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
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] ≈ 3.37
using the numbers we get
t=[tex]\frac{3.37-2.66}{\frac{0.7}{\sqrt{5} } }[/tex] ≈ 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
