The Australian Bureau of Statistics has reported that in 2015, 26.78% of the 16.8 million employees in Australia worked part-time in their main job. Suppose that you select a random sample of 250 employees from around Australia.

(a) What is the probability that more than 26.2% of those sampled work part-time in their main job?

(b) What is the probability that the proportion of part-time employees in the sample is between 27% and 29%?

(c) The probability is 77% that the sample proportion of part-time workers will be above what value?

In deciding where to invest her retirement fund, an investor recorded the weekly returns of two portfolios for one year, with the results stored in columns 1 and 2. Some of these data are shown below. Can we conclude at the 5% significance level that portfolio 2 is riskier than portfolio 1? Compare both return and risk . Do on Excel .

This is the Data :

A company claims to manufacture paper clips with a standard deviation of less than 0.12g. A
set of 71 paper clips has a sample standard deviation of 0.10 g. Is the claim supported by the test?  

Question: 2 The data below shows the amount of money in millions of shillings paid out suppliers by firm over a certain period of time 7.0 4.1 3.4 5.5 4.5 6.6 7.3 7.7 8.0 3.0 5.0 4.5 7.2 5.0 2.7 7.0 5.5 7.0 8.5 7.0 3.0 5.0 6.0 5.3 4.0 4.5 3.5 5.5 2.0 8.1 2.5 5.1 3.5 6.2 6.0 3.0 4.5 3.5 5.0 8.9 5.3 2.3 2.8 6.5 6.8 5.0 6.5 3.4 3.5 7.6

a) Starting with 2.0-2.9 ,construct a frequency distribution table for the data

b) Using the frequency table;

i. Draw a cumulative curve for the data and hence estimate the median and inter quartile range

ii. Calculate the mean and standard deviation of production

Question:5

1. What is meant by the standard error of the mean?
2. What is meant by a Type I and a Type II error?
3. What is the difference between a one and two tail test?

4. In September 2019, a Travel Agent used a random sample of 36 holiday makers to find out the average cost per person of a one week holiday in Queen Elizabeth National Park. The following information was found:

September 2019

          Mean                       $372.40

         Standard deviation $26.10

         Sample size 36

In the previous year the average cost of each holiday was $356.20.

1. Test whether the cost of a one week holiday to Queen Elizabeth National Park has increased significantly since the previous year.
2. The company based its views on customer satisfaction from the letters it receives. In the previous year, 68% of the letters it received were of a positive nature.

The company wishes to adopt a more scientific approach to estimating customer satisfaction.

What sample size would be needed to estimate the proportion of customers’ views to within 2% of the true figure at the 95% confidence limit and Interval                                  

You are working with the marketing team for a FMCG firm that produces shaving cream. The team believes that sales of some of the products are closely related to sales of other products. They want you to explore this in a little more depth for two products, SKU 123 and SKU 456. Unfortunately, all of the base sales data for these products has been destroyed. All that you have is the weekly summary data:

Now the marketing team wants to understand the potential weekly sales for these two products. Let the sales price for the two SKUs be 12.50, 7.75, respectively.

1) What is the expected weekly revenue?

2) What is the standard deviation of the weekly revenue?

3) Assuming the marketing team’s correlation of 0.79 is correct. What is the probability that weekly sales will be between 10,000 and 20,000 dollars?

Chi-square analysis compares the counts of two categorical variables to tell you if a relationship exists between the variables or not. You can apply chi-square analysis to answer important questions about factors in everyday life, and even about events like elections... or Halloween. If you are a character in a slasher film, is there a connection between your gender and your dying in some horrible manner?

1. Propose a life event or situation and two categorical variables for it. Complete a chi-square analysis of this event or situation and these variables, and share your results.

2. Do you agree with your findings? Why or why not?

(a) Using the armspanSpring2020.csv data from class, test the hypothesis that those who identify as female have a shorter armspan than those who do not so identify. Write out the null and alternative hypotheses, give the value of the test statistic and the p-value, and state your conclusion using a 5% significance level. Use R for all computations.

(b) Interpret, in your own words, the meaning of the p-value you got in part (a).

(c) Find a 95% confidence interval for the mean armspan using the data in armspanSpring2020.csv. Use R.
(d) What assumptions must you make if we wish to interpret this interval to apply to all UCLA students? Which of these assumptions do you think are met adn which are not?
(e) Find a 95% confidence interval for the difference between mean armspan and mean heights. Does it contain 0? Why is this surprising or not-surprising?

Please help with the r codes especially. It is my first time using it and I'm having a hard time. Thanks!

A corporation monitors time spent by office workers browsing the web on their computers instead of working. The computer records of a random sample of 101 workers showed that 9 of them spent at least 30 minutes per day browsing. Construct a 90% confidence interval for the proportion of workers who spend at least 30 minutes per day browsing. Do the results support the claim that over 15% of the workers do this

4) We're going to test the same hypothesis four ways. Assume the people in the dataset in armspanSpring2020.csv are a random sample of all adults. For each test, report the test statistic and the p-value. With a 5% significance level, give the conclusion of each test.
a) Test the hypothesis that the mean difference between armspan and height it not equal to 0, using the data in armspanSpring2020.csv. Do this by creating a new variable named diff = (armspan - height). Perform a one-sample t-test.
b) Test the same hypothesis, but use a two-sample t-test with paired =TRUE.
c) Test the same hypothesis, but use a two-sample t-test with paired=FALSE and var.equal=FALSE.
d) Test the same hypothesis, but use a two-sample t-test with paired=FALSE and var.equal=TRUE.
e) Which test(s) do you think are valid for this situation and why?  
hint: We almost never use the var.equal=TRUE test. Why? Because it is only valid if the population standard deviations of both populations are equal. You might be in a situation where you know this to be true. If so, fine, use it. But usually we don't, in which case (a) the var.equal=FALSE test will provide more accurate p-values if the standard deviations are not equal and (b) will provide pretty accurate p-value if they are. So you can't lose, really, with the var.equal=FALSE test, but you can lose with it the other way.

g)Data cleaning. Identify by row number which observations seem in need of cleaning and why you think so. Provide a table. (Hint: consider the "which()" and "identify()" functions.) Provide a graph to justify your identifications.

Please help with the r codes. It is my first time doing r studio and I'm having a hard time. Thanks!

Question:4

1. A new promotion by K2 has been developed with a goal of obtaining 90% new simcard connections among business firms. To evaluate the simcard sales, 20 are to be sold to business people in Kikuubo. If the 90% sales rate is correct, what is the probability that 18 or more of the sim cards will be sold to Kikuubo business men and women.

2. The life time of milk produced by Fresh Diary is normally distributed. The probability that a pack drawn from a Carton is defective is 0.1 and the success is 0.9, if a sample of 6 packs is taken, Find the probability that it will contain;

1. No defective packs   
2. 5 or 6 defective packs
3. Less than 3 defective packs

3. Using a relevant example, explain the Bernoulli trial              

There are 10 democrats and 8 Republicans on a senate committee. From this group a 5- member subcommittee is to be formed. Find the number of 5-person subcommittee that consists of:

a)Any members of the senate committee.

b)Democrats only.

c)3 Republicans and 2 Democrats.

d)At least 4 democrats.

e)John Edwards,who is a Democrat, and any 4 Republicans

f)What is the probability that the 5-members subcommittee will include 3 Republicans and 2 Democrats?

If a regression analysis was to be completed on body mass index (BMI), what could be an independent variable in that analysis? Why? If we could, what other independent variables should be included in the analysis? What statistic(s) would show the value of that regression in understanding BMI?

Alternatively, find an article that uses regression analysis to study a medical concern. In that study, what was the dependent variable and what was the independent variable(s)? Further, how would you use this study to highlight the difference between correlations and causation?

Please provide APA references

Thank You

Thinking of the many variables tracked by hospitals and doctors' offices, confidence intervals could be created for population parameters (such as means or proportions) that were calculated from many of them. Choose a topic of study that is tracked (or that you would like to see tracked) from your place of work. Discuss the variable and parameter (mean or proportion) you chose, and explain why you would use these to create an interval that captures the true value of the parameter of patients with 95% confidence.

Please provide APA reference

Thank You

Oak hill has 74806 registered automobiles. A city ordinance requires each to display a bumper decal showing that the owner paid an annual wheel tax of $50. By law, new decals need to be purchased during the month of the owner's birthday. This year's budget assumes that at least 306000 in decal revenue will be collected in November. What is the probability that the wheel taxes reported in that month will be less than anticipated and produce a budget shortfall?