Question

Question 27. The bacterial strain Acinetobacter has been tested for its adhesion properties. A sample of 20 measurements has been selected and the data is shown below. Assume that the standard deviation is known to be 0.66 dyne-cm2 5.9 2.9 2.3 3.5 3 3.6 1.8 3.2 5.1 1.6 5.7 3.8 3.1 1.7 6.2 6.1 4.1 5.6 1.7 a. The mean adhesion is 2.5 dyne-cm2 and the scientists are interested in high adhesion (at least 2.5 dyne-cm2). Write down the alternative and null hy pothe ses for the above scenario b. What is the average of the above sample? Calculate the value of the test stati stic. c. d. If you wish to draw a conclusion using the confidence interval technique, should you use a one-sided interval or a two-sided interval? Calculate an appropriate confidence interval using a confidence level of 959%? e. What is the critical value for a 5% level of significance? g. Use the confidence interval calculated in part e. to decide whether or not the mean adhesion is higher than 2.5 dyne-cm2. Explain your decision.

Answer 1

a) The null and alternative hypothesis

Η0. μΕΣ.5

Ha 2.5

b)

F =

= (5.9+2.3 +....+ 1.7) /20

= 3.795

c) Test statistic

X t Vn

  

3.795 2.5 0.66/20

= 8.77

d) We shall use one sided confidence interval as the test is one tailed .

e) The 95% one sided confidence interval for mean is

X tc s/n

For 95% confidence with df = 19

tc =1.7291(one tailed)

Therefore ,

The 95% one sided confidence interval for mean is

3.795 1.7291 0.66/V20

= (3.54, oo)

f) df =19,

0.05

critical value , tc = 1.7291

g) From the confidence interval we can see that , the lower level confidence interval > 2.5 (hypothetical mean )

We reject H0.

At 5% level , there is sufficient evidence to conclude that mean adhesion is higher than 2.5 dyne-cm2