Answer: d. −1.728
Step-by-step explanation:
As per given , we have
Null hypothesis : [tex]H_0: \mu\geq40[/tex]
Alternative hypothesis : [tex]H_a: \mu<40[/tex] (against Null hypothesis )
Alternative hypothesis is one-tailed , then the test is one-tailed test.
Sample size : n= 18
Sample mean : [tex]\overline{x}=37.8[/tex]
Sample standard deviation : s= 5.4
Population standard deviation is unknown , so we should perform t-test.
Test statistic : [tex]t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]
Substitute the values, we get
[tex]t=\dfrac{37.8-40}{\dfrac{5.4}{\sqrt{18}}}[/tex]
[tex]t=\dfrac{-2.2}{\dfrac{5.4}{4.24264068712}}[/tex]
[tex]t=\dfrac{-2.2}{1.27279220614}[/tex]
[tex]t=-1.72848324289\approx1.728[/tex]
Hence, the test statistic is −1.728.