Week 7 Lecture

B Heard,[object Object],Lecture for the Week 7 QuizStatistics For Decision Making,[object Object],Not to be used, posted, etc. without my expressed permission. B Heard,[object Object]

Be able to use Excel to do the types of problems you have seen in Weeks 5 and 6 and particularly the Week 6 Lab.Week 7 Quiz,[object Object],Not to be used, posted, etc. without my expressed permission. B Heard,[object Object]

Week 7 Quiz,[object Object],Things to Remember……….,[object Object],Not to be used, posted, etc. without my expressed permission. B Heard,[object Object]

Standard Normal Distribution,[object Object],Week 7 Quiz,[object Object]

Remember that the total area under the curve is equal to 1 or 100% (half on each side). ,[object Object],The mean of the standard normal is 0 and the standard deviation and variance are 1. ,[object Object],For any normal distribution regardless of the mean and standard deviation the area will always be 1 or 100%. ,[object Object],We see the normal distribution all around us as noted in one of our discussion topics this week. ,[object Object],Week 7 Quiz,[object Object]

Also remember that if you ever told that you have a normal distribution and then asked what the probability x = ?, it is zero. ,[object Object],We can give the probability that it is less than a value, greater than a value, or between two values – but not that it is exactly one value because it is a “continuous distribution”. ,[object Object],(For example given that mu = 9, sigma = 2.1, what is the probability that x = 7? It’s 0 because with a continuous distribution we are “slicing jello” as I like to say.,[object Object],Week 7 Quiz,[object Object]

Week 7 Quiz,[object Object],Some Questions……….,[object Object]

Central Limit Theorem,[object Object], On the Central Limit Theorem, remember that it states that given a distribution with a mean μ and variance σ², the sampling distribution of the mean approaches a normal distribution with a mean (μ) and a variance σ²/N as N, the sample size, increases. The amazing and counter-intuitive thing about the central limit theorem is that no matter what the shape of the original distribution, the sampling distribution of the mean approaches a normal distribution. Also another important fact is that any sample size is big enough when we know the population is normal. ,[object Object],Week 7 Quiz,[object Object]

Using the Central Limit Theorem answer the following question.,[object Object],Assuming you have a normal distribution for your population and you take 64 samples of size 25 each. Calculate the standard deviation of the sample means if the population’s variance is 16.,[object Object],Since the population is normally distributed with a variance of 16, then the sample means have a variance equal to 16/25 according to the Central Limit Theorem. Hence their standard deviation will be SQRT(16/25) = 4/5 = .800,[object Object],Week 7 Quiz,[object Object]

Be able to use the normal distribution to solve problems. Examples,[object Object], ,[object Object],Bob scored a 190 on his entrance exam, where the average was 165 and the standard deviation was 12. Where does he stand in relation to the rest of his class?,[object Object],He scored in the “top 2 %”, see Excel that follows.,[object Object],Week 7 Quiz,[object Object]

In a normal distribution with mu = 35 and sigma = 6 what is the z score for a value of 41?,[object Object],Z= +1, (41-35)/6 = 6/6 = 1,[object Object],Week 7 Quiz,[object Object]

In a normal distribution with mu = 35 and sigma = 6 what number corresponds to z = -2?,[object Object],23, -2 = (x-35)/6, solve algebraically by multiplying each side by 6 to get -12 = x-35, then adding 35 to each side to get x = 23,[object Object],Week 7 Quiz,[object Object]

We have an area of .4840. What z-score corresponds to this area?,[object Object], ,[object Object],Using the Standard Normal Table in your book, or the one on the attached excel you can see it is a z score of -0.04,[object Object],Week 7 Quiz,[object Object]

Find P(80 < x < 86) when mu = 82 and sigma = 4. Write your steps in probability notation.,[object Object], ,[object Object],I did this in Excel, but I still need to show my work:,[object Object], ,[object Object],The z-score corresponding to x = 86 is z = (86-82)/4 = 4/4 = 1.0. The area corresponding to z = 1.0 is .8413 The z-score corresponding to x = 80 is (80 - 82)/4 = -2/4 = -0.5. The area corresponding to z = -0.5 is .3086. Thus, P(80 < x < 86) = P(-0.5 < z < 1.0) = P(z < 1.0) - P(z < -0.5) = 0.8413 – 0.3086 = 0.5327 (my excel calculated it to be 0.5328 so I feel good about it),[object Object],Week 7 Quiz,[object Object]

Interpret a 90% confidence interval of (63.3, 83.4).,[object Object],You should note here there is a 90% probability that the interval (63.3 to 83.4) contains µ, the true population mean.,[object Object],Week 7 Quiz,[object Object]

What is the critical value corresponds to a confidence level of 96%,[object Object],100 – 96 = 4 ,[object Object],Divide 4 by 2 (tails) and get 2,[object Object],Add 2 to the original 96% and get 98% and find the critical value (z-score that corresponds to .9800 which is 2.05 (closest to it),[object Object],Week 7 Quiz,[object Object]

Compute the population mean margin of error for a 90% confidence interval when sigma is 7 and the sample size is 81.,[object Object], ,[object Object],E = z * sigma / sqrt(n) = 1.645 * 7 / sqrt(81) = 1.279 (remember +/-),[object Object],Week 7 Quiz,[object Object]

A Military entrance exam has a mean of 120 and a standard deviation of 9. We want to be 95% certain that we are within 6 points of the true mean. Determine the sample size.,[object Object], ,[object Object],n = ( z * sigma / error ) ^ 2 = (1.96*9/6)^2 = 2.94^2 = 8.6436. Round up to 9. ALWAYS ROUND SAMPLE SIZES UP!!!,[object Object],Week 7 Quiz,[object Object]

A researcher wants to get an estimate of the true mean performance measure of its product. It randomly samples 180 of its machines. The mean performance measure was 900 with a standard deviation of 60. Find a 95% confidence interval for the true mean performance measure of the machines.,[object Object],Week 7 Quiz,[object Object]

The population standard deviation is unknown and the sample size is 180. Thus, since the sample size is greater than 30, this confidence interval will use a z-value. For a 95% confidence interval, the z-value = 1.96. Sample mean = 900 and sample standard deviation = 60. ,[object Object],Population mean = 900 +/- 1.96 * 60/sqrt(180) = 900 +/- 8.765. 891.235 and 908.765,[object Object],Week 7 Quiz,[object Object]

A researcher wants to get an estimate of the true mean performance measure of its product. The researcher needs to be within 15 of the true mean. The researcher estimates the true population standard deviation is around 30. If the confidence level is 95%, find the required sample size in order to meet the desired accuracy.,[object Object],Week 7 Quiz,[object Object]

For a 95% confidence level, the z-value = 1.96. The formula for sample size is n = ( z-value * standard deviation / error ) ^ 2 = ( 1.96 * 30/ 15) ^ 2 = ( 3.92 ) ^ 2 = 15.3664. Thus, the researcher must sample at least 16 to obtain the desired accuracy.,[object Object],ALWAYS ROUND SAMPLE SIZES UP!!!!!,[object Object],Week 7 Quiz,[object Object]

A researcher wants to estimate the mean cost to develop a product. The researcher tests 18 cases and finds the mean cost to be $3000 with a standard deviation of $400. Find a 95% confidence interval for the true mean cost to develop this product.,[object Object],Week 7 Quiz,[object Object]

The population standard deviation is unknown and the sample size is 18. Thus, since the population standard deviation is unknown AND the sample is less than 30, we must use the t-value for this confidence interval. For a 95% confidence interval and degrees of freedom = 17 (from 18-1), the t-value = 2.110. Sample mean = 3000 and sample standard deviation = 400. ,[object Object],Population mean = 3000 +/- 2.110 * 400/sqrt(18) = 3000 +/- 198.93. So our bounds are 2801.07 and 3198.93,[object Object],Week 7 Quiz,[object Object]

A researcher wants to estimate what proportion of failures that are due to poor workmanship. The researcher randomly samples 50 failures and finds 18 are due to poor workmanship. Using a 95% confidence interval, estimate the true proportion of poor workmanship for all failures. ,[object Object],Week 7 Quiz,[object Object]

For a 95% confidence level, the z-value is 1.96. The sample proportion = 18/50 = 0.36, thus p hat = 0.36 and 1-p hat = 0.64 . The sample size = 50. The population proportion is between 0.36 +/- 1.96 * sqrt ( 0.36 * 0.64 / 50 ) = 0.36 +/- .13 So the bounds are .23 and .49,[object Object],Week 7 Quiz,[object Object]

Week 7 Lecture

Week 7 Lecture

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Week 7 Lecture