Respuesta :
Answer:
We conclude that [tex]\mu\neq[/tex] 50.
Step-by-step explanation:
We are given that x bar = 60, s = 12, n = 30, and alpha = 0.10
Also, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 50
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] [tex]\neq[/tex] 50
The test statistics we will use here is;
T.S. = [tex]\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, X bar = sample mean = 60
s = sample standard deviation = 12
n = sample size = 30
So, Test statistics = [tex]\frac{60-50}{\frac{12}{\sqrt{30} } }[/tex] ~ [tex]t_2_9[/tex]
= 4.564
At 10% significance level, t table gives critical value of 1.311 at 29 degree of freedom. Since our test statistics is more than the critical value as 4.564 > 1.311 so we have sufficient evidence to reject null hypothesis as our test statistics will fall in the rejection region.
P-value is given by, P([tex]t_2_9[/tex] > 4.564) = less than 0.05% {using t table}
Here also, P-value is less than the significance level as 0.05% < 10% , so we will reject null hypothesis.
Therefore, we conclude that [tex]\mu\neq[/tex] 50.
