Answer: NO.
Step-by-step explanation:
As per given , we have to test the hypothesis.
[tex]H_0:\mu=4.5\\\\ H_a:\mu\neq4.5[/tex]
∵ [tex]H_a[/tex] is two-tailed , so our test is a two-tailed test.
Also, the standard deviation is known to be 0.8 , so we use z-test.
Test statistic:[tex]z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
, where [tex]\overline{x}[/tex] = Sample mean
[tex]\mu[/tex] = population mean
[tex]\sigma[/tex] = Population standard deviation
n= Sample size
Put [tex]\overline{x}=4.6[/tex]
[tex]\mu=4.5[/tex]
[tex]\sigma=0.8[/tex]
n= 110 , we get
[tex]z=\dfrac{4.6-4.5}{\dfrac{0.8}{\sqrt{110}}}\approx1.31[/tex]
P-value for two tailed test = 2P(Z>|z|)
= 2P(Z>|1.31|) = 2(1-P(Z<1.31)) [∵ P(Z>z)=1-P(Z<z)]
=2(1- 0.9049) [By z-table]
=0.1902
Decision : ∵ P-value (0.1902) > Significance level (0.02).
It means we do not reject the null hypothesis.
[When P-values < Significance level then we reject the null hypothesis.]
Conclusion : We do not have sufficient evidence at the 0.02 level that the valve does not perform to the specifications.