Math 13 Statistics
A prospective jury pool has 51 men and 42 women. Suppose that jurors were chosen at random. Find the probability that the 12-person jury had only men. Enter your answer as a decimal, using 3 decimal places
0 or 0.0004
A biologist found that Leatherback Sea Turtles have a mean weight of 925 pounds with standard deviation 56 pounds . b) Find the probability that the mean weight of 7 Leatherback would be over 962 pounds. Round your answer to 3 decimal places
0.04
For a normal distribution, it is known that the mean is zero and the standard deviation is 1. a) Find the probability that a z score is above 1.71. Round your answer to 3 decimal places.
0.044
A biologist found that Leatherback Sea Turtles have a mean weight of 925 pounds with standard deviation 56 pounds . a) Find the probability that the weight of a Leatherback would be over 1000 pounds. Round your answer to 3 decimal places.
0.09
In China, 48.5% of the people are female. Suppose that 10 people in China were chosen at random. Find the probability that more than 6 of those people are female. Enter your answer as a decimal, using 3 decimal places.
0.148
Cards are drawn from a well-shuffled deck of 52. Find the probability for each event below. b) Drawing a Jack or a seven. Enter your answer as a decimal, using 3 decimal places.
0.154
In China, 48.5% of the people are female. Suppose that 10 people in China were chosen at random. Find the probability that 6 of those people are female. Enter your answer as a decimal, using 3 decimal places.
0.192
A biologist found that Leatherback Sea Turtles have a mean weight of 925 pounds with standard deviation 56 pounds . c) Find the probability that the weight of a Leatherback would be between 800 pounds and 880 pounds. Round your answer to 3 decimal places.
0.199
People were asked if they have cable TV at home. The people were divided by gender and the results are listed below. Yes No Male 47 57 Female 52 42 If two people are chosen, find the probability that both have cable TV at home. Enter your answer as a decimal, using 3 decimal places.
0.249
Cards are drawn from a well-shuffled deck of 52. Find the probability for each event below. a) Drawing a King or a heart. Enter your answer as a decimal, using 3 decimal places
0.308
A lottery ticket says that the chances of winning are 1 in 5.5. Suppose you buy 8 of these lottery tickets. Find the probability that exactly one of them will be a winner.
0.357
The data provided is from houses for sale in Salinas. Prices are in the $1,000. 1) Test for a linear correlation between price and square feet in South Salinas. Provide the positive critical value.
0.396
People were asked if they have cable TV at home. The people were divided by gender and the results are listed below. Yes No Male 47 57 Female 52 42 Find the probability that a person has cable TV at home. Enter your answer as a decimal, using 3 decimal places.
0.5
People were asked if they have cable TV at home. The people were divided by gender and the results are listed below. Yes No Male 47 57 Female 52 42 Find P(Female | Yes) Enter your answer as a decimal, using 3 decimal places.
0.525 or 0.712
People were asked if they have cable TV at home. The people were divided by gender and the results are listed below. Yes No Male 47 57 Female 52 42 Find P (Yes | Female) Enter your answer as a decimal, using 3 decimal places.
0.553 or 0.263
In China, 48.5% of the people are female. Suppose that 10 people in China were chosen at random. Find the probability that fewer than 6 of those people are female. Enter your answer as a decimal, using 3 decimal places.
0.659
The data provided is from houses for sale in Salinas. Prices are in the $1,000. 5) Test for a linear correlation between price and bedrooms in South Salinas. Enter the value you got for r.
0.67
People were asked if they have cable TV at home. The people were divided by gender and the results are listed below. Yes No Male 47 57 Female 52 42 If one person is chosen, find the probability that they would be male or not have cable TV at home. Enter your answer as a decimal, using 3 decimal places
0.737
The data provided is from houses for sale in Salinas. Prices are in the $1,000. 2) Test for a linear correlation between price and square feet in North Salinas. Enter the value you got for r.
0.806
The data provided is from houses for sale in Salinas. Prices are in the $1,000. 6) Test for a linear correlation between price and bedrooms in North Salinas. Enter the value you got for r.
0.819
The data provided is from houses for sale in Salinas. Prices are in the $1,000. 1) Test for a linear correlation between price and square feet in South Salinas. Enter the value you got for r.
0.897
For a normal distribution, it is known that the mean is zero and the standard deviation is 1. b) Find the probability that a z score is below 1.44. Round your answer to 3 decimal places.
0.925
A biologist found that Leatherback Sea Turtles have a mean weight of 925 pounds with standard deviation 56 pounds . d) How heavy would a Leatherback need to be in order to be in the top 7%?
1,007.88=1,008
Given the probability distribution below, find the mean. x P(x) 0 0.125 1 0.375 2 0.375 3 0.125
1.5
Billy-Bob thinks that Pepsi is cheating him! He bought 36 cans of Pepsi and found the mean amount of soda to be 11.79 ounces with a standard deviation of 0.21. Make a 95% confidence interval for the mean amount of soda in each can. Is Pepsi filling the cans with less than 12 ounces of soda?
11.719 < μ < 11.861 Yes, Pepsi is cheating Billy Bob and we can conclude that the average amount of soda in Pepsi cans is significantly less than 12 oz of soda.
Male-166,205,194,144,168,157,198,155,178,206,166,166,149,159,194,158 Female-134,154,102,122,109,161,122,133,135,141,115,123,141,115,124 Find the midrange for the women's weights.
131.5
Units Done: 49, 50, 60, 25, 35, 37, 56, 6, 75, 140, 50, 42, 30, 6, 24, 40, 30, 42,43,94, 70, 39, 35, 50, 30, 15, 34, 25, 38, 39, 68, 8, 20, 26, 40 1. d) Create a stem and leaf plot for this data.
14-0 13 12 11 10 9-4 8 7-0,5 6-0,8 5-0,0,6 4-0,0,2,2,3,9 3-0,0,0,4,5,5,7,8,9,9 2-0,4,5,5,6 1-5 0-6,6,8
Male-166,205,194,144,168,157,198,155,178,206,166,166,149,159,194,158 Female-134,154,102,122,109,161,122,133,135,141,115,123,141,115,124 For the women's weights, find the score at the 80th percentile.
141
Male-166,205,194,144,168,157,198,155,178,206,166,166,149,159,194,158 Female-134,154,102,122,109,161,122,133,135,141,115,123,141,115,124 Find the standard deviation for the women's weights. Round to 3 decimal places.
16.241
Male-166,205,194,144,168,157,198,155,178,206,166,166,149,159,194,158 Female-134,154,102,122,109,161,122,133,135,141,115,123,141,115,124 Find the median weight for men.
166
Male-166,205,194,144,168,157,198,155,178,206,166,166,149,159,194,158 Female-134,154,102,122,109,161,122,133,135,141,115,123,141,115,124 Find the mode for the men's weights.
166
Male-166,205,194,144,168,157,198,155,178,206,166,166,149,159,194,158 Female-134,154,102,122,109,161,122,133,135,141,115,123,141,115,124 Find the mean weight for men. Round to 3 decimal places.
172.688
Male-166,205,194,144,168,157,198,155,178,206,166,166,149,159,194,158 Female-134,154,102,122,109,161,122,133,135,141,115,123,141,115,124 For the men's weights, find the score at the 80th percentile.
194
The data below lists states poverty rate, gun deaths per 100,000 people, high school graduation rate and the unemployment rate. Use a 2x2 contingency table for questions 1-3. Be sure to find the test statistic and the critical values. Use the median to divide each list into 2 groups.You will need to provide the 2x2 table also. 1) Test for independence between poverty rate and gun deaths. Enter the test statistic for this problem.
2
The data below lists states poverty rate, gun deaths per 100,000 people, high school graduation rate and the unemployment rate. Use a 2x2 contingency table for questions 1-3. Be sure to find the test statistic and the critical values. Use the median to divide each list into 2 groups.You will need to provide the 2x2 table also. 3) Test for independence between unemployment and gun deaths. Enter the test statistic for this problem.
2
A class has 25 students. Each day, three students are chosen to present their work at the board. In total, how many such 3 student groups are possible?
2,300
Male-166,205,194,144,168,157,198,155,178,206,166,166,149,159,194,158 Female-134,154,102,122,109,161,122,133,135,141,115,123,141,115,124 Find the standard deviation for the men's weights. Round to 3 decimal places.
20.385
The data below lists states poverty rate, gun deaths per 100,000 people, high school graduation rate and the unemployment rate. Use a 2x2 contingency table for questions 1-3. Be sure to find the test statistic and the critical values. Use the median to divide each list into 2 groups.You will need to provide the 2x2 table also. 1) Test for independence between poverty rate and gun deaths. Enter the critical value for this problem.
3.841
A couple has 10 movies in their Netflix queue. They decide to randomly choose 2 movies from the queue. How many possible 2-movie selections are there?
45
In a poll, 82.4% of 562 people said that second hand smoke is annoying. How many people reported that second hand smoke is annoying?
463
In China, 48.5% of the people are female. Suppose that 10 people in China were chosen at random. How many of the 10 people would you expect to be female?
5
The data provided is from houses for sale in Salinas. Prices are in the $1,000. 4) Use the regression equation to predict the price of a 2000 square foot house in North Salinas. Your answer should look like 123.456
564.722
Male-166,205,194,144,168,157,198,155,178,206,166,166,149,159,194,158 Female-134,154,102,122,109,161,122,133,135,141,115,123,141,115,124 Find the range for the women's weights.
59
The data below lists states poverty rate, gun deaths per 100,000 people, high school graduation rate and the unemployment rate. Use a 2x2 contingency table for questions 1-3. Be sure to find the test statistic and the critical values. Use the median to divide each list into 2 groups.You will need to provide the 2x2 table also. 2) Test for independence between high school graduation rate and gun deaths. Enter the test statistic for this problem.
6.48
Below are the grades on an exam. b) Find the average score for the people that did not pass the exam. Round the answer to 3 decimals. 9-0 0 1 2 3 3 5 7 7 9 9 8-1 2 3 3 4 5 5 5 6 6 7 7 8 9 9 7-0 0 1 2 3 4 5 7 7 8 9 6-2 5 6 9 5-3 5
61.667 61.7
Male-166,205,194,144,168,157,198,155,178,206,166,166,149,159,194,158 Female-134,154,102,122,109,161,122,133,135,141,115,123,141,115,124 Find the range for the men's weights.
62
The data provided is from houses for sale in Salinas. Prices are in the $1,000. 3) Use the regression equation to predict the price of a 2000 square foot house in South Salinas. Your answer should look like 123.456
634.368
There were 47 people that took the last statistics exam. The standard deviation was 9.8 points. Create a 98% confidence interval for the standard deviation. Does is seem possible that the overall standard deviation is 15?
7.877 < p < 12.873 No, 15 isn't in the confidence interval so it proves it doesn't seem possible.
Below are the grades on an exam. a) What percent of the class got a C or better? 9-0 0 1 2 3 3 5 7 7 9 9 8-1 2 3 3 4 5 5 5 6 6 7 7 8 9 9 7-0 0 1 2 3 4 5 7 7 8 9 6-2 5 6 9 5-3 5
86
A 95% confidence interval is found to be 84.1<μ<96.3 . Find the best point estimate for μ.
90.2
Scores on an exam have a normal distribution with a mean of 80 and a standard deviation of 12. Find the score needed to be in the top 10% of the class.
95.36
What is a voluntary response? Is a good idea to use data collected in such a way? Explain.
A voluntary response sample is a sample in which the subjects themselves decide whether to be included in the study. A voluntary response sample is generally not suitable for a statistical study because the sample may have a bias resulting from participation by those with a special interest in the topic being studied so it is not a good idea.
1) Smokers: Out of 485 smokers, 238 have been divorced. Non-smokers: Out of 1184 non-smokers, 374 have been divorced. a) Find the 95% confidence interval for the proportion of smokers that have been divorced. Put your answer in the form: b)Find the 95% confidence interval for the proportion of non-smokers that have been divorced. Put your answer in the form: c)What do the results above indicate about smoking and divorce?
A) 0.446 < p < 0.535 B) 0.289 < p < 0.342 C) A greater proportion of smokers have been divorced, when compared with non smokers.
A science journal reported on experiments done on 5 sets of mice. There were 20 mice in each set. The report said that the percentage of success for each of the 5 sets of mice were: 53%, 58%, 63%, 46%, and 67%. What is wrong with these percentages?
All of these percentages are wrong as we can't have .2, .4, or.6 of a mouse which makes it absurd. Let me explain: 53% of 20 mice (.53)(20)= 10.6 (.58)(20)= 11.6 (.63)(20)= 12.6 (.46)(20)= 9.2 (.67)(20)= 13.4 So, the actual numbers for each percentage listed are not whole numbers.
The data below was collected from Math 13 students. They were asked how many units they were taking this semester and to rate their math anxiety level, on a scale from 0 to 10. For each hypothesis test, provide the claim, critical values, test statistic and conclusion. 1) Test the claim that men and women are taking the same number of units, on average.
Answer in Picture
The data below was collected from Math 13 students. They were asked how many units they were taking this semester and to rate their math anxiety level, on a scale from 0 to 10. For each hypothesis test, provide the claim, critical values, test statistic and conclusion. 2) Test the claim that men and women have the same level of math anxiety, on average.
Answer in Picture
The data below was collected from Math 13 students. They were asked how many units they were taking this semester and to rate their math anxiety level, on a scale from 0 to 10. For each hypothesis test, provide the claim, critical values, test statistic and conclusion. 3) Test the claim that the proportion of people taking over 12 units is the same for men and women.
Answer in Picture
The data below was collected from Math 13 students. They were asked how many units they were taking this semester and to rate their math anxiety level, on a scale from 0 to 10. For each hypothesis test, provide the claim, critical values, test statistic and conclusion. 4) Test the claim that the standard deviation for the number of units taken this semester is the same for men and women.
Answer in Picture
The data below was collected from Math 13 students. They were asked how many units they were taking this semester and to rate their math anxiety level, on a scale from 0 to 10. For each hypothesis test, provide the claim, critical values, test statistic and conclusion. 5) Divide the data into two groups: those with 0-5 math anxiety and those with 6-10 math anxiety. Test the claim that the average number of units is the same for these two groups.
Answer in Picture
Divide the data into two groups: those with under a 110 IQ and those with at least 110 IQ. b) Test the claim that the mean GPA is 3.0 for those with at least 110 IQ. For each hypothesis test, provide the claim, critical values, test statistic and conclusion.
Average124.92313.698462SD10.95800.2044 Ho =Hy=3.0 H1: Hy≠3.0 T= √13(Y2-3)/s4(2) are sample mean and s*d of GPA for those with ≥ 110 IQ. t=3.698462-3/0.2044√13=12.3206 |12.3206|>2.18 At 5% level, there is sufficient evidence that for student with ≥ 110 IQ mean GPA is not 3.0
At 9AM on a Monday at Hartnell, all people in Math 13 classes are chosen to participate in a survey. What method of sampling was used to choose these people?
Cluster
At 9AM on a Monday at Hartnell, all people in Math 13 classes are chosen to participate in a survey. What method of sampling was used to choose these people?
Cluster
The data below lists states poverty rate, gun deaths per 100,000 people, high school graduation rate and the unemployment rate. Use a 2x2 contingency table for questions 1-3. Be sure to find the test statistic and the critical values. Use the median to divide each list into 2 groups.You will need to provide the 2x2 table also. 1) Test for independence between poverty rate and gun deaths. What is the conclusion?
Fail to reject the claim
The data below lists states poverty rate, gun deaths per 100,000 people, high school graduation rate and the unemployment rate. Use a 2x2 contingency table for questions 1-3. Be sure to find the test statistic and the critical values. Use the median to divide each list into 2 groups.You will need to provide the 2x2 table also. 3) Test for independence between unemployment and gun deaths. What is the conclusion?
Fail to reject the claim
Units Done: 49, 50, 60, 25, 35, 37, 56, 6, 75, 140, 50, 42, 30, 6, 24, 40, 30, 42,43,94, 70, 39, 35, 50, 30, 15, 34, 25, 38, 39, 68, 8, 20, 26, 40 Use the data for amount Units Done to answer questions a-g. b) Create a frequency table with 5 classes. It should look like this:
Given number of classes= 5 minimum value=6 maximum value 140 class size=(140-6)/5=26.8 consider class size=30 Class Frequency 1 0-30 9 2 30-60 20 3 60-90 4 4 90-120 1 5 120-150 1
The data below lists states poverty rate, gun deaths per 100,000 people, high school graduation rate and the unemployment rate. Use a 2x2 contingency table for questions 1-3. Be sure to find the test statistic and the critical values. Use the median to divide each list into 2 groups.You will need to provide the 2x2 table also. 3) Test for independence between unemployment and gun deaths. Enter the 2x2 contingency table for this problem.
Gun Deaths Low High Total Unemployment Low 15 10 25 High 10 15 25 Total 25 25 50
The data below lists states poverty rate, gun deaths per 100,000 people, high school graduation rate and the unemployment rate. Use a 2x2 contingency table for questions 1-3. Be sure to find the test statistic and the critical values. Use the median to divide each list into 2 groups.You will need to provide the 2x2 table also. 2) Test for independence between high school graduation rate and gun deaths. Enter the 2x2 contingency table for this problem.
Gun death HS low high Total low 8 17 25 high 17 8 25 Total 25 25 50
The data below lists states poverty rate, gun deaths per 100,000 people, high school graduation rate and the unemployment rate. Use a 2x2 contingency table for questions 1-3. Be sure to find the test statistic and the critical values. Use the median to divide each list into 2 groups.You will need to provide the 2x2 table also. 5) Which factor has the biggest effect on gun deaths? Explain.
High School graduation rate has the biggest effect on gun death, as we were able to reject the null hypothesis chi-square test of independence in High school graduate rate, but we were unable to reject the null hypothesis in poverty rate or with unemployment. hence gun death and High school graduate rate was dependent
Divide the data into two groups: those with under a 110 IQ and those with at least 110 IQ. b) Test the claim that the mean GPA is 3.0 for those with at least 110 IQ. For each hypothesis test, provide the claim, critical values, test statistic and conclusion.
Ho =Hy=3.0 H1: Hy≠3.0 Under H0 is the test statistic, T=√12(y1-3)/s4(1) where, y1 is the sample mean gpa and s4(1) is sample s-d of gpa for students with < 110 IQ t=2.875-3/0.3176/√12=-1.3633 |-1.3633| ≥ 2.2 ≈ t11 (0.025) At 5% level H0 we fail to reject because at 5% level the claim of mean GPA to be for 3.0 for those with < 110 IQ is valid.
The data below was collected from Math 13 students. They were asked how many units they were taking this semester and to rate their math anxiety level, on a scale from 0 to 10. Using a 95% level of confidence, test the claim that the majority of students have at least a 3.0 GPA. At the 98% confidence level, test the claim that mean GPA for freshmen is over 3.0. Test the claim that the standard deviation for IQ scores is 15. For each hypothesis test, provide the claim, critical values, test statistic and conclusion.
Ho: σ=15 Ha: σ≠ 15 s=14.0957, n=25 Χ2=(25-1)14.0957^2/15^2 Χ2:21.194 The chi square critical value for significance level = 0.05 for two tailed test hypothesis is, Χ2 critical, lower=12.401 For 5% level of significance with df =25-1=24 Χ2critical, upper=39.364 We fail to reject H0. There is sufficient not evidence to conclude that s.d is not equal to 15 .
Divide the data into two groups: those with under a 110 IQ and those with at least 110 IQ. a) Test the claim that the mean GPA is 3.0 for those with under a 110 IQ. For each hypothesis test, provide the claim, critical values, test statistic and conclusion.
IQGPA952.91952.94982.641002.651033.221042.721042.321053.11052.751072.61083.331083.32Averagre102.66672.875SD4.63840.3176 Ho =Hy=3.0 H1: Hy≠3.0 Under H0 is the test statistic, T=√12(y1-3)/s4(1) where, y1 is the sample mean gpa and s4(1) is sample s-d of gpa for students with < 110 IQ t=2.875-3/0.3176/√12=-1.3633 |-1.3633| ≥ 2.2 ≈ t11 (0.025) At 5% level H0 we fail to reject because at 5% level the claim of mean GPA to be for 3.0 for those with < 110 IQ is valid.
Give a real-life example of a time when it would be better to use the median instead of the mean. Also, explain why the median would be a better choice for this example.
Mean, median, and mode are measures of "central tendency." They give us an overall indication of what a population looks like. It also helps to know values of range, standard deviation, variance, etc. Those tell us how "spread out" the values are. If we consider any diverse group of observations, we might prefer the median to the mean. For example, With 10,000 people, the mean salary might be $45,000, but the range is $20,000 to $3,000,000 with a mean of $100,000. In that situation, "outliers" have drastically affected the mean. The few people that earn a lot of money have distorted the overall average (mean). It is better to get the "half-way point" (median). Even the mode (the "most frequent") does not necessarily represent the population, since it could be $25,000, for example. Some more examples: grades on a test height of students (some of whom are on basketball team) age of employees in a company Often, when collecting statistics, "outliers" are discarded as either typos or invalid observations. For example, if you have collected a set of high temperatures for the month (like 70, 74, 768, 73, ...), you should consider the temperature of 768 to be invalid.
Based on a study of jockeys, researchers determine that racing horses makes people short. Does this makes sense?
No, it doesn't make sense as small people are picked to be jockeys because they don't weigh much which will make the horse run faster. The lighter the jockey the less weight on the horse and more strength a horse could produce. A jockey riding a horse will not change the riders body. When have you heard a person getting short from riding a horse?
Are the movie ratings of G, PG-13 and R ordinal, nominal, interval or ratio? Explain.
Ordinal
The data provided is from houses for sale in Salinas. Prices are in the $1,000. 1) Test for a linear correlation between price and square feet in South Salinas. Provide the conclusion.
Positive linear correlation.
In a small soda company, the manager grabs every 30th bottle of soda from the assembly line and tests it for quality. What type of sample method is being used?
Systematic
Units Done: 49, 50, 60, 25, 35, 37, 56, 6, 75, 140, 50, 42, 30, 6, 24, 40, 30, 42,43,94, 70, 39, 35, 50, 30, 15, 34, 25, 38, 39, 68, 8, 20, 26, 40 Use the data for amount Units Done to answer questions a-g. a) Is this data continuous or discrete? Explain.
The given data is a discrete data, because the units are counted.
Determine if the following is either impossible, possible but unlikely or possible and likely. Explain your choice. All nine people in a room have the same birthday.
The given event is possible but unlikely, because the event that all nine people in a room have the same birthday is possible but the probability that all nine people in a room have the same birthday is very less and that's why the given event is very unlikely. The event that all nine people in a room have the same birthday is very rare event and therefore the probability of this rare event is very less which is unlikely.
The data provided is from houses for sale in Salinas. Prices are in the $1,000. 8) Which factor has the biggest effect on the price of a house in Salinas? Explain.
The highest r value was 0.897 for a linear correlation between price and square feet in South Salinas meaning the most expensive homes are the bigger houses in that region. Also, problem number 7 confirmed the most high-priced area was in South Salinas than North Salinas.
The data below was collected from Math 13 students. They were asked how many units they were taking this semester and to rate their math anxiety level, on a scale from 0 to 10. Using a 95% level of confidence, test the claim that the majority of students have at least a 3.0 GPA. For each hypothesis test, provide the claim, critical values, test statistic and conclusion.
The null and alternative hypotheses are defined as, Ho: p=0.50 Ha:p>0.50 Q=1-0.5=0.5 From the data values, GPA>=3.017>3.08 p=17/25=0.68 z=0.68-0.50/√0.5091-0.50)/25 z=1.8 Χ2= (8-12.5)^2/12.5 + (17-12.5)^2/12.5 Χ2=3.24 For 5% level of significance with df =1 Χ2-critical = 3.841 The p-value is obtained from z-distribution table for z = 1.8 P-value=0.0359 We fail to reject H0. There is not enough evidence reject the claim that majority of students have atleast 3 GPA .
The data below was collected from Math 13 students. They were asked how many units they were taking this semester and to rate their math anxiety level, on a scale from 0 to 10. Using a 95% level of confidence, test the claim that IQ's have a mean of 100. For each hypothesis test, provide the claim, critical values, test statistic and conclusion.
The null and alternative hypothesis Ho: μ=100 Ha:µ ≠ 100 Claim: µ= 100 TRADITIONAL METHODgiven that,sample mean, x =114.24standard deviation, s =14.0957sample size, n =25I.stanadard error = sd/ sqrt(n)where,sd = standard deviationn = sample sizestandard error = ( 14.0957/ sqrt ( 25) )= 2.819II.margin of error = t ?/2 * (standard error)where,ta/2 = t-table valuelevel of significance, ? = 0.05from standard normal table, two tailed value of |t ?/2| with n-1 = 24 d.f is 2.064margin of error = 2.064 * 2.819= 5.819III.CI = x ± margin of errorconfidence interval = [ 114.24 ± 5.819 ]= [ 108.421 , 120.059 ] DIRECT METHODgiven that,sample mean, x =114.24standard deviation, s =14.0957sample size, n =25level of significance, ? = 0.05from standard normal table, two tailed value of |t ?/2| with n-1 = 24 d.f is 2.064we use CI = x ± t a/2 * (sd/ Sqrt(n))where,x = meansd = standard deviationa = 1 - (confidence level/100)ta/2 = t-table valueCI = confidence intervalconfidence interval = [ 114.24 ± t a/2 ( 14.0957/ Sqrt ( 25) ]= [ 114.24-(2.064 * 2.819) , 114.24+(2.064 * 2.819) ]= [ 108.421 , 120.059 ]-----------------------------------------------------------------------------------------------interpretations:1) we are 95% sure that the interval [ 108.421 , 120.059 ] contains the true population mean2) If a large number of samples are collected, and a confidence interval is createdfor each sample, 95% of these intervals will contains the true population mean we don't have evidence that the IQ's have a mean of 100, it is exceeded the value of 100
The data below lists states poverty rate, gun deaths per 100,000 people, high school graduation rate and the unemployment rate. Use a 2x2 contingency table for questions 1-3. Be sure to find the test statistic and the critical values. Use the median to divide each list into 2 groups.You will need to provide the 2x2 table also. 2) Test for independence between high school graduation rate and gun deaths. What is the conclusion?
The two variables are dependent.
The data below was collected from Math 13 students. They were asked how many units they were taking this semester and to rate their math anxiety level, on a scale from 0 to 10. Using a 95% level of confidence, test the claim that the majority of students have at least a 3.0 GPA. At the 98% confidence level, test the claim that mean GPA for freshmen is over 3.0. For each hypothesis test, provide the claim, critical values, test statistic and conclusion.
The single sample t test is used to compare the sample mean to hypothetical population mean The null hypothesis and the alternate hypothesis are, Ho: µ = 3 Ha: µ> 3 From the data values, IQAverage3.3032SD0.4934 t=3.032-3/0.4934√25=3.073 x̄=3.30, s=0.4934,n=25 t-critical=2.17 The P-value is obtained using the t distribution table for t = 3.073 and degree of freedom = N - 1 = 25 - 1 = 24. a=1-0.98=0.02 P-value=0.0026 From t-distribution table, for lower tail test, critical value of 't' having an area of 0.02 in the left tail for (df=24) is given by: t0.02=-2.172 Since, value of test statistic (t) is more than the critical value of 't'; we can fail to reject the null hypothesis. Hence, it can be concluded that Mean GPA for freshmen is at least 3
If temperature is measured in Fahrenheit, is the data discrete or continuous? Explain your reasoning.
When temperature is measured in Fahrenheit, the data is continuous as the temperature can take any real value in the range and the data can take uncountably infinite number of values.
Units Done: 49, 50, 60, 25, 35, 37, 56, 6, 75, 140, 50, 42, 30, 6, 24, 40, 30, 42,43,94, 70, 39, 35, 50, 30, 15, 34, 25, 38, 39, 68, 8, 20, 26, 40 f) Are there any outliers for this data? Explain.
Yes, the outlier in the data is 140 because it falls more than 1.5 times in the interquartile range above the third quartile.
A bag contains 10 blue marbles, 15 red marbles and 20 green marbles. Marbles are then chosen at random. a) Find the probability of getting a red and then a green marble. b)Find the probability of getting a red or a green marble.
a)0.152 b)0.778
Scores on an exam have a normal distribution with a mean of 80 and a standard deviation of 12. a) Find the probability that a person would score above 90. b) Find the probability that a person would score between 75 and 85. c)Find the probability that a group of 7 people would have a mean score above 84.
a)0.203 b)0.326 c)0.189
President Trump's approval rating is 42%. Suppose that 10 people were chosen at random a) Find the probability that 5 of the 10 people approve of the job President Trump is doing. b)Find the probability that at most 3 of the 10 people approve of the job President Trump is doing. c) Find the probability that at least 3 of 10 people approve of the job President Trump is doing
a)0.216 b)0.333 c)0.863
use the ages of students in a Math 13 class. 22, 19, 17, 34, 21, 19, 18, 20, 23, 27, 20, 19, 20, 21, 41, 20, 28, 23, 21, 20, 19, 23, 25, 31, 28 a)Find the mean for these ages. b)Find the standard deviation for these ages. c)List the 5 numbers needed to create the boxplot for these ages. d)Find the age at the 72nd percentile e)Create a frequency table with 4 categories. Your answer should look like: f)Based on this information, would it be unusual for a student to be 35? Find the z-score.
a)23.16 b)5.632 c)Minimum=17, Q1=19.5, Q2=21 ,Q3=26, Maximun=41 d)24 e)Ages Frequency 17-23 18 24-30 4 31-37 2 38-41 1 f)2.102
In 2016, 46.4% of voters voted for Donald Trump. There were a total of 128,838,341 votes cast. a) How many people voted for Trump? b) Is the answer to part (a) continuous or discrete? Explain.
a)59,780,990 b)The number of votes are discrete because the number of votes are countable and it can take only countable values not any value in any interval.
A bag contains 10 blue marbles, 15 red marbles and 20 green marbles. Marbles are then chosen at random. a) Find the probability of getting a red and then a green marble. b) Find the probability of getting a red or a green marble.
a. 0.152 b. 0.778
Scores on an exam have a normal distribution with a mean of 80 and a standard deviation of 12. a) Find the probability that a person would score above 90. b) Find the probability that a person would score between 75 and 85. c) Find the probability that a group of 7 people would have a mean score above 84. d) Find the score needed to be in the top 10% of the class.
a. 0.203 b. 0.326 c. 0.189 d. 95.36
Listed below are the ages of students in a Math 13 class. 22, 19, 17, 34, 21, 19, 18, 20, 23, 27, 20, 19, 20, 21, 41, 20, 28, 23, 21, 20, 19, 23, 25, 31, 28 a.Find the z-score b.Find the age at the 72nd percentile,
a. 2.102, b. 24
In 2016, 46.4% of voters voted for Donald Trump. There were a total of 128,838,341 votes cast. a) How many people voted for Trump? b) Is the answer to part (a) continuous or discrete? Explain.
a. 59,780,990 b. The number of votes are discrete because the number of votes are countable and it can take only countable values not any value in any interval.
President Trump's approval rating is 42%. Suppose that 10 people were chosen at random a) Find the probability that 5 of the 10 people approve of the job President Trump is doing. b) Find the probability that at most 3 of the 10 people approve of the job President Trump is doing. c) Find the probability that at least 3 of 10 people approve of the job President Trump is doing.
a.0.216 b.0.333 c.0.863
The data below lists states poverty rate, gun deaths per 100,000 people, high school graduation rate and the unemployment rate. Use a 2x2 contingency table for questions 1-3. Be sure to find the test statistic and the critical values. Use the median to divide each list into 2 groups.You will need to provide the 2x2 table also. 4) Divide the data into 3 groups by gun deaths: 0-9, 9.1-12 and 12.1-20. Use ANOVA to test the claim that the average poverty rate is the same for the three groups. Enter the mean for each group, the test statistic, critical value and conclusion.
data 0-99.1-1212.1-208.60%7.60%13.10%10.10%9.70%9.20%14.50%10.20%14.40%9.70%11.80%12.60%12.10%9.20%14.80%6.80%9.20%10.60%12.00%16.20%9.90%12.60%12.30%15.20%17.90%16.70%15.40%13.20%11.20%13.80%11.30%12.50%15.00%11.50%11.40%15.00%10.20%11.20%15.60%9.50%11.10%10.60%8.10%10.00%11.60%16.70%12.00%20.10%18.30%15.90% mean for each group Groupsmean0-911.43%9.1-1211.29%12.1-2014.01% Using Excel data -> data analysis -> Anova Single Factor TS = 5.6490 Critical value = 3.1951 Conclusion Since TS > critical value, we reject the null hypothesis, we conclude that there is sufficient evidence to reject the claim the average poverty rate is the same for the three groups.
The data below lists states poverty rate, gun deaths per 100,000 people, high school graduation rate and the unemployment rate. Use a 2x2 contingency table for questions 1-3. Be sure to find the test statistic and the critical values. Use the median to divide each list into 2 groups.You will need to provide the 2x2 table also. 1) Test for independence between poverty rate and gun deaths. Enter the 2x2 contingency table for this problem.
gun death poverty low high low 15 10 25 high 10 15 25 25 25 50
The data provided is from houses for sale in Salinas. Prices are in the $1,000. 7) Test the claim that average house prices are the same in South and North Salinas. Provide the claim, test statistic, critical values and conclusion.
t=2.297 P-Value=0.026 a=0.05 Df=46 critical value of t (right tailed)=1.99 critical value of t (two tailed)=2.30 Reject the null hypothesis as P-value 0.026< a(0.05). The average price is not the same in north and South Salinas .
Listed below are the ages of students in a Math 13 class. 22, 19, 17, 34, 21, 19, 18, 20, 23, 27, 20, 19, 20, 21, 41, 20, 28, 23, 21, 20, 19, 23, 25, 31, 28 Find the mean and the standard deviation for these ages. x= s=
x=23.16 s=5.632
Male-166,205,194,144,168,157,198,155,178,206,166,166,149,159,194,158 Female-134,154,102,122,109,161,122,133,135,141,115,123,141,115,124 Would it be considered unusual for a female to weigh 120 pounds? Prove your answer by using the z-score.
z score for women when x=120 is z score= (120-128.7333)/16.2413=-0.5377 It below the standard deviation of the mean hence it is not unusual
In China, 48.5% of the people are female. Suppose that 10 people in China were chosen at random. Would it be unusual if 7 of the 10 people were female? Explain.
z=7-4.85/1.580=1.361 which is not above 2, so it's not unusual.
