Demonstrate Inferences using a Single Sample Economics Math Problem

Demonstrate Inferences Using a Single Sample
Student’s Name
Institutional Affiliation
Demonstrate Inferences Using a Single Sample
Assignment 4
1 What is the confidence level for each of the following confidence intervals for µ?
From the z-tables, read for the z value below and multiply the table value by 2. So these are the answers:-
1 x±1.96δn= 95%
2 x±1.645δn= 90.1% when z is 1.65
3 x±2.575δn= 99.02% when z is 2.58
4 x±1.282δn= 79.94% when z is 1.28
5 x±0.99δn= 67.78%
2 A random sample of n measurements was selected from a population with unknown mean µ and standard deviation Ϭ. Calculate a 95% confidence interval for µ for each of the following situations:
The population mean µ and standard deviation or error Ϭ are both unknown. All sample sizes (n) are large. So we use a normal or Z-distribution. The formula is: x±z*δn
6 n= 75, x = 28, s2 = 12…………C.I= (27.22,28.784)
7 n= 200, x = 102, s2 = 22………C.I= (101.3499,102.6501)
8 n= 100, x = 15, s = 3………….C.I=(14.412,15.588)
9 n= 100, x =4.05, s = 0.83……...C.I= (3.88732, 4.21268)
10 Is the assumption that the underlying population of measurements is normally distributed necessary to ensure the validity of the confidence intervals in parts a -d? Explain.
The assumption is true. All confidence intervals are about the population mean (x). The deviation is centered about the mean and the spread is determined by the standard deviation (Hoerl & Snee, 2012).
3 Named for the section of the 1978 Internal Revenue Code that authorized them, 401(k) plans permit employees to shift part of their before-tax salaries into investments such as mutual funds. One company, concerned with what it believed was a low employee participation rate in its 401(k) plan, sampled 30 other companies with similar plans and asked for their 401(k) participation rates. The following rates (in percentages) were obtained:
807681778280856080798270
888580798375877880847275
908482777586
Descriptive statistics from SPSS for the data are as follows:
Number of Valid Observations (Listwise) = 30
Variable Mean Std Dev Minimum Maximum N Label
PARTRATE 79.73 5.96 60.00 90.00 30
* Use the SPSS information above to construct a 95% confidence interval for the mean participation rate for all companies that have 401(k) plans.
Answer is C.I = (77.5972, 81.8628)
* Interpret the interval in the context of this problem.
The confidence interval in this problem contains information about the data parameters. It indicates that 28.5 or 29 out of 30 produced confidence intervals that contain the participation parameter. It bounds the mean and the standard deviation and if this was repeated over and over, 95% of the time participation rates would fall in between 77.5972 and 81.8628.
* What assumption is necessary to ensure the validity of this confidence interval?
The assumption is that sample size (n=30) is sufficient, observations are normally distributed, i.i.d (identically independent) of each other and that they are randomly sampled from a large population of companies (Nelson, Coffin, & Copeland, 2003).
* If the company that conducted the sample has a 71% participation rate, can it safely conclude that its rate is below the population mean rate for all companies with 401(k) plans? Explain.
It is safe to conclude that its rate is below the population mean rate for all companies. This is because a 71% participation rate is a 21.3 less than the required 28.5 or 29 about 95% confidence interval. In other terms, a 71% has a p value as -1.464 which is less than the normal z (1.96) for the 95%.
* If in the data set the 60% had been 80%, how would the center and width of the confidence interval you constructed in part (a) be affected?
If 80 was used, the p value would be 0.045302 which is less than the significance level (0.05).Similarly, the margin error would be 0.04930 meaning point estimate (79.73) +/- 0.04930 would give a confidence interval = (79.6807, 79.7793). The width of the confidence interval decreased.
4 Suppose you have selected a random sample of n = 5 measurements from a normal distribution. Compare the standard normal z values with the corresponding t values if you were forming the following confidence intervals:
Using t, the degrees of freedom of a random sample n=5 is n-1=4. So using students’ –T distribution table, read t (4, α/2). Subtract the confidence level from 1 then divide by 2 i.e. (100-C.I)/2= α level. Look up the DF=4 and the result from the previous sentence to get the t-value. As for z values, subtract the confidence level then divide the result by 2.Subtract the result from the previous sentence from 1 and then look for z-score in the area up in the middle of the z-table.
80% confidence interval :- Z is 1.282 and t is 1.533
90% confidence interval:- Z is 1.645 and t is 2.132
95% confidence interval:-Z is 1.96 and t is 2.776
98% confidence interval:-Z is 2.326 and t is 3.747
99% confidence interval:-Z is 2.576 and t is 4.604
All t-values are more than the z-score values in the above question.
5 The following random sample was selected from a normal distribution: 4, 6, 3, 5, 9, 3.
The mean is 5. The standard deviation is 2.5495. The sample size (n) is 6. The z-score values for each confidence interval are 1.645, 1.96 and 2.576 for a, b and c respectively. Using the formula x±z*δn confidence intervals for each are:-
Construct a 90% confidence interval for the population mean µ.
C.I = (3.2878, 6.7122)
Construct a 95% confidence interval for the population mean µ. C.I=(2.95998,7.04002)
Construct a 99% confidence interval for the population mean µ. C.I=(2.3188,7.6812)
Assume that the sample mean x and sample standard deviations remain exactly the same as those you just calculated but that they are based on a sample of n = 25 observations rather than n = 6 observations. Repeat parts a – c.
Using n=25, the confidence intervals for the population mean of each are:-
* 90%=(4.1612, 5.8388)
* 95%=(4.0006, 5.9994)
* 99%=(3.6865, 6.3135)
What is the effect of increasing the sample size on the width of the confidence intervals? Increasing the sample size decreases the width of the confidence interval.
6 The following is a 90% confidence interval for p: (0.26, 0.54). How large was the sample used to construct this interval?
At 90% confidence interval, the value of z is 1.645. The mean of 0.26 and 0.54 is 0.14. Assuming the sample is from a normally distributed population, the variance is 1 and the mean is 0. The formula for confidence intervals is as in question 1. After equating all values, sqrt of n is 1.645/0.14. The sample size (n) is 138.0625
7 When companies employ control charts to monitor the quality of their products, a series of small samples is typically used to determine if the process is “in control” during the period of time in which each sample is selected. Suppose a concrete-block manufacturer samples nine blocks per hour and tests the breaking strength of each. During one-hour’s test, the mean and standard deviation are 985.6 pounds per square inch (psi) and 22.9 psi, respectively.
Construct a 99% confidence interval for the mean breaking strength of blocks produced during the hour in which the sample was selected.
C.I = (965.9365, 1005.2635)
The process is to be considered &...