The bacterial strain Acinetobacter has been tested for its adhesion properties, which is believed to follow a normal distribution. A sample of five measurements gave readings of 2.69, 5.76, 2.67, 1.62 and 4.12 dyne-cm. Assume that the standard deviation is known to be 0.7 dyne-cm_2 and that the scientists are interested in high adhesion (at least 2.66 dyne-cm^2)(a) Should the alternative hypothesis be one-sided or two-sided? Write down the null and alternative hypotheses.(b) Based on your answer to part (a), test the hypothesis to see if the mean adhesion is at least 2.66 dyne-cm_2 (Use the p-value approach) What can be concluded?

Answer:a) Alternative hypothesis should be one sided. Because Null and Alternative hypotheses are: [tex]H_{0}[/tex]: μ=2.66 dyne-cm. [tex]H_{a}[/tex]: μ<2.66 dyne-cm.b) the hypothesis that mean adhesion is at least 2.66 dyne-cm is true Step-by-step explanation:Let μ be the mean adhesion in dyne-cm. a) Null and alternative hypotheses are: [tex]H_{0}[/tex]: μ=2.66 dyne-cm. [tex]H_{a}[/tex]: μ<2.66 dyne-cm.b) First we need to calculate test statistic and then the p-value of it. test statistic of sample mean can be calculated as follows:t=[tex]\frac{X-M}{\frac{s}{\sqrt{N} } }[/tex] where X  is the sample mean M is the mean adhesion assumed under null hypothesis (2.66 dyne-cm) s is the standard deviation known  (0.7 dyne-cm_2)N is the sample size(5)Sample mean is the average of 2.69, 5.76, 2.67, 1.62 and 4.12 dyne-cm, that is [tex]\frac{2.69+5.76+2.67+1.62+4.12}{5}[/tex] ≈ 3.37using the numbers we get t=[tex]\frac{3.37-2.66}{\frac{0.7}{\sqrt{5} } }[/tex] ≈ 2.27The p-value is ≈ 0.043. Taking significance level as 0.05, we can conlude that sample proportion is significantly higher than 2.66 dyne-cm. Thus, according to the sample the hypothesis that mean adhesion is at least 2.66  dyne-cm is true