Answer:
The conclusion is that the average level of decibel at the library has changed since the installation of a new computer plug
Step-by-step explanation:
From the question we are told that
The population mean is [tex]\mu = 64[/tex]
The sample size is [tex]n = 47[/tex]
The sample mean is [tex]\= x = 63.2[/tex]
The sample standard deviation is [tex]\sigma = 5[/tex]
The level of significance is [tex]\alpha = 0.05[/tex]
The Null Hypothesis is
[tex]H_o : \mu = 64[/tex]
The Alternative Hypothesis is
[tex]H_a : \mu \ne 64[/tex]
The test statistics is mathematically evaluated as
[tex]t = \frac{\= x - \mu }{\frac{\sigma }{\sqrt{n} } }[/tex]
substituting values
[tex]t = \frac{63.1 - 64 }{\frac{5 }{\sqrt{47} } }[/tex]
[tex]t = -1.234[/tex]
The negative sign show that this is a left-tail test
Now the critical value of the level of significance obtained from the critical values table is
[tex]z_{0.05} = 1.645[/tex]
Now comparing the critical value of the [tex]\alpha[/tex] and the test statistics we see that critical value is greater than the test statistic which implies that the null hypothesis is rejected.
The conclusion is that the average level of decibel at the library has changed since the installation of a new computer plug
