The bacterial strain Acinetobacter has been tested for its adhesion properties. A sample of five measurements gave read- ings of 2.69, 5.76, 2.67, 1.62 and 4.12 dyne-cm2. Assume that the standard deviation is known to be 0.66 dyne-cm2 and that the scientists are interested in high adhesion (at least 2.5 dyne-cm2). a. Should the alternative hypothesis be one-sided or two-sided

If we compare the p value and the significance level assumed [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, we can conclude that the true mean is higher than 2.5 at 1% of signficance.

Step-by-step explanation:

a. Should the alternative hypothesis be one-sided or two-sided?

b. Test the hypothesis that the mean adhesion is 2.5 dyne-cm².

c. What is the P-value of the test statistic?

Data given and notation

Data: 2.69, 5.76, 2.67, 1.62 ,4.12

We can calculate the sample mean with this formula:

[tex]\bar X= \frac{\sum_{i=1}^n X_i}{n}[/tex]

[tex]\bar X=3.372[/tex] represent the sample mean

[tex]\sigma=0.66[/tex] represent the population standard deviation

[tex]n=5[/tex] sample size

[tex]\mu_o =2.5[/tex] represent the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

Part a: State the null and alternative hypotheses.

We need to conduct a one tailed test

We need to conduct a hypothesis in order to check if the mean is higher than 2.5, the system of hypothesis would be:

Null hypothesis:[tex]\mu \leq 2.5[/tex]

Alternative hypothesis:[tex]\mu > 2.5[/tex]

If we analyze the size for the sample is < 30 but we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Part b: Calculate the statistic

We can replace in formula (1) the info given like this:

[tex]z=\frac{3.372-2.5}{\frac{0.66}{\sqrt{5}}}=2.954[/tex]

Part c :P-value

Since is a one sided test the p value would be:

[tex]p_v =P(z>2.954)=0.0016[/tex]

Conclusion

If we compare the p value and the significance level assumed [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, we can conclude that the true mean is higher than 2.5 at 1% of signficance.