average from here forward, all other things being equal. The months of data collection can be seen as a sample of all the months to come, and even arguably months from the past too. The same process can be undertaken with any kind of data in your life: tire pressure, spending on video games (and video scores), rainfall, time to drive to work or school, prices of grocery items, volume of music heard from next door (in db's), electric bills, prices of gasoline, time spent studying, amounts spent for lunches, times running the 100-meters, time spent with a BFF, and on and on.
Statistics...?
1.For the interval estimation of m when s is known and the sample is large, the proper distribution to use is a. the normal distribution b. the t distribution with n degrees of freedom c. the t distribution with n + 1 degrees of freedom d. the t distribution with n + 2 degrees of freedom 2.A two-tailed test is performed at 95% confidence. The p-value is determined to be 0.09. The null hypothesis a. must be rejected b. should not be rejected c. could be rejected, depending on the sample size d. has been designed incorrectly
Why will an interval estimate most likely fall around the population mean?
Interval estimation involves taking many samples from the population and finding the means of each of these samples. Then you can compute a confidence interval that is centered around the mean of the sets of samples. The interval estimate will most likely fall around the population mean because when you take a lot of samples they will fall on either side of the normally distributed population mean almost equally. The more samples and the more people in each sample, the better your estimate of the population mean will be. It's a bit confusing but I hope that helped you a bit