Statistics Final Exam Problems and formulas used
In a random sample of 16 residents of the state of Washington, the mean waste recycled per person per day was 2.8 pounds with a standard deviation of 0.24 pounds. Determine the 80% confidence interval for the mean waste recycled per person per day for the population of Washington. Assume the population is approximately normal. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Construct 80% confidence Interval
1-0.8=0.2/2=0.1 2nd Vars invT area:0.10 df= n-1= 15 t= 1.341 STAT, Tests TInterval -> Stats n=16 x=2.8 s=0.24 c-level:0.80 (2.7196, 2.8804)
Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places. P(X≥9), n=11, p=0.6
P(X≥9)= 1-(X<8) binomialcdf to calculate (x<8)
Suppose 55% of politicians are lawyers. If a random sample of size 633 is selected, what is the probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by greater than 5%? Round your answer to four decimal places.
P=55% p hat=5% σ of p hat= √[P(1-P)/n] z of p hat= (P-p hat)-P/σ of p hat normalcdf multiply by 2
x:-4, -3, -2, -1, 0 P(X=x): 0.3, 0.1, 0.1, 0.3, 0.2 Find E(x) Find Variance Find Standard deviation Find value of P(X≤-3) Find Value of P(X<-2)
u=E(X=x)=∑[xi · P(X=xi)] σ^2forp(X=x)= ∑[xi^2 · P(X=xi)]-u^2 σ=√σ^2 P(X≤-3)= P(X=-3)+P(X=-4) P(X<-2)= P(X=-3)+P(X=-4)
Variance of data set
variance= ∑(xi-u)^2 ---------- n
The electric cooperative needs to know the mean household usage of electricity by its non-commercial customers in kWh per day. Assume that the population standard deviation is 2.4 kWh. The mean electricity usage per family was found to be 1717 kWh per day for a sample of 944 families. Construct the 80% confidence interval for the mean usage of electricity. Round your answers to one decimal place.
x= 17 σ=2.4 n=944 C=0.80 STAT, Tests 7: ZInterval -> Stats σ:2.4 x:17 n: 944 C-level: 0.8 (16.9, 17.1)
Suppose the horses in a large stable have a mean weight of 1467lbs, and a standard deviation of 93lbs. What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable? Round your answer to four decimal places.
"mean weight of sample" z= x-u/ (σ/√n) normalcdf multiply by 2 and subtract from 1
A quality control inspector has drawn a sample of 11 light bulbs from a recent production lot. If the number of defective bulbs is 2 or more, the lot fails inspection. Suppose 20% of the bulbs in the lot are defective. What is the probability that the lot will fail inspection? Round your answer to four decimal places.
n=11 P=0.20 p hat= 2 P(x≥2)= 1-P(x<1) binomialcdf
A parenting magazine reports that the average amount of wireless data used by teenagers each month is 10Gb. For her science fair project, Ella sets out to prove the magazine wrong. She claims that the mean among teenagers in her area is less than reported. Ella collects information from a simple random sample of 16 teenagers at her high school, and calculates a mean of 9.1Gb per month with a standard deviation of 1.3Gb per month. Assume that the population distribution is approximately normal. Test Ella's claim at the 0.05 level of significance. state null and alt find test statistic and draw conclusion
n=16 sigma=1.3 x=1.9 STAT, Test Z-Test u1=10 sigma=1.3 x=9.1 n=16 u: >u1 z=-2.769 p=0.0028 Since alpha is 0.05 and p is 0.002, we reject the null, there is sufficient evidence
There are 14 people in an office with 5 different phone lines. If all the lines begin to ring at once, how many groups of 5 people can answer these lines?
nCr= n! / r!(n-r)! 14 MATH, PRB, nCr, enter, 5
A coordinator will select 4 songs from a list of 10 songs to compose an event's musical entertainment lineup. How many different lineups are possible?
nPr= n!/(n-r)! 10 MATH, PRB, nPr, enter, 4
Suppose that shoe sizes of American women have a bell-shaped distribution with a mean of 8.36 and a standard deviation of 1.49. Using the empirical rule, what percentage of American women have shoe sizes that are no more than 3.89? Please do not round your answer.
z= x-u/ σ =-3 1SD= 68% 2SD= 95% 3SD= 99.7% We want to find the area to the left of 3.89 because we want percentage of women that shoe's are NO MORE THAN 3.89. Since we have 3SD, 99.7% of our data is already accounted for, so we are left with an area of 1-0.997= 0.003. This area is the sum of both tails of the ND curve, and we only want the left tail. 0.003/2 = 0.0015, or 0.15%.
The number of hours per week that the television is turned on is determined for each family in a sample. The mean of the data is 30 hours and the median is 26.2 hours. Twenty-four of the families in the sample turned on the television for 15 hours or less for the week. The 12th percentile of the data is 15 hours. Based on the given information, determine if the following statement is true or false. The 55th percentile is less than 25 hours. How any families are in the sample? Based on the given information, determine if the following statement is true or false. Approximately 100 families turned on their televisions for less than 26.2 hours. What is the value of the 50th percentile? Based on the given information, determine if the following statement is true or false. The first quartile is less than 15 hours.
Since the median is the 50th percentile, and is equal to 26.2, the 55th percentile would be greater than 26.2. the statement is false. total number of data values= (number of data values related to x / percentile of x) (100) total families = (24/12)(100) =200 Because half of 200 is 100, and the median, or half of the data watches tv for less than 26.2 hours, the statement is true. The 50th percentile is 26.2. Since the 12th percentile is 15hrs, the first quartile, 25th percentile must be greater than 15hrs. statement is false.
Suppose that the walking step lengths of adult males are normally distributed with a mean of 2.5 feet and a standard deviation of 0.5 feet. A sample of 43 men's step lengths is taken. Find the probability that an individual man's step length is less than 2.1 feet. Round your answer to 4 decimal places, if necessary. Find the probability that the mean of the sample taken is less than 2.1 feet. Round your answer to 4 decimal places, if necessary.
individual: z= x-u/ σ normalcdf sample: z= x-u/ (σ/√n) normalcdf
finding position of a quartile formula
l = n (P/100) ex finding first quartile: 26 (25/100)=6.5 the first quartile values is the 7th value of the data.
For the following type of data set, would you be more interested in looking at the mean, median, or mode? State your reasoning. The salary of actors on TV
median; The salaries of actors are quantitative data with outliers.
σ is known; n=13; population normally distributed.
Even though n<30 is a t distribution quality, σ is know so it is a Z distribution. Normal Z Distribution
An SAT prep course claims to increase student scores by more than 60 points, on average. To test this claim, 9 students who have previously taken the SAT are randomly chosen to take the prep course. Their SAT scores before and after completing the prep course are listed in the following table. Test the claim at the 0.01 level of significance assuming that the population distribution of the paired differences is approximately normal. Let scores before completing the prep course be Population 1 and let scores after completing the prep course be Population 2. Before Prep: 1040, 1200, 1270, 1210, 1330, 1320, 1150, 1220, 1010 After Prep: 1330, 1240, 1290, 1310, 1380, 1460, 1240, 1500, 1280 State Null and Alt Compute Test statistic and draw conclusions
H0: ud=60 Ha: ud > 60 STAT, Edit input X1 values into L1 input X2 values into L2 input X2-X1 into L3 STAT, Tests T-Test-> Data ud=60 List=L3 Freq=1 u>ud enter t=2.2569 p=0.0269 since alpha is 0.01, and p is 0.02, we fail to reject the null; insufficient evidence.
A researcher is interested in exploring the relationship between calcium intake and weight loss. Two different groups, each with 29 dieters, are chosen for the study. Group A is required to follow a specific diet and exercise regimen, and also take a 500-mg supplement of calcium each day. Group B is required to follow the same diet and exercise regimen, but with no supplemental calcium. After six months on the program, the members of Group A had lost a mean of 10.4 pounds with a standard deviation of 1.2pounds. The members of Group B had lost a mean of 10.6 pounds with a standard deviation of 1.9pounds during the same time period. Assume that the population variances are not the same. Construct a 90% confidence interval to estimate the true difference between the mean amounts of weight lost by dieters who supplement with calcium and those who do not. Let Population 1 be the amount of weight lost by Group A, who took a 500-mg supplement of calcium each day, and let Population 2 be the amount of weight lost by Group B, who did not take a calcium supplement. Round the endpoints of the interval to one decimal place, if necessary.
NOT POOLED STAT, Test 2-SampleTInterval x1: 10.4 Sx1: 1.2 n1: 29 x2: 10.6 Sx2: 1.9 n2: 29 C-Level: 0.90 Pooled: No (-0.9001, 0.50012)
A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective. Suppose a sample of 766 floppy disks is drawn. Of these disks, 84 were defective. Using the data, construct the 90% confidence interval for the population proportion of disks which are defective. Round your answers to three decimal places. Use this interval when you have a proportion based on data from sample, and you want to estimate the true proportion in the population.
STAT, Tests 1-PropZInt -> Stats x=84 n=766 C-Level=0.9 (0.091, 0.129) p hat: 0.109
Steve believes that his wife's cell phone battery does not last as long as his cell phone battery. On twelve different occasions, he measured the length of time his cell phone battery lasted and calculated that the mean was 19.6 hours with a standard deviation of 7.3 hours. He measured the length of time his wife's cell phone battery lasted on eight different occasions and calculated a mean of 18.1 hours with a standard deviation of 4.7 hours. Assume that the population variances are the same. Let Population 1 be the battery life of Steve's cell phone and Population 2 be the battery life of his wife's cell phone. Not looking specifically to compare means n<30 Step 1 of 2: Construct a 95% confidence interval for the true difference in mean battery life between Steve's cell phone and his wife's. Round the endpoints of the interval to one decimal place, if necessary.
STAT, Tests 2-SampleTInterval -> Stats x1: 19.6 Sx1: 7.3 n1: 12 x2: 18.1 Sx2: 4.7 n2: 8 C-Level: 0.95 Pooled: Yes (-4.7, 7.7) Contains 0, data does not provide evidence.
Aaliyah wants to know if there is a difference between the proportions of customers who just order water to drink at two popular restaurants in town. She collected the data in the following table. Construct a 95% confidence interval for the true difference between the proportions of customers who just order water to drink at Restaurants A and B. Let Population 1 be the customers of Restaurant A and Population 2 be the customers of Restaurant B. Round the endpoints of the interval to three decimal places, if necessary. Restaurant A: Water: 45 Other: 81 Restaurant B: Water: 83 Other:58
STAT, Tests B: 2-PropZInterval x1= 45 n1= 126 x2= 83 n2= 141 C-Level= 0.90 (-0.3294, -0.1337)
A professor is concerned that the two sections of college algebra that he teaches are not performing at the same level. To test his claim, he looks at the mean exam score for a random sample of students from each of his classes. In Class 1, the mean exam score for 17 students is 80.3 with a standard deviation of 1.6. In Class 2, the mean exam score for 19 students is 82.1 with a standard deviation of 2.9. Test the professor's claim at the 0.01 level of significance. Assume that both populations are approximately normal and that the population variances are equal. Let Class 1 be Population 1 and let Class 2 be Population 2. Used to test whether the unknown population means of two groups are equal or not. State null and alt Compute test stat and draw conclusion
Data is pooled because variances are equal. H0: u1=u2 Ha: u1 does not =u2 2-SamptTest is used because we are interested in the mean. STAT, Test, 2-SamptTest -> Stats x1: 80.3 Sx1: 1.6 n1: 17 x2: 82.1 Sx2: 2.9 n2: 19 u1 does not =u2 Pooled: Yes z= -2.2668 p= 0.023 Since alpha is 0.01, and p is 0.02, p is greater than alpha and we fail to reject the null with insufficient evidence.
Problem given table of totals at the bottom and the side, and there are categories on the bottom and the side. How do you calculate E(X) Compute Value of Test Statistic and draw conclusion
E(X)= [(row total)(column total)/n] 2nd X^-1 (matrix) Edit, enter Adjust matrix to table size and enter given values Stat, Test C: x^2-Test enter Test stat= x^2= 8.607 p= 0.4743 Since alpha is 0.1, and P is 0.5, P is greater than alpha so we fail to reject the null with insufficient evidence.
A local fast-food restaurant serves buffalo wings. The restaurant's managers notice that they normally sell the following proportions of flavors for their wings: 20% Spicy Garlic, 20% Classic Medium, 10%Teriyaki, 20% Hot BBQ, and 30% Asian Zing. After running a campaign to promote their nontraditional specialty wings, they want to know if the campaign has made an impact. The results after 1010 days are listed in the following table. Is there sufficient evidence at the 0.005 level of significance to say that the promotional campaign has made any difference in the proportions of flavors sold? BUFFALO WING SALARIES Spicy Garlic: 113 sold Classic Medium: 114 sold Teriyaki: 75 sold Hot BBQ: 122 sold Asian Zing: 142 sold What's the expected value for Teriyaki? What's the value of the test statistic?
E(xi)=nPi E(xi)= 566(0.10) STAT, Edit L1: X values L2: E(xi) values STAT, TESTS D: x^2 GOF-Test observed: L1 Expected: L2 df: n-1
An engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 270 engines and the mean pressure was 6.0 pounds/square inch (psi). Assume the population standard deviation is 0.6. If the valve was designed to produce a mean pressure of 6.1 psi, is there sufficient evidence at the 0.05 level that the valve does not perform to the specifications? State Null and Alt Hypothesis Calculate test statistic and draw conclusions
Given: n, sigma, x H0: u1=6.1 Ha: u2 does not = 6.1 STAT, Test Z-Test-> Stats u=6.1 sigma: 0.6 x=6 n=270 u1 does not = u2 enter z= -2.74 p=0.006 Since alpha is 0.05, and p is 0.006, we reject the null with sufficient evidence.
University officials hope that changes they have made have improved the retention rate. Last year, a sample of 2028 freshmen showed that 1628 returned as sophomores. This year, 1689 of 2059 freshmen sampled returned as sophomores. Determine if there is sufficient evidence at the 0.05 level to say that the retention rate has improved. Let last year's freshmen be Population 1 and let this year's freshmen be Population 2. State Null and Alt. Hypotheses Compute test statistic and draw conclusion. given only n1, x1, n2, x2, and level of confidence
H0: p1=p2 Ha: p1<p2 sigma unknown u is unknown STAT, Tests 2-PropZTest x1: 1628 n1: 2028 x2: 1689 n2: 2059 p1: <p2 enter Test statistic= z= -1.43 p=0.07 Since alpha =0.05, and p=0.07, p is greater than alpha so we fail to reject the null; insufficient evidence.
Fran is training for her first marathon, and she wants to know if there is a significant difference between the mean number of miles run each week by group runners and individual runners who are training for marathons. She interviews 34 randomly selected people who train in groups, and finds that they run a mean of 44.2 miles per week. Assume that the population standard deviation for group runners is known to be 2.5 miles per week. She also interviews a random sample of 49 people who train on their own and finds that they run a mean of 45.1 miles per week. Assume that the population standard deviation for people who run by themselves is 1.1 miles per week. Test the claim at the 0.01 level of significance. Let group runners training for marathons be Population 1 and let individual runners training for marathons be Population 2. State Null and Alt hypotheses Compute test statistic and draw conclusion
H0: u1=u2 Ha: u1 doesn't =u2 STAT, Tests 2-SampZTest Stats n1=34 u1=44.2 σ1=2.5 n2=49 u2=45.1 σ2=1.1 u1 does not equal u2 enter z, test statistic=-1.9709 p=0.05 Since alpha=0.01, and p=0.05, p is greater than alpha so we fail to reject the null; insufficient evidence.
To test the effect of a physical fitness course on one's physical ability, the number of sit-ups that a person could do in one minute, both before and after the course, was recorded. Ten individuals are randomly selected to participate in the course. The results are displayed in the following table. Using this data, find the 99% confidence interval for the true difference in the number of sit-ups each person can do before and after the course. Assume that the numbers of sit-ups are normally distributed for the population both before and after completing the course. Sit ups before: 30, 35, 52, 34, 50, 27, 42, 41, 41, 25 Sit ups after: 42, 46, 58, 44, 56, 45, 57, 48, 46, 37
STAT, Edit Enter x1 into L1 Enter x2 into L2 Enter x2-x1 into L3 STAT, Tests 8: TInterval ->Data List: L3 Freq: 0.99 Interval: (5.8184, 14.582) Point Estimate: x: 10.2 Standard Deviation: Sx: 4.263 Margin of error used in constructing confidence interval: (UB-LB) / 2
