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[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 a = 0.1, we reject the null hypothesis if Z < -1.282, where:

eq001

and q = 1-p. In our case, we have:

eq002

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:

eq003

In our case, we have:

eq004

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.

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[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

eq001

where q = 1 - p. We have:

eq002

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.

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[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 a = 0.05.

Solution: The manufacturer is concerned if there are too many knots in the door. Ergo, we set up our hypothesis as:

H0: m = 1.2
H1: m > 1.2

Because of the sample size of 500, we can use Z because of the Central Limit Theorem. With a = 0.05, we reject the null hypothesis if Z > 1.645. Our test statistic is:

eq001

where

eq001

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.

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[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: m = 1.02 against H1: m ą 1.02 at a = 0.05, what would be your conclusion using only the confidence interval?

Solution part a): Our formula for the confidence interval is:

eq001

For our 95% confidence interval we need Z0.025 = 1.96 because 1 - a = 0.95, and so a/2 = 0.025.

We have:

eq002

Plugging these values into the equation we get 1.025 - 0.0127 < m < 1.025 + 0.0127 or 1.0123 < m < 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).

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[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 - a which is the level of confidence we are looking for.

Our formula is:

eq001

Solving for Z, we have:

E = 1; s = 5; n = 100

Plugging these numbers into the formula, we get Za/2ł 2. Since P(Z > 2) = 0.0227, that means a/2 = 0.0227. Solving for our level of confidence, we get 1 - a = 1 - (2)(0.0227) = 0.9546. So, our minimum level of confidence is 95.46%

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