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Data Classification and Descriptive Statistics
Probability, Discrete and Normal Probability
Confidence Intervals and Hypothesis Testing
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Mystery Box
100
Identify the data as Qualitative or Quantitative, and then Nominal, Ordinal, Interval or Ratio: The temperatures of 25 refridgerators
1) Quantitative 2) Interval
100
A single six-sided die is rolled. a.) What is the sample space? b.) Find the probability of rolling a number less than 6
a.) {1,2,3,4,5,6} b.) 5/6
100
Find the critical value zc that corresponds to a 94% confidence level. Give a positive value.
Invnorm(.03) = -1.88 So the critical value is 1.88.
100
Which correlation coefficient below is stronger?........... -0.85 or 0.5?
-0.85.......... Take the absolute value; the larger number is stronger.
100
A probability P(x) is always between what two values?
0 and 1
200
Send someone to the board to draw: Draw a bar graph that is skewed right. Identify where the Median, Mean, and Mode are.
Tail Determines where the skew is (Tail on Right.) Mode is always tallest bar, Median in the middle, Mean closest to the tail.
200
How many different codes of 4 digits are possible if the first digit must be a 2 or 5 and if the code may not end in 0?
(2)*(10)*(10)*(9)= 1800 codes possible
200
A random sample of 40 students has a mean annual earnings of $3120. Assume the population standard deviation is $677. Construct the confidence interval for the population mean, ยต if c = 0.95.
Population standard deviation is known, so use normal distribution (z). zinterval(677,3120,40,.95) = (2910.2,3329.8)
200
Coordinate data for Linear Regression! See #2 in Attached tables.
a) Use LinReg to get y = -2.75x + 91.14 b) y = -2.75*7 + 91.14 = 71.89
200
Find the 10% trimmed mean of the data: 15, 20, 25, 25, 30, 40, 5
23
300
In the sample below, identify the Mean, Median, and Standard Deviation: 73 76 69 73 66 74 (Use 2 decimal places.)
Put Data in L1, use 1-VAR STats Mean = 71.83 Sample Standard Deviation = 3.66 Median = 73 (scroll down to get median from 1-Var Stats.)
300
Assume that adults have IQ scores that are normally distributed with a mean of 100 and standard deviation of 20. Find the 15th percentile. (Round to nearest hundredth)
79.20
300
A manufacturer of golf equipment wishes to estimate the number of left-handed golfers. How large a sample is needed in order to be 95% confident that the sample proportion will not differ from the true proportion by more than 4%? A previous study indicates that the proportion of left-handed golfers is 9%.
At 95% confidence, z-critical = 1.96. p-hat = .09, q-hat = .91, E = .04. Use n = (.09)(.91)*(1.96/.04)^2 = 196.6 Round up, need a sample of at least 197 people.
300
Contingency Table! See Tables, #3.
a) Ho: Blood pressure and exercise are independent...(claim).......... Ha: Blood pressure and exercise are dependent......... b) Use chi-square distribution to find critical value. D.F. = (2-1)(3-1) = 2. chi-square critical = 9.210. (Always a right tail test.) Enter data in Matrix A and do chi-square test. Test statistic = 3.473.............. c) Fail to Reject. There is not enough evidence to reject claim that blood pressure is independent of exercise.
300
What does NormalCDF do on your calculator? What does InvNorm do?
Put z values into NormalCDF to calculate a probability or area under the curve................ Put an area into InvNorm to calculate the z value.
400
The population mean for a group of test scores is 78 and the standard deviation is 4....... Is a test score of 68 normal, unusual, or very unusual? Calculate the z-score and explain why.
z-score = -2.5 This means that the score of 68 is more than 2 standard deviations away from the mean. It is unusual.
400
You observe the gender of the next 100 babies born at a local hospital. You count the number of girls born. a) Identify the values of n, p, and q, and list the possible values of the random variable x. b) What is the probability of getting exactly 42 girls? c) What is the probability of getting more than or equal to 60 girls? (round to 3 decimals places.)
a) n=100, p=1/2, q=1/2, x={0,1,2,3,.....100} b) binompdf(100,0.5,42) = 0.022 c) Less than 60 = binomcdf(100,0.5,59) So 60 or more girls = 1 - binomcdf(100,0.5,59) = 0.028
400
A trucking firm suspects that the mean life of a certain tire it uses is less than 35,000 miles. To check this, they sample 18 tires and get a mean lifetime of 34,350 miles with a sample standard deviation of 1200 miles. Use level of significance 0.05 to test the claim. a) List hypothesis b) Give level of significance and test statistic c) Interpret
a) Ho: u = 35,000 H1: u less than 35,000 (claim) b) Use t-distribution table to find critical value -1.740 Use t-test(35000,34350,1200,18,
400
Coordinate Data! See #4 in Tables
a) Ho: p = 0 (no correlation) Ha: p not equal to 0 (significant correlation) ...Claim........ b) Critical value uses t-distribution table with degrees freedom = n-2 so D.F = 8. Always a two-tail test for correlation coefficient............. t-critical = -2.306, 2.306.................. For test statistic, put data in L1, L2. Do LinRegTtest. t = 4.497................. c) Reject Ho. There is enough evidence to support claim that there is a significant correlation.
400
Calculate the area under the curve to the right of z = 1.2 (round to 4 decimal places.)
NormalCDF(1.2, 10000) = 0.1151
500
The mean SAT verbal score is 480, with a standard deviation of 40......................................................... a) Use the Empirical Rule to determine what percent of the scores lie between 480 and 520...................... b) Assuming 250 individuals took the test, how many of these would have a score between 480 and 520?
a) 34% b) 85
500
Salaries of elementary school teachers are normally distributed with a mean of $37,000 and a standard deviation of $5000. a) What is the probability that a teacher earns between $40000 and $60000? b) What is the cutoff salary for teachers in the bottom 25%?
a) Normalcdf(40000,60000,37000,5000) = 0.2743 b) $33627.55
500
In a survey of 300 T.V. viewers, 120 said they watch network news programs. Find a 95% confidence interval for the population proportion of TV viewers that watch network news programs.
0.345 < p < 0.455 There is 95% confidence that the population proportion of TV viewers that watch network news programs is between 34.5% and 45.5%
500
Table Data! See #5 in Tables.
a) Ho: u1 = u2 = u3 = u4 (claim)............ Ha: At least one mean is different from the others.......... b) Use F-distribution to find critical value with D.F numerator = 4-1 = 3 and D.F. denominator = 24 - 4 = 20. F critical = 4.94. (Always right tail test.) Put data in L1,L2,L3,L4. Do ANOVA. Test statistic = 8.357........... c) Reject Ho. There is enough evidence to reject claim that fertilizer makes no difference in mean number of raspberries.
500
In a class of 50 students, 18 take chorus, 26 take band, and 2 take both chorus and band. How many students in the class are not enrolled in either chorus or band?
8 students are not enrolled in either chorus or band.