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?

a) Alternative hypothesis should be one sided. Because Null and Alternative hypotheses are:

: μ=2.66 dyne-cm.

: 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:

: μ=2.66 dyne-cm.

: 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= 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 ≈ 3.37

using the numbers we get

t= ≈ 2.27

The 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

40 because it’s just a little longer than 32 feet

Which of the following values are in the range is the function graphed below? check all that applya.3b.1c.0d.-2e.5f.2a