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 n1 t60 1.671 (Using the table) * 90% CI : x t n1 (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