Mr. Rogers - AP Statistics, Second Quarter Objectives
1st Semester Exam  Practice Test Answers

First Semester Exam Practice Test Answers

Note:

Questions with numbers highlighted in blue such as 3) have material in them that at this point in time may not have been fully discussed in class. Although you may not fully understand all the thinking behind them, by studying the examples provided most students will be able to answer similar questions on the first semester exam.

Questions similar to the ones with numbers highlighted in orange such as 6) were answered incorrectly more than 60% of the time on an actual AP Statistics exam. They are important questions to understand.

1) What is meant by a 5% margin of error? The difference between the estimate based on the sample percentage  and the actual population percentage is likely to be less than 5%

2) Bob claims that his apples are bigger than Martha's apples.  The mean of Martha's apples is 27. If you perform a significance test, what would the null and alternative hypothesis be? Ho: μ = 27, Ha μ >27

Note Ho is the null hypothesis representing "no difference". Ha is the alternative hypothesis representing "there is a difference".

3)  A consultant takes a random sample and gets a mean size of 28 for Bob's apples based on a sample size of 16. If the standard deviation of Martha's apples is 4, find the p-value and answer the question are Bob's apples larger.
Find the p-value on the TI-83 as follows:

Note that in this case we use

s  = σ / (16)0.5

= 4 / 4

= 1

p-value =

2nd

DISTR  normalcdf ( 28, 10000, 27, 1)

 p-value =  15.9%

 p-value is always a tail area and is defined as the probability of getting a test statistic (in this case 28) as far from the mean or further from the mean than the one obtained.
 Conclusion: There is not statistically significant (p = 15.9%) evidence to indicate that Bob's apples are larger than Marta's.
4) What is the best way to establish a cause and effect relationship between two variables? A controlled experiment.

5) Given the equation: y-hat = 4x +10, if x is increased by 5 what is the change in y? 4 x 5 = 20

6) 1,000,000 people belong to Martha's Fitness Emporium. 600,000 are women and 400,000 men. 50% of the women work out every week and 40% of the men. In a sample of 1000, what number would be expected to work out (wo) every week?

7) If Ho: μ = 50, Ha μ >50 and the mean of a sample were 52 with a z score = 2 and a p-value = 0.023 what would be the likely conclusion? There is significant reason to conclude that Ha μ >50. Explanation: there is only a 2.3% chance of getting a sample mean as extreme or more extreme than 52. Therefore it's reasonable to reject the null hypothesis and accept the alternative hypothesis. The null hypothesis is usually rejected if p< 5%.

8) Since she wants a proportion answering a yes or not yes question (not yes includes undecided), we model the problem binomial. Since she will be surveying a large sample size we can use a normal approximation for the binomial. All this is a complicated way of saying that we should use the following equation:

Confidence interval = p ± Z* [ p (1 - p ) / n ]^0.5

Z* = 1.96

M = Z* [ p (1 - p ) / n ]^0.5

Unfortunately, we have one equation and two unknowns (shown in red, knowns are shown in blue), so we're forced to estimate p. To do so, we want to select the most conservative possible value and this would be the one giving the highest value of n. It turns out that p = 50% always gives the highest value of n and will always be the one selected when p is unknown.

M2  = (Z*)2  [p (1 - p ) / n ]

M2 / (Z*)2 = p (1 - p ) / n

 n = p (1 - p ) M2 / (Z*)2

 n = 0.5 (1 - 0.5 ) 0.022 / (1.96)2

 n = 2401   Note: always round fractions upward

9) Bob's TocoRama has a deal. Every Monday customers can spin a wheel that gives them a discount of  up to \$1.00 on their next purchase. The mean is \$0.50 and standard deviation is \$0.15. If a customer participates for 40 weeks, what is the probability that he will have a savings greater than \$21?

In 40 weeks the mean savings will be:

μ = 40 * \$0.50

= \$20

Now we can draw a picture of the problem in order to visualize
it, but before calculating the area we have to first determine the standard deviation of the distribution we've just drawn.
The standard deviation for 40 weeks is as follows:

σ = [40 * (0.15)2]0.5

= 0.9487

Note: standard deviations are not additive, however, variances (the standard deviation squared) are. Hence, the standard deviations must be squared before being multiplied by 40. This gives the total variance for 40 weeks. Taking the square root of this this variance give the 40 week standard deviation.

probability =

2nd

DISTR  normalcdf ( 21, 10000, 20, .9487)

 probability savings greater than \$21 =  14.6%
10) 10% of Juan's fruit special cakes weigh less than 20 grams and 15% weigh more than 25 grams. Find the mean and standard deviation.

First, draw a picture as shown at left. From this we can find z1 = - 1.28 and z2 = 1.04 using a z-table. Next it's time to write an equation:

z = ( xi -  μ ) / σ

Note that since the above equation has 2 unknowns in it ( μ and σ), we have to write a version of it for each side as shown below. This gives 2 equations and 2 unknowns which we can solve as simultaneous equations.

Left Side   Right Side
- 1.28 = ( 20 -  μ ) / σ

σ  = ( - 20 +  μ ) / 1.28

1.04 = ( 25 -  μ ) / σ

σ  = ( 25 -  μ ) / 1.04

next merge the two equations as follows:

( μ - 20 ) / 1.28 = ( 25 -  μ ) / 1.04

( μ - 20 ) 1.04 = ( 25 -  μ ) 1.28

1.04 μ - 20.8 = 32 - 1.28 μ

1.04 μ + 1.28 μ = 20.8 + 32

2.32 μ = 52.8

μ = 52.8 / 2.32

 μ = 22.76

Substituting μ into the left side's equation gives

σ  = ( - 20 +  22.76 ) / 1.28

= 2.95 / 0.25

 σ  = 2.16