Question

The production of a nationally marketed detergent results in certain workers receiving prolonged exposures to a Bacillus subtilis enzyme. Nineteen workers were tested to determine the effects of those exposures, if any, on various respiratory functions. One such function, airflow rate, is measured by computing the ratio of a person’s forced expiratory volume (FEV) to his or her vital capacity (VC). (Vital capacity is the maximum volume of air a person can exhale after taking as deep a breath as possible; FEV is the maximum volume of air a person exhale in one second.) In persons with no lung dysfunction, the “norm” for FEV/VC ratios is 0.80. Assume that the FEV/VC ratios are known to be normally distributed. For the 19 workers in the study, the mean FEV/VC ratio was 0.766 with standard deviation 0.0859. a. Based on your result, is it believable that exposure to Bacillus subtilis enzyme has no effect on the FEV/VC ratio? Please compute the p-value for the test in two different ways: (i) assuming σ = 0.09 and (ii) assuming σ is unknown. b. Based on this data, is it believable that σ = 0.09 ? Conduct a hypothesis test to answer this question.

Answer 1

a) Let be the true FEV/VC ratios in workers having exposure to Bacillus subtilis enzyme.

We want to test if exposure to Bacillus subtilis enzyme has no effect on the FEV/VC ratio, the “norm” for FEV/VC ratios being 0.80. That is we want to test if is equal to 0.80

The hypotheses are

We have the following sample information

n=19 is the sample size

is the sample mean FEV/VC ratio among 19 workers

is the sample standard deviation of  FEV/VC ratio among 19 workers

is the hypothesized value of mean FEV/VC ratio among the exposed (from the null hypothesis)

i) We know the population standard deviation of  FEV/VC ratio

The standard error of mean is

Since we know the population standard deviation, we will use normal distribution as the distribution of sample mean (even though the sample size is less than 30). That is we will use z-test.

The test statistics is

This is a 2 tailed test (The alternative hypothesis has "not equal to ")

The p-value is

We will reject the null hypothesis if the p-value is less than the significance level.

Here, let us assume a significance level . (5% significance)

The p-value of 0.099 is not less than the significance level of 0.05. Hence we do not reject the null hypothesis.

We conclude that at 5% level of significance, exposure to Bacillus subtilis enzyme has no effect on the FEV/VC ratio.

ii) We do not know the population standard deviation of  FEV/VC ratio

is unknown

We will estimate using the sample standard deviation

The estimated standard error of mean is

Since we do not know the population standard deviation and the sample size is less than 30, and assuming that the FEV/VC ratios are known to be normally distributed, we will use t distribution as the distribution of sample mean. That is we will use t-test.

The sample statistics is

The test statistics is

This is a 2 tailed test (The alternative hypothesis has "not equal to "). The degrees of freedom for t is n-1=19-1=18.

The p-value is

We can get an approximate p-value using the t tables. Using df=18, we can see that for t=1.734, we get the area under the right tail is 0.05 and total area under both the tails is 0.1. That means the value of P(T>1.734) = 0.05 and P(T>1.725) >0.05.

Hence the p-value is >0.10

(Using excel function =T.DIST.2T(1.725,18) we can get more accurate p-value=0.1017)  

We will reject the null hypothesis if the p-value is less than the significance level.

Here, let us assume a significance level . (5% significance)

The p-value of >0.1 is not less than the significance level of 0.05. Hence we do not reject the null hypothesis.

We conclude that at 5% level of significance, exposure to Bacillus subtilis enzyme has no effect on the FEV/VC ratio.

b) Let be the true value of standard deviation of FEV/VC ratio among the exposed.

We want to test the hypotheses

The test statistics is

the degrees of freedom for chi-square is n-1=19-1=18

Let us assume a significance level of alpha=0.05

this is a 2 tailed test (The alternative hypothesis has "not equal to"). This means that the area under each tail to get the critical value is alpha/2=0.05/2=0.025

Using the chi-square table for df=18, we get the right tail critical value using the area under right tail=0.025 as 31.526

and we get the left tail critical value using the area under right tail=1-0.025=0.975 as 8.231

The acceptance region is 8.231 to 31.526. That is we will reject the null hypothesis if the test statistics is out side the range 8.231 to 31.526.

Here, the test statistics is 16.40 and it lies with in 8.231 to 31.526.

We do not reject the null hypothesis.

We conclude that at 5% level of significance, it is believable that σ = 0.09