Question

1. A study of 99 dental hygienists were found to have an exposure to mercury vapors at an average of 0.08 mg/m³ (σ=0.005) over a 10-hour work shift. The regulatory limit for mercury vapors is 0.05 mg/m³ over a 10-hour work shift. Preform a one-mean hypothesis test to see if the exposures of the dental hygienists are above that allowable by the regulatory limit. Make sure to follow all 4 steps.

2. Construct a 95% confidence interval for the information in question 1.

Answer 1

1)

Let X be the amount of dental exposure to mercury vapors              
Data Summary

Step 1: Identify null and alternative hypothesis

The null and alternative hypotheses are              
Ho : μ = 0.05       μ is the population mean amount of dental exposure to mercury vapors      
Ha : μ > 0.05              
              
Let α = 0.05                ...level of significance

We use T test since population standard deviation is unknown, and sample size is small              
         

Step 2: Calculate t-statistic    

Using the below formula, we get the test statistic             

Degrees of Freedom              
df = n - 1 = 98              

t-statistic = 59.6992              
       

Step 3: Calculate p-value     
For t = 59.6992 df = 98 we find the Right Tailed p-value using Excel function t.dist               
p-value = t.dist.rt( 59.6992, 98)              
p-value = 0              
              
Step 4: Decision and conclusion
0 < 0.05              
that is p-value < α              
Hence we REJECT Ho              
              
Conclusion              
There exists enough statistical evidence at α = 0.05 to show that exposures of the dental hygienists are above the allowable regulatory limit.

            
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2)

95% confidence interval for the population mean

Since the population standard deviation is unknown, we use the t-distribution

Degrees of Freedom = df = n - 1 = 99 - 1 = 98

For 95% Confidence interval

α = 0.05,      α/2 = 0.025
From t tables of Excel function T.INV.2T (α, degrees of freedom) we find the t value
t = T.INV.2T (0.05, 98) = 1.984
We take the positive value of t

Confidence interval is given by


= (0.079, 0.081)

95% Confidence interval is mg/m³

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