# StatCrunch Statistical Tool in Hypothesis Testing Research Paper

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Updated: Apr 29th, 2021

## Introduction

According to our professor, the average proportion of clothesperweek population is ten centimeters.

## Research Question

Based on a random sample of thirty clothesperweek, is our professor’s claim that the average proportion of clothesperweek population is ten centimeters true at a significance level (α) of 0.05? To prove or disapprove of our professor’s claim, I carried out research on the average proportion of clothesperweek population and analyzed the results/findings using StatCrunch. Below are the methodology questions and theory of how I carried out the analysis using StatCrunch. According to Berenson, Levine, Szabat, and Krehbiel (2012, p.123), the Statcrunch software tool is easier to use in analysis than other statistical tools of analysis.

What is the variable of interest?

The variable of interest is the average proportion of clothesperweek population.

What confidence interval are you going to use?

For this analysis, I will be using ninety-five percent for the average proportion of clothesperweek population.

State both the null hypothesis and alternative hypothesis

My null and alternative hypotheses are –

• Ho: μ1 = 10 cm the average proportion of clothesperweek is ten centimeters
• HA: μ1 ≠10 cm the average proportion of clothesperweek is not ten centimeters.

Which kind of sample test are you using between- sample test and two-sample test?

According to my analysis for the proportion of the clothesweek population, the one-sample test remains the applicable option. This has been evident in the manner of comparison of measurements from the data set.

According to you, did you use a two-tail test, right (an upper), or left (a lower)-tail test?

Since the inequality sign in the alternative hypothesis is ≠, the test used is right (an upper) or left (lower)-tail test. At the same time, the region of rejecting the null hypothesis is only one end (Berenson et al., 2012, p. 133).

What is the significance level you are using? Give your reasons for using that level.

I used a significant level of 0.05; it coincides with the ninety-five percent interval. In addition, it is appropriate according to consequential type I error.

Do you think the conditions necessary to satisfy the hypothesis and confidential interval population proportion are met? Explain your answer.

As per my random sample size of n equal to thirty (n = 30), there is a guarantee of normal population distribution. The sample was randomly selected from a population of over five thousand. From my samples, there were at least ten successes and eleven failures, which make my null hypothesis valid. In my case, since the samples were picked without returning or replacing, the population proportion should be more than ten times the samples taken (Berenson et al., 2012, p. 128). Notably, this condition is satisfied since the real size of the sample used is more than five thousand times while the sample size used is only thirty. Another fact that makes the conditions satisfied is that the sample I used was independently selected. At the same time, I assumed that the null hypothesis is true.

According to your analysis, what are the upper and lower limits of your confidence interval?

From my analyzed results, the lower bound and the upper bound are 3.916861 and 28.377306 respectively.

The confidence interval error term is given by (3.8161-2.7012)/2= 0.5575

From the assessment of the question you posed, what is the mean of the confidence interval?

According to my analysis, I am ninety-five percent confident that the proportion of the clothesperweek population is between forty-five thousand eight hundred and five thousand nine hundred and fifty.

What p-value did you get in your hypothesis test? Explain the meaning of this p-value as per the hypothesis question you posted.

From my analysis, the p-value is 0.1709. This value indicates that if the null hypothesis, which states that the average proportion of clothesperweek population is ten centimeters, is true, the chance of finding a sample proportion of 14.2823 or bigger than that in a sample of size thirty is 28.452.

According to your analyzed results, did you reject or accept the null hypothesis? Give your reasons.

I did not reject the null hypothesis since my p-value, 0.1709, lies in the range of the significant value of 0.05.

What conclusion did you make in accordance with the question you posted?

The analyzed results favored the hypothesis question, implying that there is evidence that the average proportion of clothesperweek population is ten centimeters (Berenson et al., 2012, p. 123).

Comparing the hypothesis test results and the confidence interval results, do you think the two results agree? Explain.

Since I did not reject the hypothesis, which states that the average proportion of clothesperweek population is ten centimeters with a ninety-five percent confidence limit of 3.916861 to 28.377306 and a significance level of 0.05, it is true to say that the hypothesis test results and the confidence interval results agree (Berenson et al., 2012, p. 123). If the rejection of the hypothesis could have taken place, the above values could not have agreed upon.

## Reference

Berenson, M., Levine, D., Szabat, K. A., & Krehbiel, T. C. (2012). Basic business statistics: Concepts and applications. Boston, MA: Pearson/Prentice Hall.

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