Answer:
The condition are
The Null hypothesis is [tex]H_o : \mu = 5[/tex]
The Alternative hypothesis is [tex]H_a : \mu < 5[/tex]
The check revealed that
There is sufficient evidence to support the claim that in a small city renowned for its music school, the average child takes at least 5 years of piano lessons
Step-by-step explanation:
From the question we are told that
The population mean is [tex]\mu = 5 \ year[/tex]
The sample size is n = 20
The sample mean is [tex]\= x = 4.6 \ years[/tex]
The standard deviation is [tex]\sigma = 2.2 \ years[/tex]
The Null hypothesis is [tex]H_o : \mu = 5[/tex]
The Alternative hypothesis is [tex]H_a : \mu < 5[/tex]
So i will be making use of [tex]\alpha = 0.05[/tex] level of significance to test this claim
The critical value of [tex]\alpha[/tex] from the normal distribution table is [tex]Z_\alpha = 1.645[/tex]
Generally the test statistics is mathematically evaluated as
[tex]t = \frac{\= x - \mu}{ \frac{\sigma }{\sqrt{n} } }[/tex]
substituting values
[tex]t = \frac{ 4.6 - 5}{ \frac{2.2}{\sqrt{20} } }[/tex]
[tex]t = -0.8131[/tex]
Looking at the value of t and [tex]Z_{\alpha }[/tex] we see that [tex]t < Z_{\alpha }[/tex] so we fail to reject the null hypothesis
This implies that there is sufficient evidence to support the claim that in a small city renowned for its music school, the average child takes at least 5 years of piano lessons.
