Let [tex]\mu[/tex] be the population mean.
For the given claim , we have
Null hypothesis : [tex]H_0: \mu=35.9[/tex]
Alternative hypothesis : [tex]H_a: \mu\neq35.9[/tex]
Since alternative hypothesis is two-tailed , so the test is a two-tailed test.
Given : Sample size : n=220 ;
Sample mean: [tex]\overline{x}=35.6[/tex] ;
Standard deviation: [tex]s=2.2[/tex]
Test statistic for population mean:
[tex]z=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]
[tex]z=\dfrac{35.6-35.9}{\dfrac{2.2}{\sqrt{220}}}\approx-2.02[/tex]
By using the standard normal distribution table of z , we have
P-value ( two tailed test ) : [tex]2P(Z>|z|)=2(1-P(Z<|z|))[/tex]
[tex]=2(1-P(z<|-2.02|))=2(1-0.9783083)=0.0433834[/tex]
Since , the P-value is greater than the significance level of [tex]\alpha=0.02[/tex] , it means we do not have sufficient evidence to reject the null hypothesis.