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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:

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.

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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

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.

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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:

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.

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[2004] In order to maintain quality control, the angle measurement of a gear tooth must be maintained at 1.02^o. A sample of 60 gears is taken. The average angle is 1.025^o with a standard deviation of 0.05^o.
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 ¹ 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:

For our 95% confidence interval we need Z_0.025 = 1.96 because 1 - a = 0.95, and so a/2 = 0.025.

We have:

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 5^o. 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 1^o, 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:

Solving for Z, we have:

E = 1; s = 5; n = 100

Plugging these numbers into the formula, we get Z_a/2 ³ 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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