On a Separate Piece of Paper, Answer the Following Questions

Solutions Topics 15 – 20

1. An educational testing corporation has designed a standard test of mechanical aptitude. Scores on this test are normally distributed with a mean of 75 and a standard deviation of 15. a. If a subject is randomly selected and tested, find the probability that his score will be between 75 and 100. 100  75 Z   1.67 1 15 75  75 Z   0 2 15 P(75  x  100)  P(0  Z  1.67)  .453 b. If a subject is randomly selected and tested, find the probability that his score will be below 85. 85  75 Z   0.67 15 P(x  85)  P(Z  0.67)  .748 c. If a subject is randomly selected and tested, find the probability that his score will be above 80. 80  75 Z   0.33 15 P(x  80)  P(Z  0.33)  .37 d. Find the score a subject must score in order to be considered better than 60 percent of the population (this is also called the 60th percentile) P(Z*  Z)  0.60 Z*  .25 x  75 .25  15 x  78.8 e. If we were to look at samples of 100 people taking this test, what would the Standard deviation of the sample means be?

 x  15 15  x   1.5 100 f. What would the probability be of a sample mean being at least 80 if we have a sample of 100 people taking the test? 80  75 Z   3.33 1.5 P(x  80)  P(Z  3.33)  .00043 g. We sampled 100 people and found a mean of 72. What is the probability that a sample at most this average occurs by chance? 72  75 Z   2.00 1.5 P(x  72)  P(Z  2.00)  .0228 Solutions Topics 15 – 20

2. In one county, the conviction rate for speeding is 85%. Suppose we look at 100 summonses. (20 pts) a. What is the parameter of interest? The conviction rate for speeding tickets in the given county b. What is π? π = .85 c. What is σ for the 100 summonses picked? .85(1 .85)    .0357 pˆ 100 d. What is the probability that there will be at least 90 convictions in the sample? .9  85 Z   1.4 .0357 P( pˆ  90%)  P(Z  1.4)  ..0807 e. What is the probability that we will have between 75 and 80 convictions in the sample? .75  85 Z   2.80 1 .0357 .80  85 Z   1.40 2 .0357 P(75%  pˆ  80%)  P(2.80  Z  1.4)  .078 Solutions Topics 15 – 20

3. In a randomly selected paragraph of text that had 67 words in it from a Charles Dickens novel, the average length of each word was found to be 3.87 characters, with a sample standard deviation of 1.95. We are interested in the value of mean of the length of all words in a Dickens novel. For this problem assume that a paragraph is a reasonable representation of the entire text (30 pts) a. Find a 90% confidence interval for the mean of the length of all words of a Dickens novel. Be sure to show your work and include the following.  State the variable and parameter of interest and whether the variable is quantitative or categorical  Find all the necessary statistical information needed to answer this question  Show all pertinent technical conditions are satisfied  Use proper notation  Include a sketch of your results.  State the confidence interval in the context of the problem. b. The true mean for a certain Dickens novel is 4.23, is this in the interval you created?

Take inventory:

n  67 df  67 1  66 x  3.87

S x  1.95 (Since we do not know  we will use a t  interval) 1.95 SEx   .238 67 x = the number of characters in a particular word in a Dickens Novel μ = the average number of characters per word in the entire text The variable is quantitative

Technical Conditions: 67 ≥ 30 so n is large enough, but it is questionable that we use a single paragraph to represent the text – though this was explained away in the problem declaration. Therefore, technical conditions are met.

Finding the 90% Confidence interval for the value of μ: * t n1  t60  1.671 (Using the table) * 90% CI : x  t n1 (SEx ) 3.87 1.671(.238) 3.87  .398 (3.472,4.268)

We are 90% confident that the true mean number of characters in a word in this particular Dickens novel is within the interval 3.472 and 4.268 characters.

The value 4.23 is contained in our confidence interval. Solutions Topics 15 – 20

4. The school’s cafeteria wants to know the milk preference of lunch purchasers at FHS. A random sample of students was taken. When asked their preference, it was found that of the 33 sampled 23 preferred chocolate milk. (30 pts) a. Find the 90% confidence interval of the proportion of students that prefer Chocolate milk. Be sure to include the following  State the variable and parameter of interest and whether the variable is quantitative or categorical  Find all the necessary statistical information needed to answer this question  Show all pertinent technical conditions are satisfied  Use proper notation  Include a sketch of your results  State the confidence interval in the context of the problem. b. Would we be surprised if the true proportion of Chocolate milk drinkers is 50%, 55% or 60%?

Take Inventory: x = 23 n = 33 23 pˆ = = .697 33 p = ? .697(1- .697) SE = = .080 pˆ 33 Variable of interest: (x) A students preference for milk in the school cafeteria Parameter of interest: (π) the true proportion of students that prefer chocolate milk Create a 90% Confidence interval for π:

* pˆ  Z (SE pˆ ) .697 1.645(.080) (.5654,.8286) We are 90% confident that the true proportion of students in our cafeteria that prefer chocolate milk is in the interval (.5654, .8286) Would we be surprised? π=50%: We would be very surprised since this not close to our interval π=55%: We would be fairly surprised since it falls just outside our interval π=60%: We would not be surprised since this value is well within our interval