Solution Code: 1AFAC
This assignment falls under Statistics which was successfully solved by the assignment writing experts at My Assignment Services AU under assignment help through guided sessions service.
RESEARCH/TUTORIAL PAPER A
A manufacturer of DVD players purchases a particular microchip, called the DV-6, from three suppliers; Perth Electronics, Sound Sales and Critical Components. Thirty percent of the DV-6 chips are purchased from Perth Electronics, 20 percent from Sound Sales and the remaining 50 percent from Critical Components. The manufacturer has extensive histories on the three suppliers and knows that 3 percent of the DV-6 chips from Perth Electronics are defective, 5 percent of chips from Sound Sales are defective and 4 percent of the chips purchased from Critical Components are defective.
When the DV-6 chips arrive at the manufacturer, they are placed directly in a bin and not inspected or otherwise identified by the supplier.
Note: You may wish to construct a joint probability table or a tree diagram to analyse this problem and use a power-point presentation for the class in your presentation.
RESEARCH/TUTORIAL PAPER B
This tutorial paper demonstrates the relationship between the population parameters and those derived for the sampling distribution of the mean.
For simplicity’s sake, we will imagine that there exists only N = 5 business executives in an industry. There salaries are given below.
Executive Annual Salary ( thousands of dollars)
A 195 B 205 C 125 D 275 E 200
(i) Compute the mean and standard deviation of annual salary for this population.
(ii) Now list the 10 possible samples of size 3 that could be drawn for this population without replacement.
(iii) Compute the sample mean of annual salary for each sample and show a graph of the sampling distribution of X .
(iv) Using the 10 values of X found in part (iii), compute the mean ( x? ) and standard deviation of the sampling distribution ( x? ) of X .Use the 10 values of X as the data.
(v) Use the population mean and population standard deviation found in part (i) in order to compute the mean and standard deviation of X . Note that with n/N = 3/5 = .60, you must use the finite population correction factor (fpc).
(vi) Compare the results in part (iv) with the results in part (v). Which approach – listing and using all possible X values (part iv) OR the use of the equations for x? and x? (part v) – provided the easier way to determine the mean and standard deviation of the X values?
Your power-point presentation should outline initially how this problem was tackled, the methodology used, justification for the solution reached and an interpretation of your result.
RESEARCH/TUTORIAL PAPER C
The attached data set relates to Real Estate. It reports information on homes sold in the Eastern suburbs area last year.
Your task is to select the correct variables and the answer the questions below.
The variables are defined as:
Data Set 1 – Real Estate
X1 = Selling price in $000 X2= Number of bedrooms X3= Size of the home in square feet X4= Pool (1=yes, 0=no) X5= Distance from the centre of the city X6= Township X7= Garage attached (1=yes, 0=no) X8= Number of bathrooms
What is the p-value?
The data set Real Estate is available on the Bb site
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Section C
The 95% confidence interval for mean selling price of the homes is calculated by using the formula given below
x-t2,(n-1)*sn,x+t2,(n-1)*sn
The table given below shows the workings of 95% confidence interval for mean selling price of the homes
Data | |
Sample Standard Deviation | 47.10996606 |
Sample Mean | 221.1005567 |
Sample Size | 105 |
Confidence Level | 95% |
Intermediate Calculations | |
Standard Error of the Mean | 4.597461932 |
Degrees of Freedom | 104 |
t Value | 1.9830 |
Interval Half Width | 9.1169 |
Confidence Interval | |
Interval Lower Limit | 211.98 |
Interval Upper Limit | 230.22 |
From the above table, we see that, the 95% confidence interval for mean selling price of homes is (211.98, 230.22). This indicates that, when more number of samples are extracted from the same population, then 95% (95 out of 100 times) of the times the true mean selling price of homes will fall between 211.98 and 230.22
In order to determine whether the mean size of the homes was more than 2,100 square feet, we perform one sample t test. The null and alternate hypotheses are given below
Null Hypothesis: H0: µ ? 2100
That is, the mean size of the homes was not significantly more than 2,100 square feet
Alternate Hypothesis: Ha: µ > 2100 (Right tailed test)
That is, the mean size of the homes was significantly greater than 2,100 square feet
Level of Significance: Let the level of significance be ? = 0.01
Critical Region
The critical value of t corresponding to 1% level of significance and at 105 degrees of freedom is
t0.01,105 = 2.363 (by referring t distribution table)
The critical region is
CR = {X: t > 2.363}
Test Statistic
The t test statistic is
t=x-?sn
The table given below shows the workings of
Data | |
Null Hypothesis m= | 2100 |
Level of Significance | 0.01 |
Sample Size | 105 |
Sample Mean | 2231.408571 |
Sample Standard Deviation | 249.3150999 |
Intermediate Calculations | |
Standard Error of the Mean | 24.3307 |
Degrees of Freedom | 104 |
t Test Statistic | 5.4009 |
Upper-Tail Test | |
Upper Critical Value | 2.3627 |
p-Value | 0.0000 |
Reject the null hypothesis |
Critical Value approach
From the above table, we see that the value of t test statistic is 5.4009 and its corresponding p – value is 0.000. Since the calculated value of t test statistic is greater than the critical value of t, we reject the null hypothesis at 1% level of significance. Therefore, there is sufficient evidence to conclude that the mean size of the homes was significantly greater than 2,100 square feet
P – Value Approach
From the above table, we see that the p – value of t test statistic is 0.000. Since the calculated value of t test statistic is less than 0.01, we reject the null hypothesis at 1% level of significance. Therefore, there is sufficient evidence to conclude that the mean size of the homes was significantly greater than 2,100 square feet
Totally, there are 106 homes included in this study. 71 out of 106 homes has attached garage in it. That is, 66.98% of the sample homes had garage in it
In order to determine whether more than 60 percent of the homes have an attached garage, we perform one proportion z test. The null and alternate hypotheses are given below
Null Hypothesis: H0: P ? 0.6
That is, the proportion of homes having an attached garage is not greater than 60%
Alternate Hypothesis: Ha: P > 0.6 (Right tailed test)
That is, the proportion of homes having an attached garage is greater than 60%
Level of Significance: Let the level of significance be ? = 0.05
Critical Region
The critical value of z corresponding to 5% level of significance is
Z0.05 = 1.645 (by referring normal distribution table)
The critical region is
CR = {X: z > 1.645}
Test Statistic
The t test statistic is
t=p-PP*(1-P)/n
The table given below shows the workings of
Data | |
Null Hypothesis p = | 0.6 |
Level of Significance | 0.05 |
Number of Items of Interest | 71 |
Sample Size | 106 |
Intermediate Calculations | |
Sample Proportion | 0.669811321 |
Standard Error | 0.0476 |
Z Test Statistic | 1.4671 |
Upper-Tail Test | |
Upper Critical Value | 1.6449 |
p-Value | 0.0712 |
Do not reject the null hypothesis |
Critical Value approach
From the above table, we see that the value of z test statistic is 1.4671 and its corresponding p – value is 0.0712. Since the calculated value of z test statistic is less than the critical value of z, we reject the null hypothesis at 1% level of significance. Therefore, there is no sufficient evidence to conclude that the proportion of homes having an attached garage is greater than 60%
P – Value Approach
From the above table, we see that the p – value of t test statistic is 0.0712. Since the calculated value of z test statistic is greater than 0.05, we reject the null hypothesis at 5% level of significance. Therefore, there is no sufficient evidence to conclude that the proportion of homes having an attached garage is greater than 60%
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