Wk4 DQ1

Ans: 

In the fourth week of our course, we covered and discussed the chapters 10,11 and 12 where we dealt with various topics like One-Sample Tests of Hypothesis, Two-Sample Tests of Hypothesis and Analysis of Variance . Hypothesis Testing is a statistical test used to determine whether the hypothesis assumed for the sample of data stands true for the entire population or not. This week 4 DQ 1 is mainly focused on how a point estimate is a single value used to estimate a population value along with a numerical solution.

A point estimate is a single value (statistic) used to estimate a population value (parameter), explain this statement with example.

It is a true statement that “A point estimate is a single value (statistic) used to estimate a population value (parameter)”.The point estimate is the statistic calculated from sample data used to estimate the true unknown value in the population called the parameter. We take samples from the population in order to estimate the true value for a population and use the statistics obtained from the samples to estimate the parameter (Payne, Study.Com). The sample mean X̅ is a point estimate of the population mean μ. Similarly, the sample proportion p is a point estimate of the population proportion P.

For example, if we want to study the average number of tourists visiting in the western part of Nepal yearly, it will be very difficult or maybe impossible for to collect data for every tourist. For this, we can use point estimator to estimate the unknown value of the population parameter.

Numerical Illustration:

Given Information:

Stress Level of Scores: 94, 78, 83, 90, 78, 99, 97, 90, 97, 90, 93, 94, 100, 75, 84

1. Numerical Answer (a):

Mean (X̅) = ∑Xi/n = 1342/15 = 89.466 = 89.47 (Approx.)

Therefore, the mean stress level for this sample is 89.4 and the point estimate of the population mean is also 89.47 because the point estimator is used to estimate the unknown value of population parameter.

1. Numerical Answer (b):

For 95% Confident intervals of the population mean,

Standard deviation (S) = √∑ (Xi- X̅)² /(n-1)

                                    = √913.7335/ (15-1)

                                        = 8.079

  Degree of freedom (df) = n – 1 = 15 – 1 = 14

Level of Significance = 5% = 2.5%  , Z= 2.145

Then,

Confidence interval   = X̅±Z S/√n = 89.47±2.145* 8.079/√15 = 89.47±4.47

• Lower confidence limit = 89.47 – 4.47 = 85
• Higher confidence limit = 89.47 + 4.47 = 93.94

Therefore, the Confidence level for the population mean lies in between 85 to 93.94 at 95% level of Significance.

1. Numerical Answer(c):

According to Dr. Benner’s test, it would not be reasonable to conclude that the mean stress level of Internet executives is 80 as the value doesn’t lie within the range of population parameter (85 to 93.94).

References

Payne, T. (Study.Com). Point Estimate in Statistics. Retrieved from https://study.com/academy/lesson/point-estimate-in-statistics-definition-formula-example.html

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I am Smirti Bam, an enthusiastic edu blogger with a passion for sharing insights into the dynamic world of business and management through this website.

I hold a MBA degree from Presidential Business School, Kathmandu, and a BBA degree with a specialization in Finance from Apex College,

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