Answer :
Answer:
The conclusion is that the researcher was correct
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 13[/tex]
The sample mean is [tex]\= x = 9.63[/tex]
The standard deviation is [tex]s = 0.585[/tex]
The significance level is [tex]\alpha = 0.05[/tex]
The Null Hypothesis is [tex]H_o : \mu = 0[/tex]
The Alternative Hypothesis is [tex]H_a = \mu < 10[/tex]
The test statistic is mathematically represented as
[tex]t = \frac{\= x - \mu }{\frac{s}{\sqrt{n} } }[/tex]
Substituting values
[tex]t = \frac{9.63 - 10 }{\frac{0.585}{\sqrt{13} } }[/tex]
[tex]t = - 2.280[/tex]
Now the critical value for [tex]\alpha[/tex] is
[tex]t_{\alpha } = 1.645[/tex]
This obtained from the critical value table
So comparing the critical value of alpha and the test value we see that the test value is less than the critical value so the Null Hypothesis is rejected
The conclusion is that the researcher was correct