Question 1
You are advised to first perform the appropriatehypothesistest using pencil and paper, along with a calculator and statistical tables, and then use your working to answer the questions below.
An administrator at a medium-sized hospital tells the board of directors that, among patients received at the Emergency room and eventually admitted to a ward, the average length of time between arriving at Emergency and being admitted to the ward is 4 hours and 15 minutes. One of the board members believes this figure is an underestimate and checks the records for a sample of 25 patients. The sample mean is 6 hours and 30 minutes. Assuming that the population standard deviation is 3 hours, and that the length of time spent in Emergency is normally distributed, use the sample data to determine whether there is sufficient evidence at the 5% level of significance to assert that the administrator's claim is an underestimate?
(NOTE: Convert all times to hours (as a decimal) before completing calculations. For example 3 hours and 20 minutes = 3 hours + 20/60 = 2.33 hours )
(a) According to the null hypothesis, the value of the population mean isAnswerhours. (Give your answer in hours to 3 decimal places x.xxx (Answer in hoursnotin hours and minutes)
(b) Is thetestone-tailed or two-tailed? Answer:Answer(answer 1 or 2)
(c) To 3 decimal places the critical value of the test statistic for this hypothesis test is:Answer(Answer in the form x.xxx or -x.xxx)
(d) To 3 decimal places the test statistic calculated from the sample is:Answer (Answer in the formx.xxx or -x.xxx)
(e) Is the null hypothesis rejected at the 5% level? Answer:Answer(Answer YES or NO)
(f)Does this mean that the hospital administrator provided accurate information to the board? Answer:Answer(Answer YES or NO)
(g) Is the assumption that the number of beds filled is normally distributed necessary in order to perform this test?? Answer(Answer YES or NO)
Question 2
You are advised to first perform the appropriatehypothesistest using pencil and paper, along with a calculator and statistical tables, and then use your working to answer the questions below.
According to the Australian Bureau of Statistics, the quantity of waste ending up in municipal landfills amounts to 0.80 kg per person per day. Some claim that because of recycling and greater emphasis on the environment, this figure is now lower. Others contend that constant increases in packaging and other life-style developments have pushed this figure higher in spite of recycling efforts. To test whether the amount of garbage per person has changed, a random sample of 54 Australians was taken and they were asked to keep a log of their garbage for a day. The sample mean was 0.92 kg.
Assuming that the population standard deviation is 0.5 kg, and that the amount of garbage per person per day is normally distributed, use the sample data to determine whether there is sufficient evidence at the 5% level of significance to assert that the amount of garbage per person per day has changed.
(a) According to the null hypothesis, the value of the population mean isAnswerkg. (Answer in the form x.xx)
(b) Is thetestone-tailed or two-tailed? Answer:Answer(answer 1 or 2)
(c) To 3 decimal places the upper tail critical value of the test statistic for this hypothesis test is:Answer(Answer in the form x.xxx)
(d) To 3 decimal places the upper tail test statistic calculated from the sample is:Answer (Answer in the formx.xxx)
(e) Is the null hypothesis rejected at the 5% level? Answer:Answer(Answer YES or NO)
(f)Does this mean that the amount of garbage per person per day has changed? Answer:Answer(Answer YES or NO)
(g) Is the assumption that the amount of garbage is normally distributed necessary in order to perform this test? Answer(Answer YES or NO)
Qestion 3
In order to determine the p-value, which of the following items of information isnotneeded?
Select one:
a.The level of significance
b.Whether the test is one- or two-tailed
c.The value of the test statistic
d.All of the above are needed
Question 4
If we reject the nullhypothesis, we conclude that
Select one:
a.there is enough statistical evidence to infer that the null hypothesis is true
b.there is enough statistical evidence to infer that the alternative hypothesis is true
c.there is not enough statistical evidence to infer that the alternative hypothesis is true
d.thetestis statistically insignificant at whatever level of significance the test was conducted at
Question 5
If we reject the nullhypothesis, we conclude that
Select one:
a.there is enough statistical evidence to infer that the null hypothesis is true
b.there is enough statistical evidence to infer that the alternative hypothesis is true
c.there is not enough statistical evidence to infer that the alternative hypothesis is true
d.thetestis statistically insignificant at whatever level of significance the test was conducted at
Question 6
You are advised to first perform the appropriatehypothesistest using Excel and pencil and paper, along with a calculator and statistical tables where appropriate, and then use your working to answer the questions below.
Many overseas sellers using online auction platforms such as Ebay provide an estimate for the number of days a package will take to arrive at a particular destination country from the time of order. A random sample of the time taken for a package to arrive in Australia from the time of ordering from a seller based in Hong Kong was obtained. The data in the Excel fileA4_S2_2014.xlsrepresent the number of days taken for a package to arrive from the overseas seller.
The seller claims that the mean number of days for a package to arrive in Australia is 13 days or less. Does the sample data obtained provide evidence that the mean arrival time for packages is as the seller claims at the 5% level of significance? Formulate null and alternative hypotheses regarding the data, and explain why the choice of null and alternative hypotheses is appropriate. (Remember we are seeking evidence that the claim made by the seller isnot true).
Give the following statisitcs for this data. Complete required working using Excel first.
Population mean (µ): Answer . (Give your answer as a whole number in the form xx).
Sample mean (x_bar): Answer . (Give your answer as a whole number in the form xx).
Sample standard deviation (s): Answer . (Give your answer to three decimal places in the form x.xxx).
Sample size (n): Answer . (Give your answer as a whole number in the form xxx).
Level of significance (alpha α): Answer . (Give your answer as a proportion to two decimal places in the form xxx).
Complete the two hypotheses which describe this situation
Null HypothesisHo: µ = Answer . (Give your answer here as a whole number in the form xx).
Alternative HypothesisHa: µ Answer Answer . (In the first box insert the appropriate symbol <,> OR ≠. For the second box give your answer here as a whole number in the form xx)
Whatdistributiondo the sample means follow in this case? Answer . (Type Z or t).
Is this distribution associated with a number ofdegrees of freedom? Answer . (If Yes, give the appropriate number of degrees of freedom as a whole number in the form xxx or else type NO)
Calculate the value of theteststatisticfor this set of data: Answer . (Give your answer to three decimal places in the form x.xxx).
Give thecritical valuefor the test at 5% level of signifcance: Answer . (Use Excel to find this value. Give your answer to three decimal places in the form x.xxx}
What is thep-valuefor the test? Answer .(Use Excel to find this value. Give your answer to three decimal places in the form x.xxx. If its in scientifc notation don't forget to change it to a decimal first}
What is your decision about the hypotheses:
Answer the Null Hypothesis. (Type ACCEPT or REJECT).
And hence Answer the Alternative Hypothesis. (Type ACCEPT or REJECT).
A series of statements ragarding the evidence you used to make this decision are given below. For each statement type TRUE or FALSE.
We rejected the null hypothesis because the test statisitc was greater than the critical value. Answer .
We did not reject the null hypothesis because the test statistic was greater than the critical value. Answer .
The p-value is less than the level of signifcance so we reject the null hypothesis. Answer .
The p-value is greater than the level of significance so we do not reject the null hypothesis. Answer .
We conclude that the claim by the seller that the mean number of days taken for apackage to arrive is 13 days or less Answer been supported by the data. (Type HAS or HAS NOT).