Answer:
Yes we reject the null hypothesis
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 9[/tex]
The population mean is [tex]\mu = 40[/tex]
The sample mean is [tex]\= x = 33[/tex]
The standard deviation is [tex]\sigma = 9[/tex]
The level of significance is [tex]\alpha = 0.05[/tex]
For a two-tailed test
The null hypothesis is [tex]H_o : \mu = 40[/tex]
The alternative hypothesis is [tex]H_a : \mu \ne 40[/tex]
Generally the test statistics is mathematically represented as
[tex]t = \frac{\= x - \mu }{ \frac{\sigma}{\sqrt{n} } }[/tex]
=> [tex]t = \frac{ 33 - 40 }{ \frac{9}{\sqrt{9} } }[/tex]
=> [tex]t = -2.33[/tex]
The p-value for the two-tailed test is mathematically represented as
[tex]p-value = 2 P(z > |-2.33|)[/tex]
From the z-table
[tex]P(z > |-2.33|) = 0.01[/tex]
[tex]p-value = 2 * 0.01[/tex]
[tex]p-value = 0.02[/tex]
Given that [tex]p-value < \alpha[/tex] Then we reject the null hypothesis
