Statistics Note Sheet Essay Sample

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Practice Question #1: A gambler claims he can predict the roll of a die more often than chance would predict. To test this, a die is rolled 100 times and the gambler guesses the number that comes up twenty times and guesses incorrectly the other eighty times. Is this strong evidence that the gambler’s claim is true? a. Specify the null and alternative hypothesis for this problem. p = probability gambler guess correctly for an individual roll null (Ho): p = 1/6 alternative (Ha): p > 1/6

b. Find the test statistic and calculate the p-value. What do you conclude? pˆ = 0.2 so z = (0.2-0.1667)/ .1667(.8333) /100 = 0.894. From the table the p-value is between 18% and 19%. The null hypothesis is a reasonable explanation of this data so we do not have strong evidence that the gambler can predict the outcome of the die. c. If the gambler had guessed correctly on forty of the rolls, would the p-value go up or down? Since this would be farther away from what is wexpected under the null, the p-value would get smaller.

Practice Question #2: Does the person paying the bill order more expensive meals, or less expensive meals, at a restaurant? The bills from 100 parties of two are examined and it turns out that the person paying the bill ordered a meal costing an average of \$0.50 more than their companion with a standard deviation of \$2.00 a. Clearly define the parameter of interest in this situation and then state the null and alternative hypotheses as statements about this parameter. μ = the average amount the bill-payer orders more than their companion for the whole population…null (Ho): μ = 0 alternative (Ha): μ ≠ 0

b. Find the test statistic and calculate the p-value. What do you conclude? x = 0.5 and z = (0.5-0)/(2.00/100 ) =2.5. From the tables the p-value ≈ 2(100-99.38) = 1.24%. It is unlikely that average differences of this size would occur for 100 pairs of customers. We have evidence that the person paying the bill and their companion do not order the same priced meals on average. c. In lecture we learned that statistical significance does not imply practical significance. Explain how this is illustrated by this example. Even though the bill-payer spent significantly more, the actual difference of 50 cents is very small and not likely to be of any practical importance. Practice Question #3: Which of these statements are true and which are false. Explain. A) If the p-value is 100% then the null hypothesis must be true. False – for a two-sided test this just says the data came out exactly as expected under the null hypothesis. But this might also happen under the alternate (though the chances would be lower). B) If the p-value is 1% then the alternative hypothesis has a 99% chance of being true.

False – the p-value is calculated assuming the null hypothesis is true so it doesn’t tell you about the chances under the alternative. C) If the p-value is 50% this says that 50% of the time the test statistic would be this far from what is expected when the null hypothesis is true. True D) For the same data, a one-tailed significance test will give a lower p-value than a two-tailed significance test. True – since the p-value for the two-sided test includes the chance of getting both high and low values under the null. E) It is important to look at the results before deciding on whether to use a one-tailed or two tailed test. False – the null and alternative hypothesis are created before the data are collected. F) If you conduct 100 hypothesis tests you are likely to find some significant results even when the data are all just due to chance. True – for example about 5 of the tests will give p-values less than 5%.

Sampling Distributions and Confidence Interval Practice Problems. (note : you will need to use Table 21.1 and the table on page 435 of the text) Practice Question #1: A researcher wants to know the percentage of Columbus residents who would favor a two cent increase in the gasoline tax to fund road repairs.A random sample of 900 residents is chosen and 278 favor the increase. a. Specify the parameter and statistic for this problem.

Parameter = percentage of Columbus residents who would favor a two cent increase in the gasoline tax to fund road repairs = p Statistic = 278/900 (30.9%) =pˆ b. Find an 80% confidence interval for the parameter

Estimated st. dev. Of pˆ = 100% .309(.691) /900 ≈ 1.54%
Confidence interval is 30.9% ±1.28(1.54%) or 30.9% ± 1.97% Practice Question #2: Suppose 50% of all OSU students drive to school every day. We randomly select 36 OSU students. a. What is the standard deviation of the sampling distribution of the sample proportion who drive to school in this problem? The standard deviation is .5(.5) /36= 0.083 b. Describe the sampling distribution of the sample proportion from a sample like this. Specifically explain what shape it would have and where it will be centered. The sampling distribution will look like the normal curve centered at 0.5 with a standard deviation of 0.083. c. What is the probability that in a random sample of 36 OSU students, more than 52% drive to school?

Standard score = z = (.52-.50)/0.083 = 0.24
Using Table B the answer is about 100% – 59% = 41% (okay if students don’t bother to interpolate answer from table) Practice Question #3: A random sample of 400 OSU students are asked how much money they spent on entertainment in the past week. These 400 spent an average of \$28 with a standard deviation of \$18. a. Find a 90% confidence interval for the parameter of interest in this problem. Estimated st. dev. Of x is \$18/400 = \$0.9 so the confidence interval is \$28 ± \$1.64(0.9) or \$28 ± \$1.48 b. How would the interval change if you made a 70% confidence statement instead? With 70% confidence we would have a shorter interval (smaller margin or error). In fact the new interval would be \$28 ± \$1.04(0.9) or \$28 ± \$0.94 Practice Question #4: The books in the OSU library contain an average of 425 pages with a standard deviation of 100 pages. a. Is it more likely that a sample of 20 books will have an average of more than 500 pages or is this more likely in a sample of 200 books? Explain. It is more likely for a sample of 20 books.

The Law of Averages says that being away from what you expected for an average is more likely with a smaller sample (here you expect 425 and are being asked about over 500) b. What is the probability that the average number of pages in the next 200 books checked out of the library will be between 410 and 440 pages? Show your calculations and describe any assumptions you are making in carrying out these calculations.You must assume that the books being checked out have an independent number of pages so this is like a random sample of 200 books making the sampling distribution of this average approximately follow the normal curve. Here, the standard deviation of the sampling distribution is 100 / 200 ≈7.07. The standard units are (440-425)/7.07≈2.1 and (410-425)/7.07 ≈ -2.1 From the tables the answer is 98%-2% = 96%

1) A test statistic is used to measure the difference between the observed sample data and what is expected when the null hypothesis is true.True 2) If a value has a standard score of –1.8 then it must be below the mean. True 3) A boxplot will quickly show if a distribution is bimodal. False 4) Most of the students in last quarter’s statistics 135 class, that responded to a survey, did not smoke, although one student smoked three packs a day. True or False: For this group of students the mean number of cigarettes smoked per day was larger than the median.True 5) Two newspapers report the results of the same Gallup Poll regarding the percentage of people who favor term limits for Senators. The first newspaper uses the poll to present a 95% confidence interval for this percentage, while the second newspaper presents an 80% confidence interval. True or False: The confidence interval presented by the first newspaper will be wider than the interval reported by the second paper. True 6) If everyone who works at a restaurant is given a \$500 holiday bonus then the correlation between their individual incomes and hours worked would not change. True 7)

If the P-value is 99.999% then the null hypothesis does not provide a plausible explanation for the data. False8) If a list of numbers has a mean of 0, then the SD of the list will also be zero.False 9) You are more likely to get heads on between 40% and60% of the tosses when you toss a coin 800 times than when you toss the coin 8 times. True 10) One thousand students are randomly selected from the list of those currently registered at Ohio State and they each report how many miles they rode in a COTA bus during the past week. Since many students did not ride a bus at all, a histogram of the values does not look like the normal curve.

True or False: Even though the histogram did not follow the normal curve, it is still possible to use the normal curve to make a confidence interval for the average number of miles that all OSU students rode on COTA buses last week. True B 11) One morning a pet store weighed each rabbit it has for sale. The SD of these weights is 2 pounds. A collar that weighs 0.1 pounds is then put on each of these rabbits. The SD of their weights at this point (including the collars) is then: A) 2.1 pounds B) 2 pounds C) 1.9 pounds D) changed by a factor corresponding to the standard units of 0.1 years.

C 12) Below is a histogram of the size (length of longest dimension in mm) of ocular tumors seen at the OSU ophthalmology clinic over the past three years. One patient had a tumor that was 1mm in size. Its standard units in this data set

A) would follow the normal curve. B) would be bigger than the median. C) would be a negative number. D) would be a positive number. 13) Boxes of birthday candles in a shipment weigh an average of 4 ounces with a standard deviation of 0.2 ounces. A histogram ofthese weights followed the Normal distribution quite closely. Approximately what percentage weighed between 3.78 ounces and 4.22 ounces? A) 4.7% B) 9.3% C) 54.7 % D) 72.9% Answer – D

z = (3.78-4)/0.2 = -1.1; z = (4.22-4)/0.2 = 1.1
86.43% – 13.57% = 72.86%
14) A spinner can land in eight possible positions so that all eight have the same probability of coming up. Each position must have probability A) between 0 and 1, but can’t say more. B) between -1 and 1, but can’t say more. C) 1/2. D) 1/8. Answer: D they must all add to 1 so if they are the same then they each must have probability 1/8 15) A sample is about to be taken from the 900 students taking Statistics 135 this quarter. Which of the following would most closely follow the normal distribution? A) The histogram of the weights of 2 randomly selected students B) The histogram of the weights of 20 randomly selected students. C) The sampling distribution of the average of the weights of 2 randomly selected students D) The sampling distribution of the average of the weights of 20 randomly selected students Answer – D. The normal approximation works better for larger samples 16) Match the summary statistics with the histograms

i) mean = 6.6, median = 6.8, standard deviation = 1.3
ii) mean = 6.6, median = 6.0, standard deviation = 8.65
iii) mean = 6.6, median = 3.75, standard deviation = 7.4
17) Royals and Connor studied the relationship between the pH of the water in 53 lakes in Florida and the levels of mercury (in parts per million) found in the lakes’ fish. (note: pH measures the acidity of the water). One lake had a pH level of 5.3. You would expect the fish in this lake to have a mercury content of __________. About __________ percent of the variability in the mercury content of the fish is explained by the pH of the lake. Answers:

estimated mercury content = 1.53092 – 0.152301(5.3)≈ .72 ppm R2 = 33.1% = %of variability in Y (mercury content) explained by X (pH) 18) The depth (in feet below sea level) of the deepest trench in the Pacific Ocean is measured 100 times independently using the same process and the results are shown in the histogram below. If one more measurement is taken using this same process, we would expect it to come out around __________________ give or take a standard deviation of about __________________. Fill-in the blanks from the choices 20) A random sample of 1000 people who signed a card saying they intended to quit smoking on November 20, 1995 (the day of the “Great American Smoke-out”) were contacted in June, 1996. It turned out that 210 (21%) of the sampled individuals had not smoked over the previous 6 months. a) Specify the population of interest, the parameter of interest,the sample and the sample statistic in this problem. • Population: everyone who signed the card • Parameter: % of card signers who didn’t smoke for 6 months after that • Sample: the 1000 people contacted in June 1996 • Statistic: 21% b) Make a 90% confidence interval for the percentage of all people who had stopped smoking for at least six months after signing the nonsmoking pledge.

pˆ ± z * pˆ ( 1− pˆ ) /n

0.21±1.64 0.21(1− 0.21) /1000
0.21 ± 1.64(0.0129)
0.21 ± 0.0211
21) A genetic theory says that a cross between two pinkflowering plants will produce red flowering plants 25% of the time. To test the theory, 100 crosses are made and 31 of them produce a red flowering plant. Is this strong evidence that the theory is wrong? Carry out the appropriate hypothesis test. Be sure to write down the null and alternative hypotheses, find the test statistic and the Pvalue, and state your conclusions. Null hypothesis: p=0.25 (where p = the expected proportion of red-flowering plants) Alternative hypothesis: p≠0.25 Test statistic: P-value ≈16% (Table – be sure to include the region in both tails) The null hypothesis is a reasonable explanation of the data. Such a large P-value indicates that we don’t have strong evidence that the genetic model is wrong. 22) Customers using a self-service soda dispenser take an average of 12 ounces of soda with an SD of 4 ounces. What is the chance that the next 100 customers will take an average of less than 12.24 ounces? The sampling distribution of x-bar with n=100 is normally distributed with a mean of μ=12 and a standard deviation of = 0.4. Standard score = (12.24 – 12)/0.4 = 0.6 From Table B, the answer is 72.6% 23)

A researcher at The Ohio State University believes that a certain component of ant venom can be used to lessen the amount of swelling in the knuckles of people suffering from arthritis. The ant venom treatment has been made into a capsule form that can be swallowed. Explain how you would design an experiment to investigate whether this new treatment when taken orally each day for one week causes a lower degree of swelling in arthritis sufferers. (You may suppose that 200 people suffering from arthritis have already volunteered to be experimental subjects). Identify the explanatory variable and the response variable in your experiment. Randomly assign people to ant venom capsule or placebo capsule. At the end of one week check for an increase or decrease in swelling (be sure the person checking doesn’t know which group the subject is in. This would be an example of a randomized, doubleblind, comparative experiment. Explanatory variable: whether the subject got ant venom or placebo Response variable: the degree of change in swelling. 24) Historical Polling example. Perot voters were less likely to be watching the debate since their candidate wasn’t involved.

Thus the polls that only questioned people who had watched the debates would provide biased estimates of the preferences of all voters. The pivotal idea: Always check whether what brings people into the sample might be related to the response being measured. 25) A paint manufacturer fills cans of paint using a machine that has been calibrated to fill the cans to contain an average of 1 gallon (128 ounces) each. To test whether their machine has come out of calibration, and will tend to overfill the cans, the manufacturer takes a random sample of 100 cans and finds that they average 128.2 ounces with an standard deviation of 4 ounces. Is this strongevidence that the can-filling machine is set too high? Carry out the appropriate hypothesis test. Be sure to write down the null and alternative hypotheses, find the test statistic, the P-value, and state your conclusions. Null hypothesis: μ = 128 (where μ = long run average put in can) Alternative hypothesis: μ > 128 Test statistic:= 0.5 P-value = 30.85% The null hypothesis provides a reasonable explanation of the data. With such a high P-value, we don’t have strong evidence that the machine is out of calibration. z = (128.2 −128)

4 / 100 0.5
26) Customers at a grocery store pay with a credit card, with cash, or with a check. Sixty percent of the customers pay with cash. Eighty percent pay don’t use a credit card. What is the probability that a randomly picked customer pays with a check? Explain. Since 80% don’t use a credit card, 20% do 100% (all customers) – 60% (chance of cash use) – 20% (chance of credit card use) = 20% (chance of check use) 27) Trucks are weighed at a Truck Scale to establish the amount owed in road taxes. Someone complains that the weighing procedure has three problems. Problem I: Sometimes the driver is sitting in the truckwhen it is weighed. Problem II: When the same truck is weighed independently more than once, the truck scale will give different values. Problem III: When the legislature established the road tax,they intended to tax according to the value of the goods being shipped, not according to the weight.

Which of the above indicates (Explain each briefly)
a)  a problem with bias? b) a problem with reliability? c) a problem with validity? Answers: a) I b) II c) III Lurking variable-a variable that has an important effect on the response variable but is not included as an explanatory variable. Confounded variables- two variables are confounded if their effects on the response variable can’t be distinguished from each other. Variables that are confounded with each other might be explanatory variables or they might be lurking variables. “explanatory – explains – response” Facts about Standard Deviation

1. the SD of a population is denoted by “sigma” σ. The SD of a sample is
denoted by s. 2. The SD is NEVER NEGATIVE. It is positive if there is any variation at all in the data. 3. The SD is zero only if all values in our data are exactly the same. The bigger the SD, the flatter the curve. Small SD- pointy curve. Standard score “z” = observed value-mean/SD

Shifting-adding or subtracting the same number to all of the values in a distribution Scaling- multiplying or dividing all of the values in a distribution by the same number Measurements:
1. A measurement is RELIABLE if repeated measurements on the same individual give the same results. 2. A measurement is BIASED if it consistently deviates from the true value in the same direction. (a measurement is UNBIASED if it doesn’t deviate in one direction more than the other) 3. A measurement is VALID if it is relevant or appropriate as a representation of the property it is meant to meas. Improving measurements?

* To improve RELIABILITY, we can take several measurements and average them. * Averages tend to be less variable than individual measurements. * To improve BIAS, we typically need to find a new method of measurement. * Repeating a biased procedure does not fix bias.

* *To improve VALIDITY, we need to find a more valid measure. * Repetition doesn’t help with validity either.
Simpson’s Paradox- an observed association b/ 2 variables can be misleading or even reverse direction when there’s another variable that interacts strongly with both variables.*look out for lurking variables-bc of SP they can completely change Scatterplots:

1. Describe the form; linear or nonlinear
2. Describe the direction; positive (up to the R), negative (down to the R) 3. Describe the strength; strong (bunched up around the line) or weak (can’t distinguished a line, scattered all over) 4. Outliers; deviation from the overall pattern, can fall on or off line Regression Output: Y=A(m)X+B

Beers: X
BAC: Y
Coefficient: -0.0127006 = B
Coefficient: 0.0179638 = M

BAC= -0.0127006 + 0.0179638

Probability:
60% of the customers at a gas station fill up their takes, which is more likely? A) Between 58% and 62% of the next 100 customers fill up their tanks B) Between 58% and 62% of the next 1000 customers fill up their tanks Tossing a fair coin, which is more likely?

A) Between 40 & 60 of the next 100 tosses are heads. B) Between 90 & 110 of the next 200 tosses are heads. Thirty percent of the cars passing through a turnpike toll area pay with a state issued “Fast Pass”. Which is more likely? A) That between 25% & 35% of the next 100 cars pay with fast pass. B) That between 25% & 35% of the next 800 cars pay with fast pass. Thirty percent of the cars passing through a turnpike toll area pay with a state issued Fast Pass. Which is more likely? A) That exactly 30 of the next 100 care pay with fast pass B) That exactly 240 of the next 800 cars pay with fast pass. The Law of Averages State that:

1. Averages or proportions are likely to be more stable with more trials. 2. Sums or counts are likely to be more variable when there are more trials. *This does not happen by compensation for a bad run of luck since independent trials have no memory.

Which is more likely?
A) The avg amnt spent by the next 10 couples is between 40 & \$60. B) The avg amnt spent by the next 40 couples is between \$40 & \$60. more trials Which is more likely
A) The avg amnt spent by the next 10 couples is over \$60 less trials B) The avg amnt spent by the next 40 couples is over \$60

A sampling distribution tells the distribution of a statistic. It tells us what values the statistic can take on if different samples are chosen, and how often the statistic would take on each value if many samples were taken.

Determining the hypothesis:
* The null hypothesis is always the “equals” statement. Ho: µ=___ or Ho: p=___ * What would we like to conclude? Ha: µ > ___ or Ha: µ ≠ ___ Always the same numbers for both hypotheses.
Alternative hypothesis is what we want to know based on question.

* Researchers studied the behavior of drivers on a rural interstate highway in Maryland where the speed limit was 55 miles per hour. They found that 5690 out of 12,931 vehicles were exceeding the speed limit. Is this good evidence that (in this location) fewer than half of all drivers are speeding?

We have enough evidence to conclude that fewer than half of all drivers are speeding. * A tobacco company claims that the amount of nicotine in its cigarettes is normally distributed with an average of 2.2 mg with a standard deviation of 0.3 mg. A random sample of 100 cigarettes had a mean nicotine content of 3.1 mg. Perform an appropriate hypothesis test.

We have enough evidence to conclude that the average nicotine content is greater than 2.2 mg.

A recent Gallup poll found that 55% of Americans worry a great deal or a fair amount about global warming. Results for this Gallup poll are based on telephone interviews conducted March 8-11, 2012, with a random sample of 1,024 adults, aged 18 and older, living in all 50 U.S. states and the District of Columbia. One can say with 95% confidence that the maximum margin of sampling error is ±4 percentage points. •Population: American adults •Parameter: proportion of all Americans that worry a great deal or a fair amount about global warming •Sample: 1024 American adults •Statistic: 55% •Confidence Level: 95% confidence •MoE: +/- 4%

•What would happen to the margin of error if we decided to use… –90%
confidence? decrease
–99% confidence? increase
•What would happen to the margin of error if we sampled… –500 American adults? increase

* The Bureau of Labor Statistics uses 90% confidence intervals in presenting unemployment results from the monthly Current Population Survey. The January 2008 survey interviewed an SRS 134,163 people. Of these, 65,481 were employed and 3293 were unemployed. Give a 90% confidence interval for the proportion of those surveyed who were unemployed. (0.0238, 0.0252)

* Suppose that the return of the S&P 500 each year, measured in percentage gained/lost is normally distributed with a standard deviation of 5%. Suppose that a random sample of 36 years averaged 8%. Test to see if the average return of the S&P 500 is more than 7%, on average.

We do not have enough evidence to conclude that the average return is more than 7%. •Give a 95% confidence interval.
(6.4, 9.6)

* The 18th century French naturalist Count Buffon tossed a coin 4040 times. He got 2048 heads. Give a 95% confidence interval for the probability that Buffon’s coin lands heads up. (.492, .522)

•Are you confident that this probability is not 1/2? No, the confidence interval contains 1/2

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