Let me know if you find anything you think is a mistake.
1)
H0: µ = 60
H1: µ 60
Assumptions: sample must be large enough (>30) to use CLT or the data is distributed normally.
Distribution: t-distribution. Two-tail. Degrees of freedom: 11
Critical values: ±2.20
α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though it is true.
test statistic = -.874
Since the test statistic (-.874) is not more extreme than the critical value (-2.20), we fail to reject the null hypothesis
Since the p-value (.401) is greater than the level of significance (α=.05), we fail to reject the null hypothesis
In English: there is insufficient evidence to conclude that the mean reading score for this fifth grade class is different than the national average.
2)
H0: σ = 15
H1: σ 15
Assumptions: data is distributed normally.
Distribution: χ2. Two-tail. Degrees of freedom: n-1 = 11
Critical values: 3.82, 21.92
α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though it is true.
Get the standard deviation using 1-Var Stats, and it’s 16.837.
test statistic = (11 * 16.8372) / 152= 13.86
Since the test statistic (13.86) is not more extreme than the critical values (3.82 or 21.92), we fail to reject the null hypothesis
No p-value
In English: there is insufficient evidence to conclude that the standard deviation of reading scores for this fifth grade class is different than the national average.
3)
H0: µO = µN
H1: µO < µN
Assumptions: samples (both) must be large enough (>30) to use CLT or the data is distributed normally.
Distribution: t-distribution. One-tail. Since I’ve got the old first (on the left above and will be list 1), this is a left tail test. Degrees of freedom: (from calculator) 21
Critical values: -1.72
α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though it is true.
test statistic = -1.11
Since the test statistic (-1.11) is not more extreme than the critical value (-1.72), we fail to reject the null hypothesis
Since the p-value (.139) is greater than the level of significance (α=.05), we fail to reject the null hypothesis
In English: there is insufficient evidence to conclude that the mean reading score for the new teaching program are lower than the older program.
4)
H0: σO = σN
H1: σOσN
Assumptions: the data is distributed normally.
Distribution: F-test. Two-tail. Degrees of freedom: 11 & 11.
Critical values: .288 & 3.474
α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though it is true.
Calculate the two standard deviations using 2-Var stats. They are 19.41 & 12.02.
test statistic = 19.412/12.022 = 2.61
Since the test statistic (2.61) is not more extreme than the critical values (3.474), we fail to reject the null hypothesis
In English: there is insufficient evidence to conclude that the new variant of the reading program has a different standard deviation than the old variant of the program.
5)
H0: µ1 = µ2 = µ3
H1: at least one mean (µ) is different than the others.
Assumptions: data are distributed normally, independent, with the same standard deviation.
Distribution: F-test. One-tail. Right tail. Degrees of freedom: 2 & 33
Critical values: 3.28
α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though it is true.
test statistic = 1.01
Since the test statistic (1.01) is not more extreme than the critical value (3.28), we fail to reject the null hypothesis
Since the p-value (.373) is greater than the level of significance (α=.05), we fail to reject the null hypothesis
In English: there is insufficient evidence to conclude that at least one mean reading score is different than the others.
Multiple Choice Section
1)c. Since the scores look like they’re only whole numbers, then
2)c.
3)c.
4)b.
5)b. (this is a sample, so the estimate is s, not σ)
6)b.
7)a. (we’re testing if they are different)
8)d.
9)a.
10)c.
11)a.
12)b.
Practice Test, Final, Part 2
1) The next set of questions concerns the following: the probability that a customer who has walked into a chocolate shop buys something is 62%. This rate is constant, and the outcome of any customer is independent of other customers.
What is the sample space if 20 people walk into the store?
{0, 1, 2, …. 20}
If 20 customers enter the store, what is the mean number who will make a purchase?
.62 * 20 = 12.4
If 20 customers enter the store, what is the probability that exactly 15 will make a purchase?
binompdf( 20, .62, 15) = 9.4%
If 20 customers enter the store, what is the probability that more than 12 will make a purchase?
Write out the Event list for the previous question. Also write out the complement.
Event = { 13, 14, … 20}
Complement = {0, 1, 2, …12}
P(Event) = 1 – P(Complement) = 1 – bionomcdf( 20, .62, 12) = 48.9%
2) The next set of questionsconcerns the following: This question concerns the number of customers who enter a chocolate shop. The average rate of customers coming into a chocolate shop during operating hours is 5 per hour. This rate is constant, and the occurrence of any customer entering the store is independent of other customers entering.
What is the sample space of how many customers will walk into a store in an hour.
{0, 1, 2, …. ∞}
What’s the probability of exactly 2 customers entering the store in an hour?
poissonpdf(5*1,2) = 8.4%
What’s the probability of at least 4 customers entering the store in an hour?
1 – poissoncdf(5*1, 3) = 73.5%
What’s the probability of exactly 7 customers entering the store in two hours?
poissonpdf(5*2, 7) = 9.0%
What’s the mean number of customer who will enter the store in eight hours?
5*8 = 40
3) The next set of questions concerns the following box of assorted chocolates. The box has twelve dark chocolates (with the following fillings: Earl Grey, dark truffle, vanilla, caramel, lemon, cherry, nougat, almond, cardamom, hazelnut, ancho chile, and espresso), eight milk chocolates (with the following fillings: mint, peanut brittle, caramel, marzipan, praline, vanilla, peanut butter, dark truffle), and 5 white chocolates (with the following fillings: key lime, dark truffle, macadamia nut, chai tea, vanilla malt).
If a person randomly chooses three pieces, how many different outcomes are possible if order matters?
25*24*23 = 25 P 3 = 13800
If a person randomly chooses three pieces, how many different outcomes are possible if order does not matter?
25*24*23 / 3*2*1 = 25 C 3 = 2300
Note: the 3*2*1 is the number of different ways you could arrange 3 pieces.
If a person randomly chooses two pieces, what is the probability they’re both dark chocolates?
= .22 = 22.0%
If a person randomly chooses three pieces, what is the probability that one is dark, one is milk, and one is white chocolate, order does not matter?
= .209 = 20.9%
If a person randomly chooses three pieces, what is the probability that one of the pieces chosen is their favorite: white chocolate with key lime filling?
= .12 = 12%
4) The next set of questionsconcerns the following: for customers who make a purchase at the chocolate store, the amount they spend is distributed normally with a mean $55 and a standard deviation of $20.
What is the probability of a customer spending more than $50?
=normalcdf(50, 1e99, 55, 20) = 59.9%
What is the probability of a customer spending between $30 and $60?
=normalcdf(30, 60, 55, 20) = 49.3%
What is the probability of a customer spending less than $40?
=normalcdf(-1e99, 40, 55, 20) = 22.7%
If 20 customers shop in the store, what is the probability that the average purchase is less than $50?
=normalcdf(-1e99, 50, 50, 20/√20) = 13.2%
5) The next set of questions concerns the following categorization of the results of a test for a rare disease. All questions concern a single person chose at random.
Tested Negative / Tested PositiveDoes not have the rare disease / 9845 / 95
Has the disease / 3 / 57
What is the probability that the person has the disease?
(3+57) / 10,000 = 0.6%
What is the probability that the person has the disease or tests positive?
(3+57+95) / 10,000 = 1.55%
What is the probability that the person tests positive and does not have the disease?
95 / 10,000 = .95%
What is the probability that the person tests negative?
(9845+3) / 10,000 = 98.58%
What is the probability that the person tests negative given that they don’t have the disease? (Test accuracy for people without the disease).
9845 / (9845+95) = 99.9%
What is the probability that the person tests positive given that they have the disease? (Test accuracy for people with the disease).
57 / (57+3) = 95.0%
What is the probability of the person having the disease given that they test positive?
57 / (57+95) = 37.5%
6) The next set of questions concerns the following: you work at a chocolate shop and develop a new flavor, dark chocolate with a cabernet wine filling. You want to estimate how many people “like” the new flavor. You survey a random sample of customers after offering them a free sample of the new flavor. You sample 40 customers, and 27 secretly mark on a survey card that they “like” the flavor.
In choosing the sample size, n, what do you want to make sure of?
There are at least 10 who like the flavor and 10 who don’t.
What is the point estimate of the proportion of people who “like” the new flavor.
27/40 = 67.5%
Construct a 95% confidence interval for the proportion of customers that like the new flavor
(53.0%, 82.0%)
What is the margin of error?
14.5%
How many customers would you have to sample to make the margin of error 2%?
= 2106.9 = 2107
At least 60% of customers must “like” the flavor to justify the cost of production. Can you conclude that at least 60% of customers like the new flavor (α = .05)? (Include all steps expected for a hypothesis test)
**** Note: I should have worded the question as “more than 60% of customers must ‘like’ the flavor to justify..” instead of “at least 60%”.
H0: p = .60
H1: p > .60
Assumptions: at least 10 people in each category and population is 20 times sample size.
Distribution: Normal (z test) distribution. One-tail, right tail. Degrees of freedom: NA
Critical values:1.64
α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though it is true.
test statistic = .968
Since the test statistic (.968) is not more extreme than the critical value (1.64), we fail to reject the null hypothesis
Since the p-value (.166) is greater than the level of significance (α=.05), we fail to reject the null hypothesis
In English: there is insufficient evidence to conclude that sufficient people (at least 60%) like the new flavor to justify cost of production.
7) The next set of questions is a continuation of the previous question. But another flavor is tried, dark chocolate with a port wine filling. 60 customers are sampled and 44 customers “liked” the new flavor.
Can you conclude that more customers like the port filling than the cabernet (α = .05)? (Include all steps expected for a hypothesis test)
H0: pP = pC
H1: pPpC
Assumptions: at least 10 people in each category and population is 20 times sample size for both samples
Distribution: Normal (z test) distribution. One-tail, left tail. Degrees of freedom: NA
Critical values:-1.64
α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though it is true.
test statistic = -.630
Since the test statistic (-.630) is not more extreme than the critical value (-1.64), we fail to reject the null hypothesis
Since the p-value (.264) is greater than the level of significance (α=.05), we fail to reject the null hypothesis
In English: there is insufficient evidence to conclude that more people like the port flavor than the cabernet.
8) Consider the following types of sales the chocolate shop make
Can you conclude that the types of sales have changed in 2017(α = .05)? (Include all steps expected for a hypothesis test).
H0: p1 = .17, p2=.32, p3=.11, p4=.40
H1: some or all of those percentages are different
Distribution: χ2distribution. One-tail, right tail. Degrees of freedom: 3
Critical values: 7.81
α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though it is true.
test statistic = 210.9
Since the test statistic (210.9) is more extreme than the critical value (7.81), we reject the null hypothesis and accept the alternative
Since the p-value (1.9x10-45) is less than the level of significance (α=.05), we reject the null hypothesis
In English: there is sufficient evidence to conclude that the types of sales in 2017 are different than the historical distribution.
9) The next set of questions concerns the following: A mature almond tree produces an average quantity of 7 pounds of almonds. The quantity of almonds produced is normally distributed. A farmer plants a grove of 20 trees. These trees averaged 6.4 pounds of almonds with a standard deviation of 2.1 pounds.
Can you conclude that the grove is performing below average (α = .05)? (Include all steps expected for a hypothesis test)
H0: µ = 7
H1: µ < 7
Assumptions: sample must be large enough (>30) to use CLT or the data is distributed normally.
Distribution: t distribution (don’t have population standard deviation). One-tail, left tail. Degrees of freedom: 19
Critical values: -1.73
α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though it is true.
test statistic = -1.28 (note: use T Test function, but for input, click on “STATS” not “DATA”
Since the test statistic (-1.28) is not more extreme than the critical value (-1.73), we fail to reject the null hypothesis
Since the p-value (.108) is greater than the level of significance (α=.05), we fail to reject the null hypothesis
In English: there is insufficient evidence to conclude that the grove produces below average.
10) The next set of questions uses the following data. A regression will be run on the quantity of boxes of chocolate sold regressed on the price. That is, the regression you will run is:
Qty = α + β Price
Price / $35 / $37 / $40 / $42 / $45 / $47 / $50 / $53 / $55 / $60 / $65Quantity Sold / 43 / 42 / 45 / 40 / 39 / 34 / 39 / 27 / 34 / 24 / 26
Run the regression, state the values of a & b and interpret all results (a, b, r, r2, x-intercept)
qty = 68.8 - .688 Price
a: if the price is $0, people will “buy” 68.8
b: for every $1 increase in price, people will purchase .688 fewer goods
x-intercept: 100. It would take a price of $100 for people to buy 0.
r: -.900 There is a strong negative relationship between price and quantity
r2: .810. 81.0% of the variance in quantity can be explained by the variance of price.
Make a prediction, if the shop charged $49, how many would they sell?
35.1
Be able to calculate predicted Y’s and residuals
For example, for the first data pair, $35 and 43:
Predicted Qty =44.7
Error = 43 – 44.7 = -1.7
Conduct hypothesis test to see if price reduces quantity sold. Include all steps of a hypothesis test.
H0: β = 0
H1: β < 0
Assumptions: errors (or residuals) are independent and distributed normal.
Distribution: t distribution. One-tail, left tail. Degrees of freedom: 9 (n – 2)
Critical values: -1.83
α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though it is true.
test statistic = -6.20
Since the test statistic (-6.20) is more extreme than the critical value (-1.73), we reject the null hypothesis
Since the p-value (.00008) is less than the level of significance (α=.05), we reject the null hypothesis
In English: there is sufficient evidence to conclude that higher prices reduce the quantity sold.
Other Issues:
What type (classification) of variables are there?
What is a simple random sample, and why are they used?
What is the difference between an observation study and an experiment?
What creates bad samples?
What is the Central Limit Theorem and why is it important?
You’ll have to research these yourself.