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Recently a large academic medical center, determined that 9 of 21 employees in a particular position were male, whereas 57% of the employees

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Asked: Jan 23,2019 In: Hypothesis testing
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name data Genius Punditsdkoslkdosdkoskdo
Asked: Jan 23,2019 In: Hypothesis testing

Recently a large academic medical center, determined that 9 of 21 employees in a particular position were male, whereas 57% of the employees

Recently, a large academic medical center determined that 9 of 21 employees in a particular position were male, whereas 57% of the employees for this position in the general workforce were male. At the 0.01 level of significance, is there evidence that the proportion of males in this position at this medical center is different from what would be expected in the general workforce?

What are the correct hypotheses to test to determine if the proportion is different?

A. H0: pi <= 150.57; H1: pi != 0.57
B. H0: pi != 0.57; H1: pi = 0.57
C. H0: pi >= 0.57; H1: pi < 0>

D. H0: pi = 0.57; H1: pi != 0.57

Calculate the test statistic.
ZSTAT = ___ ?
What is the p-value?


State the conclusion of the test.

Do not reject the null hypothesis. There is insufficient evidence to conclude that the proportion of males in this position at this medical center is different from the proportion in the general workforce.
 


statistical-hypothesis hypothesis-testing test-of-proportions
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i sixsigma
i sixsigma Added an answer on Jun 04,2019

Assume the normal distribution even if the sample is less 30, the problem states that the organization is large, hence we would assume that the population is sufficiently large. 

Note that the alternative hypothesis is a two-tailed test, which means D is the appropriate hypothesis. 

Using the z-statistic formula: Z  = (p^  - p)/sqrt((p * (1 - p))/n)

z = (9/21  - 0.57)/sqrt(0.57 * (1- 0.57)/21)  =  -1.31

The z-statistic is negative which means the sample statistic is less than the population parameter. 

A p-value for a z - statistic can be obtained in Excel, using the build in normal distribution functions. 

=NORM.S.DIST(-1.31,TRUE)  =  0.0951,  multiply this by 2 because the alternative is a two-tailed test. 

p-value = 2 * 0.0951  =  0.1901,  this p-value is greater than tthe value of alpha set, hence fail to reject the null hypothesis

 

 

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