Name:________________________________________. Math 110 Lab PDF on One Sample T-Test Confidence Interval and Hypothesis Test of Significance Lab Textbook Page 478 #32, Do the inferences (remove an...

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I need to do these 2 assignments using statcrunch and copy the graph in the 4 step process method.


Name:________________________________________. Math 110 Lab PDF on One Sample T-Test Confidence Interval and Hypothesis Test of Significance Lab Textbook Page 478 #32, Do the inferences (remove an outlier). ******Plus PDF below: 1. BK,a large shoe company that specialized in light-up shoes, is interested in estimating the population mean length of fourth grader’s feet, their largest sales group. Historical data has been shown to be roughly normal. A random sample of 20 fourth graders produced the following foot lengths in centimeters: 22.6 22.3 23.9 23.1 22.4 23.7 23.6 21.6 20.3 22.8 21.6 22.8 20.4 23.6 20.3 21.8 24.5 23.1 24.5 24.2 Construct a 90% confidence interval for ?, the population mean length of all fourth grader’s feet. State: Plan: Solve: Conclude: 2. The College Board reported that the mean tuition and fees for community college at the national level was $2364 per year. You think that California community colleges are cheaper than the national average. The average tuiton for 49 random California Community colleges is �̅� =$2000 with a standard deviation ?�̅� = $400. Do you have enough evidence to prove your claim at . 05 significance level? State: Plan: Solve: Conclude: 3. A university of 25,000 students is trying to attract more commuting students from the local community. School researchers decided to ask the following question around campus:” How far do you commute from home to school every day?” Data in previous years has shown to be normal. The following data was collected as an SRS: 10 9 8 11 14 12 10 6 10 10 10 15 13 11 13 15 11 9 8 10 11 13 13 11 10 10 11 13 12 12 Construct the 90%, confidence intervals for the data provide above. Using a 4 step process. State: Plan: Solve: Conclude: 4. The average height of 18-year-old American women is 64.2 inches. You wonder whether the mean height of this year’s female graduates from your local high school is different from the national average. You collect an SRS of 78 female graduates and find that �̅� = 63.1 inches with a standard deviation ?�̅� = 2.1. Do you have enough information to within ? = .05 to prove your claim? State: Plan: Solve: Conclude: Lab Textbook Page 464 For #37, remove an outlier. ******Plus PDF Name_______________________________ Math 110 Chapter 21. Practice Resources Comparing Two Means 1. Which gender watches more Television? Men or Women? A local University wants to find out which gender watches more TV. The research team gathers 170 students and measures the amount of time each subject watches television on a typical day. The data is shown in the table shown at right. Conduct a Hypothesis test to find if there is any difference in which gender watches more Television at the ? = .1 level. Use the 4-step Process. State Plan Solve Conclude Sex Sample Size Mean Std. Dev. Male 57 2.29 1.85 Female 113 1.21 1.48 2. Which gender watches more Television? Men or Women? A local University wants to find out which gender watches more TV. The research team gathers 170 students and measures the amount of time each subject watches television on a typical day. The data is shown in the table shown at right. Create a 90% confidence interval for the Mean Difference in the TV watching time between men and women. Use the 4-step Process. State Plan Solve Conclude Sex Sample Size Mean Std. Dev. Male 57 2.29 1.85 Female 113 1.21 1.48 3. A physical education director claims by taking a special vitamin, a weight lifter can increase his strength. Eight athletes are selected and given a test of strength, which is the maximum number of pounds the athlete can bench press. After 2 weeks of regular training, supplemented with the vitamin, they are tested again. The table below is the data generated. Test the effectiveness of the vitamin regimen at α = 0.05. Assume the variable is approximately normally distributed State: Plan: Solve: Conclude: 4. A physical education director claims by taking a special vitamin, a weight lifter can increase his strength. Eight athletes are selected and given a test of strength, which is the maximum number of pounds the athlete can bench press. After 2 weeks of regular training, supplemented with the vitamin, they are tested again. The table below is the data generated. Create a 95% confidence interval for the effectiveness of the drug (the difference between before and after treatments) State: Plan: Solve: Conclude: 5. We know that the mean weight of men is greater than the mean weight of women and men tend to be taller than women. The true measure of health in any person is their BMI (Body Mass Index). To the right are BMI Statistics for a random sample of males and females. Is there enough evidence to prove that the male population has a higher BMI than the female population? Conduct a Hypothesis test to find if there is any difference in BMI between genders. Use the 4-step Process. State: Plan: Solve: Conclude: 6. A sample of 10 college students in a class were asked how many hours per week they watch TV and how many hours a week they used a computer. Is there a difference in the mean number of hours a college student spends on a computer versus watching TV at α = 0.01? Assume the population of differences is approximately normally distributed. State: Plan: Solve: Conclude: Chapter 22: Inference about a Population Proportion Chapter 22: Inference about a Population Proportion Math 34 Page 1 1. A large research team wants to know which soft drink people prefer: FizzyPop or Bubbles. The team gathers 652 people at a local mall and performs taste tests; 316 of which preferred FizzyPop. Can we determine, with 90% confidence, the preference of the general population? State Plan Solve Conclude 2. A university is contemplating switching from the quarter system to a semester system. The Administration conducts a survey of a random sample of 450 students and finds that 268 of them prefer to remain on the quarter system. Can we determine, with 99% confidence, the student population’s preference? State Plan Solve Conclude Chapter 22: Inference about a Population Proportion Math 34 Page 2 3. The Affordable Care Act was criticized thoroughly during President Obama’s two terms. In 2013, the percentage of people without health insurance in the US was 18%. You decide to find out if the Affordable Care Act has decreased the percentage of people without health insurance in the US in the Bay point/Pittsburg/Antioch Areas. A small survey of 122 people at LMC revealed that only 14 people are uninsured. Did the Affordable Care Act decrease the rate of uninsured in the US? ???? ?? .05 ????? ?? ????????????. State Plan Solve Conclude Chapter 22: Inference about a Population Proportion Math 34 Page 3 4. The National Transportation Safety Board publishes statistics on the number of automobile crashes that people in various age groups. In 2003, a study concluded that young people aged 18- 24 had an accident rate of 12%. A more recent in 2015 study gathered 1000 young drivers and found that 134 of them were involved in accidents. Conduct a Hypothesis test to find if there has been an increase in accident rates of people aged 18-24 ( ???? ?ℎ? ????? ?? .01 ????? ?? ????????????.). Use the 4-step process. State : Plan Solve Conclude Chapter 22: Inference about a Population Proportion Math 34 Page 4 5. The National Center for Health Statistics reported in 2004 that 22% of the American Population used some form of tobacco products (cigarettes, chewing tobacco, etc). A statistics class decides to find out if current measures have decreased tobacco use in the overall population. Of 400 people polled, 78 used some form of tobacco product. Conduct a Hypothesis test to find if there has been a decrease in tobacco use (? = .1). Use the 4- step process. State: Plan Solve Conclude
Answered Same DayMay 05, 2021

Answer To: Name:________________________________________. Math 110 Lab PDF on One Sample T-Test Confidence...

Monali answered on May 07 2021
151 Votes
1. Which gender watches more Television? Men or Women?
A local University wants to find out which gender watches more TV. The research team gathers 170 students and measures the amount of time each subject watches television on a typical day. The data is shown in the table shown at right. Conduct a Hypothesis test to find if there is any difference in which gender watches more Television at the ? = .1 level. Use the 4-step Proce
ss.
Step1 - State: Large number of data is collected with sample size of 170 college students; therefore, we can say that sample is based on law of large number. With this principle sample parameter estimated will be closet to actual population parameter and distribution likely to be symmetric. But to use data in better way, segregation of higher television watching time is required. In this case sample collected data is surmise as below.
    Sex
    Sample Size
    Mean
    Std Dev
    Male
     n1 = 57
    µ(S) = 2.29
    s = 1.85
    Female
    n2 = 113
    µ = 1.21
    σ = 1.48
Step 2 - Plan:
Hypothesis testing is used to statistically check one group watches more television than other. In this case, data is examined for mean of male more than female is used. Mean of female is used as hypothetical mean and standard deviation of male is used as standard deviation of estimated value. Sample size of male is used as estimated sample size.
First starting with hypothesis writing;
Null hypothesis: Mean of female watching television is less than mean of male, that is µ(S) > µ
Alternative hypothesis: Mean of male watching television is not less than mean of female, µ(S) ≤ µ
Test calculated = (Estimated value - Hypothesis Value) ÷ Standard Error of Estimated value
Standard Error = σ÷√n
Step 3 - Solve: Substituting values in equation;
Standard Error = σ÷√n = 1.85÷√57 = 0.245
Test calculated = (Estimated value - Hypothesis Value) ÷ Standard Error of Estimated value
t = 2.29-1.21 / 0.245 = 4.40
t-critical value = 1.65 at ? = .10 for one-tail hypothesis.
Now proceed to compare Z values.
t calculated = 4.40 > Z critical = 1.65
Therefore, reject null hypothesis that mean of female group watching television is higher compared to male.
Step 4- Conclude:
Sufficient statistical test evidence there to reject null hypothesis. Mean television watching time for female is not greater than male group. Therefore, male group with higher television watching time is segregated from female group.
2. Which gender watches more Television? Men or Women?
A local University wants to find out which gender watches more TV. The research team gathers 170 students and measures the amount of time each subject watches television on a typical day. The data is shown in the table shown at right. Create a 90% confidence interval for the Mean Difference in the TV watching time between men and women. Use the 4-step Process.
Step1 - State: Data of 170 college student is collected for assessing differences in watching television. Sample is large as more than 100 participants are included in study. Data of male and female is not equal making sample as random trial. There are two sets of data and therefore, segregated group with same mean is required to be examined to understand characteristics of each gender. Average provides summary data statistic. Below is statistical summary;
    Sex
    Sample Size
    Mean
    Std Dev
    Male
     n1 = 57
    ḿ1 = 2.29
    σ1 = 1.85
    Female
    n2 = 113
    ḿ 2 = 1.21
    σ2 = 1.48
Step 2 - Plan:
Hypothesis testing is statistical mechanism of checking difference between two groups. Check for difference in mean equal to 0 else otherwise. Hypothesis is constructed to check if there is difference in mean for watching television between male and female. Since sample size is more than 100, use z-statistics with following equation.
Null hypothesis: mean of male – mean of female = 0, that is µ1 - µ2 = 0
Alternative hypothesis: mean of male – mean of female ≠ 0, that is µ1 - µ2 ≠ 0
Then solve for equation;
(σ1^2/n1 + σ2^2/n2)
Step 3 - Solve: Substituting values in equation;
Z = (2.29-1.21) – (0) ÷ √ 1.85^2/57 + 1.48^2/113
Z = 1.08 ÷ 0.28 = 3.85 and corresponding P value = 0.99994
Comparing the ? = .1 and P-value calculated = 0.99994
P-value test calculated (0.99994) ≤ 0.10. Even though the difference in calculated test p-value and ? is very less, we reject...
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