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http://books.google.com.ph/books?id=Aw6ZA-LJ6jQC&pg=SA3PA77&dq=free+advanced+statistics+book&hl=tl&ei=SvhbTYb6M4eqvQPrq_3QDA& sa=X&oi=book_result&ct=result&resnum=1&ved=0CCcQ6AEwAA#v=onepage&q& f=false http://books.google.com.ph/books? id=3LhPwUhrVIcC&pg=PA205&lpg=PA205&dq=hypothesis+testing+problems+ +in+advance+statistics&source=bl&ots=Zj94UJmobs&sig=pChNykwdhE5PZ9stSMT 9ioc0oHQ&hl=tl&ei=IQJcTZf1G4GevgP8lLjZCw&sa=X&oi=book_result&ct=result&re snum=5&ved=0CEQQ6AEwBDgK#v=onepage&q&f=false http://books.google.com.ph/books? id=Sf_rPhW7rEQC&printsec=frontcover&dq=free+advanced+statistics&hl=tl&ei=D _lbTdigGI3UvQPkn4TUDA&sa=X&oi=book_result&ct=result&resnum=2&ved=0CCw Q6AEwAQ#v=onepage&q=free%20advanced%20statistics&f=false http://www.mathwizz.com/statistics/choosehyp.htm http://sites.google.com/site/fundamentalstatistics/unit-6-anova-and-advancedhypothesis-testing http://faculty.ksu.edu.sa/69424/EXAM%20324%20frist%2014281429/ex2.pdf http://cnx.org/content/m11271/latest/
.) a.)A large university provides housing for 10% of their students a survey done shows that 63 out of the 481 who responded will be looking for housing the upcoming year, does this show evidence that they should buy more housing? If needed use alpha of .05 b.) Of the above survey, 19 students did not respond, if they did, would it have changed the outcome? Explain. 2.) a.)A company claims to have a average of 2% defective products, a test is taken and determines that 3 out of 100 parts are defective. Determine if this give evidence that the 2% claim is not true. b.) How large of a sample must be taken in order to get a 90% confidence interval within .1% Answers: Let X be the number of success in n independent and identically distributed Bernoulli trials, i.e., X ~ Binomial(n, p) To test the null hypothesis of the form H0: p = p0, or H0: p ≥ p0, or H0: p ≤ p0 Assuming that n*p0 > 10 and n * (1-p0) > 10 (some will say the necessary condition here is > 5, I prefer this more conservative assumption so that the approximations in the tail of the distribution are more accurate) then
find the test statistic z = (pHat - p0) / sqrt(p0 * (1-p0) / n) where pHat = X / n The p-value of the test is the area under the normal curve that is in agreement with thealternate hypothesis. H1: p ≠ p0; p-value is the area in the tails greater than |z| H1: p < p0; p-value is the area to the left of z H1: p > p0; p-value is the area to the right of z If the p-value is less than or equal to the significance level α, i.e., p-value ≤ α, then we reject the null hypothesis and conclude the alternate hypothesis is true. If the p-value is greater than the significance level, i.e., p-value > α, then we fail to reject the null hypothesis and conclude that the null is plausible. Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible. The hypothesis test in this question is: H0: p ≤ 0.1 vs. H1: p > 0.1 The test statistic is: z = ( 0.1309771 - 0.1 ) / ( √ ( 0.1 * (1 - 0.1 ) / 481 ) z = 2.264605 The p-value = P( Z > z ) = P( Z > 2.264605 ) = 0.01176846 Since the p-value is less than the significance level of 0.05 we reject the null hypothesis and conclude the alternate hypothesis p > 0.1 is true.
if the 19 who did not respond did and all answered that they did not need housing you'd have:
H0: p ≤ 0.1 vs. H1: p > 0.1 The test statistic is: z = ( 0.126 - 0.1 ) / ( √ ( 0.1 * (1 - 0.1 ) / 500 ) z = 1.937926 The p-value = P( Z > z ) = P( Z > 1.937926 ) = 0.02631615 Since the p-value is less than the significance level of 0.05 we reject the null hypothesis and conclude the alternate hypothesis p > 0.1 is true. So no change in the conclusion.
2) Hypothesis Test for proportions: Let X be the number of success in n independent and identically distributed Bernoulli trials, i.e., X ~ Binomial(n, p) To test the null hypothesis of the form H0: p = p0, or H0: p ≥ p0, or H0: p ≤ p0 Assuming that n*p0 > 10 and n * (1-p0) > 10 (some will say the necessary condition here is > 5, I prefer this more conservative assumption so that the approximations in the tail of the distribution are more accurate) then find the test statistic z = (pHat - p0) / sqrt(p0 * (1-p0) / n) where pHat = X / n The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis. H1: p ≠ p0; p-value is the area in the tails greater than |z| H1: p < p0; p-value is the area to the left of z H1: p > p0; p-value is the area to the right of z If the p-value is less than or equal to the significance level α, i.e., p-value ≤ α, then we reject the null hypothesis and conclude the alternate hypothesis is true. If the p-value is greater than the significance level, i.e., p-value > α, then we fail to reject the null hypothesis and conclude that the null is plausible. Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible. The hypothesis test in this question is: H0: p0 = 0.02 vs. H1: p0 ≠ 0.02 The test statistic is: z = ( 0.03 - 0.02 ) / ( √ ( 0.02 * (1 - 0.02 ) / 100 ) z = 0.7142857 The p-value = P( Z > |z| ) = P( Z < -0.7142857 ) + P( Z > 0.7142857 ) = 2 * P( Z < -0.7142857 ) = 0.4750505 Since the p-value is greater than the significance level of 0.01 we fail to reject the null hypothesis and conclude p = 0.02 is plausible.
large sample confidence intervals are used to find a region in which we are 100 (1-α)% confident the true value of the parameter is in the interval. For large sample confidence intervals for the proportion in this situation you have: pHat ± z * sqrt( (pHat * (1-pHat)) / n) where pHat is the sample proportion z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α n is the sample size here we are only concerned with the error term and the width w. z * sqrt( (pHat * (1-pHat)) / n) = w n = pHat * (1 - pHat) (w/z) ^ -2 we have, w = 0.001 pHat = 0.02 from the above information z = 1.645 for a 90% CI n = 0.02 * (0.98) * (0.001 / 1.645) ^ -2 n = 53038.09 n must be integer valued. always take the ceiling to make sure you have the correct size for the interval n = 53039.
[2003] Joe Smith was running for mayor. His campaign manager estimated that he needed at least 40% of the popular vote in order to win. A poll of 400 people was taken and 150 indicated they would vote for Smith. a) If we use a 10% level of significance, does he have sufficient support? b) Construct a 90% confidence interval on the percentage who will vote for Smith. Solution part a): The campaign manager is concerned if support falls below 40%. So, we set up our hypotheses as: H0: p = 0.4 H1: p < 0.4
Because α = 0.1, we reject the null hypothesis if Z < -1.282, where:
and q = 1-p. In our case, we have:
Plugging these values into the formula for Z, we get Z = -1.0206. Since -1.0206 > -1.282 we do not reject the null hypothesis and conclude that Smith has enough support to win the election. Solution part b): The formula for our confidence interval for p is:
In our case, we have:
Plugging these numbers into the formula, we have: 0.375 - 0.0398 < p < 0.375 + 0.0398 0.3352 < p < 0.4148 So the support for Smith lies between 33.52% and 41.48% based on the evidence. Back to top [2005] A certain intersection had an accident rate of 1 accident per 1000 vehicles. A study suggested that partial obstruction of the stop sign by a tree branch could be partly to blame. The branch was trimmed and the traffic was monitored for 10,000 vehicles. During this period, there were 8 accidents. Did trimming the tree branch help? Use a p-value to reach your conclusion. Solution: If trimming the tree branch helped, we would expect fewer accidents. So we set up our hypotheses as:
H0: p = 0.001 H1: p < 0.001 Because of the sample size, we can use Z because of the Central Limit Theorem. We will reject the null hypothesis for low p-values where p-value = P(Z < test statistic). Our formula for Z is
where q = 1 - p. We have:
Plugging these numbers into the equation we get Z = -0.6328. Our p-value is P(Z < -0.6328) = 0.2634 Because of the high p-value, we do not reject the null hypothesis and conclude that trimming the tree branch made no significant difference. Back to top [2001] A wood door manufacturer wants to minimize the number of knots in each door. On average, he wants no more than 1.2 knots per door. A random sample of 500 doors had, on average, 1.4 knots with a standard deviation of 0.26 knots. Are the manufacturer's standards being met? Test at α = 0.05. Solution: The manufacturer is concerned if there are too many knots in the door. Ergo, we set up our hypothesis as: H0: µ = 1.2 H1: µ > 1.2
Because of the sample size of 500, we can use Z because of the Central Limit Theorem. With α = 0.05, we reject the null hypothesis if Z > 1.645. Our test statistic is:
where
Plugging these numbers into the formula, we get Z = 17.2. Since 17.2 > 1.645, we reject H0 and conclude that the standards are not being met. Back to top [2004] In order to maintain quality control, the angle measurement of a gear tooth must be maintained at 1.02o. A sample of 60 gears is taken. The average angle is 1.025o with a standard deviation of 0.05o. a) Construct a 95% confidence interval of the average gear angle. b) If you were to test the hypothesis H0: µ = 1.02 against H1: µ ≠ 1.02 at α = 0.05, what would be your conclusion using only the confidence interval? Solution part a): Our formula for the confidence interval is:
For our 95% confidence interval we need Z0.025 = 1.96 because 1 - α = 0.95, and so α /2 = 0.025. We have:
Plugging these values into the equation we get 1.025 - 0.0127 < µ < 1.025 + 0.0127 or 1.0123 < µ < 1.0377 Solution part b):Because 1.02 is contained in the 95% confidence interval, we would conclude at a 5% level of significance that the
standards are being met (in other words, we would not reject the null hypothesis). Back to top [2002] Joe's Ceramics wants to keep the standard deviation of the oven temperature to within 5o. He samples 100 baked ceramic pieces and measures (somehow) the oven temperature when he takes the piece out. If he wants the bound on the error to be at most 1o, what should be the minimum level of confidence? Solution: This is a two step problem: Step 1: Solve for Z. Step 2: Solve for 1 - α which is the level of confidence we are looking for. Our formula is:
Solving for Z, we have: E = 1; σ = 5; n = 100 Plugging these numbers into the formula, we get Zα /2 ≥ 2. Since P(Z > 2) = 0.0227, that means α /2 = 0.0227. Solving for our level of confidence, we get 1 - α = 1 - (2)(0.0227) = 0.9546. So, our minimum level of confidence is 95.46%
.) a.)A large university provides housing for 10% of their students a survey done shows that 63 out of the 481 who responded will be looking for housing the upcoming year, does this show evidence that they should buy more housing? If needed use alpha of .05 b.) Of the above survey, 19 students did not respond, if they did, would it have changed the outcome? Explain. 2.) a.)A company claims to have a average of 2% defective products, a test is taken and determines that 3 out of 100 parts are defective. Determine if this give evidence that the 2% claim is not true. b.) How large of a sample must be taken in order to get a 90% confidence interval within .1% Answers: Let X be the number of success in n independent and identically distributed Bernoulli trials, i.e., X ~ Binomial(n, p) To test the null hypothesis of the form H0: p = p0, or H0: p ≥ p0, or H0: p ≤ p0 Assuming that n*p0 > 10 and n * (1-p0) > 10 (some will say the necessary condition here is > 5, I prefer this more conservative assumption so that the approximations in the tail of the distribution are more accurate) then
find the test statistic z = (pHat - p0) / sqrt(p0 * (1-p0) / n) where pHat = X / n The p-value of the test is the area under the normal curve that is in agreement with thealternate hypothesis. H1: p ≠ p0; p-value is the area in the tails greater than |z| H1: p < p0; p-value is the area to the left of z H1: p > p0; p-value is the area to the right of z If the p-value is less than or equal to the significance level α, i.e., p-value ≤ α, then we reject the null hypothesis and conclude the alternate hypothesis is true. If the p-value is greater than the significance level, i.e., p-value > α, then we fail to reject the null hypothesis and conclude that the null is plausible. Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible. The hypothesis test in this question is: H0: p ≤ 0.1 vs. H1: p > 0.1 The test statistic is: z = ( 0.1309771 - 0.1 ) / ( √ ( 0.1 * (1 - 0.1 ) / 481 ) z = 2.264605 The p-value = P( Z > z ) = P( Z > 2.264605 ) = 0.01176846 Since the p-value is less than the significance level of 0.05 we reject the null hypothesis and conclude the alternate hypothesis p > 0.1 is true.
if the 19 who did not respond did and all answered that they did not need housing you'd have:
H0: p ≤ 0.1 vs. H1: p > 0.1 The test statistic is: z = ( 0.126 - 0.1 ) / ( √ ( 0.1 * (1 - 0.1 ) / 500 ) z = 1.937926 The p-value = P( Z > z ) = P( Z > 1.937926 ) = 0.02631615 Since the p-value is less than the significance level of 0.05 we reject the null hypothesis and conclude the alternate hypothesis p > 0.1 is true. So no change in the conclusion.
2) Hypothesis Test for proportions: Let X be the number of success in n independent and identically distributed Bernoulli trials, i.e., X ~ Binomial(n, p) To test the null hypothesis of the form H0: p = p0, or H0: p ≥ p0, or H0: p ≤ p0 Assuming that n*p0 > 10 and n * (1-p0) > 10 (some will say the necessary condition here is > 5, I prefer this more conservative assumption so that the approximations in the tail of the distribution are more accurate) then find the test statistic z = (pHat - p0) / sqrt(p0 * (1-p0) / n) where pHat = X / n The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis. H1: p ≠ p0; p-value is the area in the tails greater than |z| H1: p < p0; p-value is the area to the left of z H1: p > p0; p-value is the area to the right of z If the p-value is less than or equal to the significance level α, i.e., p-value ≤ α, then we reject the null hypothesis and conclude the alternate hypothesis is true. If the p-value is greater than the significance level, i.e., p-value > α, then we fail to reject the null hypothesis and conclude that the null is plausible. Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible. The hypothesis test in this question is: H0: p0 = 0.02 vs. H1: p0 ≠ 0.02 The test statistic is: z = ( 0.03 - 0.02 ) / ( √ ( 0.02 * (1 - 0.02 ) / 100 ) z = 0.7142857 The p-value = P( Z > |z| ) = P( Z < -0.7142857 ) + P( Z > 0.7142857 ) = 2 * P( Z < -0.7142857 ) = 0.4750505 Since the p-value is greater than the significance level of 0.01 we fail to reject the null hypothesis and conclude p = 0.02 is plausible.
large sample confidence intervals are used to find a region in which we are 100 (1-α)% confident the true value of the parameter is in the interval. For large sample confidence intervals for the proportion in this situation you have: pHat ± z * sqrt( (pHat * (1-pHat)) / n) where pHat is the sample proportion z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α n is the sample size here we are only concerned with the error term and the width w. z * sqrt( (pHat * (1-pHat)) / n) = w n = pHat * (1 - pHat) (w/z) ^ -2 we have, w = 0.001 pHat = 0.02 from the above information z = 1.645 for a 90% CI n = 0.02 * (0.98) * (0.001 / 1.645) ^ -2 n = 53038.09 n must be integer valued. always take the ceiling to make sure you have the correct size for the interval n = 53039.
[2003] Joe Smith was running for mayor. His campaign manager estimated that he needed at least 40% of the popular vote in order to win. A poll of 400 people was taken and 150 indicated they would vote for Smith. a) If we use a 10% level of significance, does he have sufficient support? b) Construct a 90% confidence interval on the percentage who will vote for Smith. Solution part a): The campaign manager is concerned if support falls below 40%. So, we set up our hypotheses as: H0: p = 0.4 H1: p < 0.4
Because α = 0.1, we reject the null hypothesis if Z < -1.282, where:
and q = 1-p. In our case, we have:
Plugging these values into the formula for Z, we get Z = -1.0206. Since -1.0206 > -1.282 we do not reject the null hypothesis and conclude that Smith has enough support to win the election. Solution part b): The formula for our confidence interval for p is:
In our case, we have:
Plugging these numbers into the formula, we have: 0.375 - 0.0398 < p < 0.375 + 0.0398 0.3352 < p < 0.4148 So the support for Smith lies between 33.52% and 41.48% based on the evidence. Back to top [2005] A certain intersection had an accident rate of 1 accident per 1000 vehicles. A study suggested that partial obstruction of the stop sign by a tree branch could be partly to blame. The branch was trimmed and the traffic was monitored for 10,000 vehicles. During this period, there were 8 accidents. Did trimming the tree branch help? Use a p-value to reach your conclusion. Solution: If trimming the tree branch helped, we would expect fewer accidents. So we set up our hypotheses as:
H0: p = 0.001 H1: p < 0.001 Because of the sample size, we can use Z because of the Central Limit Theorem. We will reject the null hypothesis for low p-values where p-value = P(Z < test statistic). Our formula for Z is
where q = 1 - p. We have:
Plugging these numbers into the equation we get Z = -0.6328. Our p-value is P(Z < -0.6328) = 0.2634 Because of the high p-value, we do not reject the null hypothesis and conclude that trimming the tree branch made no significant difference. Back to top [2001] A wood door manufacturer wants to minimize the number of knots in each door. On average, he wants no more than 1.2 knots per door. A random sample of 500 doors had, on average, 1.4 knots with a standard deviation of 0.26 knots. Are the manufacturer's standards being met? Test at α = 0.05. Solution: The manufacturer is concerned if there are too many knots in the door. Ergo, we set up our hypothesis as: H0: µ = 1.2 H1: µ > 1.2
Because of the sample size of 500, we can use Z because of the Central Limit Theorem. With α = 0.05, we reject the null hypothesis if Z > 1.645. Our test statistic is:
where
Plugging these numbers into the formula, we get Z = 17.2. Since 17.2 > 1.645, we reject H0 and conclude that the standards are not being met. Back to top [2004] In order to maintain quality control, the angle measurement of a gear tooth must be maintained at 1.02o. A sample of 60 gears is taken. The average angle is 1.025o with a standard deviation of 0.05o. a) Construct a 95% confidence interval of the average gear angle. b) If you were to test the hypothesis H0: µ = 1.02 against H1: µ ≠ 1.02 at α = 0.05, what would be your conclusion using only the confidence interval? Solution part a): Our formula for the confidence interval is:
For our 95% confidence interval we need Z0.025 = 1.96 because 1 - α = 0.95, and so α /2 = 0.025. We have:
Plugging these values into the equation we get 1.025 - 0.0127 < µ < 1.025 + 0.0127 or 1.0123 < µ < 1.0377 Solution part b):Because 1.02 is contained in the 95% confidence interval, we would conclude at a 5% level of significance that the
standards are being met (in other words, we would not reject the null hypothesis). Back to top [2002] Joe's Ceramics wants to keep the standard deviation of the oven temperature to within 5o. He samples 100 baked ceramic pieces and measures (somehow) the oven temperature when he takes the piece out. If he wants the bound on the error to be at most 1o, what should be the minimum level of confidence? Solution: This is a two step problem: Step 1: Solve for Z. Step 2: Solve for 1 - α which is the level of confidence we are looking for. Our formula is:
Solving for Z, we have: E = 1; σ = 5; n = 100 Plugging these numbers into the formula, we get Zα /2 ≥ 2. Since P(Z > 2) = 0.0227, that means α /2 = 0.0227. Solving for our level of confidence, we get 1 - α = 1 - (2)(0.0227) = 0.9546. So, our minimum level of confidence is 95.46%
