In an effort to promote healthy lifestyles, health screenings are given to emplo

Question

In an effort to promote healthy lifestyles, health screenings are given to employees of a large corporation. In running a promotional trial, 94 out of the 165 people who work in one office for the corporation participate in the health screening. a. Is the above information sufficient for you to be completely certain that more than 50% of all employees of the corporation will participate in the health screening? Why or why not? b. In establishing a statistical hypothesis testing of this situation, give the required null and alternative hypotheses for such a test, if it is desired that more than 50% of the employees participate in the health screening. H0: H1: c. Based on your answer in part b, should you use a right-tailed, a left-tailed, or a two-tailed test? Briefly explain how one determines which of the three possibilities is to be used. d. Describe the possible Type I error for this situation--make sure to state the error in terms of the percent of employees in the corporation who will participate in the health screenings. e. Describe the possible Type II error for this situation--make sure to state the error in terms of the percent of employees in the corporation who will participate in the health screenings. f. Determine the appropriate critical value(s) for this situation given a 0.05 significance level. g. Determine/calculate the value of the sample's test statistic. h. Detemine the P-value. i. Based upon your work above, is there statistically sufficient evidence in this sample to support that more than 50% of employees will participate in the health screening? Briefly explain your reasoning. In an effort to promote healthy lifestyles, health screenings are given to employees of a large corporation. In running a promotional trial, 94 out of the 165 people who work in one office for the corporation participate in the health screening. a. Is the above information sufficient for you to be completely certain that more than 50% of all employees of the corporation will participate in the health screening? Why or why not? b. In establishing a statistical hypothesis testing of this situation, give the required null and alternative hypotheses for such a test, if it is desired that more than 50% of the employees participate in the health screening. H0: H1: c. Based on your answer in part b, should you use a right-tailed, a left-tailed, or a two-tailed test? Briefly explain how one determines which of the three possibilities is to be used. d. Describe the possible Type I error for this situation--make sure to state the error in terms of the percent of employees in the corporation who will participate in the health screenings. e. Describe the possible Type II error for this situation--make sure to state the error in terms of the percent of employees in the corporation who will participate in the health screenings. f. Determine the appropriate critical value(s) for this situation given a 0.05 significance level. g. Determine/calculate the value of the sample's test statistic. h. Detemine the P-value. i. Based upon your work above, is there statistically sufficient evidence in this sample to support that more than 50% of employees will participate in the health screening? Briefly explain your reasoning.

Explanation / Answer

a)

No, because it is just a sample, not the population.

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b)

Formulating the null and alternatuve hypotheses,          
          
Ho:   p   <=   0.5
Ha:   p   >   0.5 [ANSWER]

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c)

As we can see, this is a RIGHT TAILED TEST. [ANSWER]

We can see this because we used a > sign for Ha.

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d)

A type I error is incorrectly concluding that more than 50% of all employees of the corporation will participate in the health screening, when in fact, it is not more than 50%.

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e)

A type I error is incorrectly concluding that not more than 50% of all employees of the corporation will participate in the health screening, when in fact, it is more than 50%.

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f)

As it is right tailed, 0.05 level,

zcrit = 1.645 [ANSWER]

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g)

As we see, the hypothesized po =   0.5      

Getting the point estimate of p, p^,          
          
p^ = x / n =    0.56969697      
          
Getting the standard error of p^, sp,          
          
sp = sqrt[po (1 - po)/n] =    0.038924947      
          
Getting the z statistic,          
          
z = (p^ - po)/sp =    1.790547572   [ANSWER, TEST STATISTIC]

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h)  
          
As this is a    1   tailed test, then, getting the p value,  
          
p =    0.036682963   [ANSWER, P VALUE]

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i)  

As P < 0.05, we   REJECT THE NULL HYPOTHESIS.      

There no sufficient evidence in this sample to support that more than 50% of employees will participate in the health screening. [CONCLUSION]

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