In: Statistics and Probability
Find the value of the standard score, z, and determine whether to reject the null hypothesis at a 0.05 significance level. Is the alternative hypothesis supported? Upper H 0 : mu equals18.2 meters, Upper H Subscript a : mu not equals18.2 meters, nequals 144, x overbar equals17.5 meters, sigma equals1.4 meters. Two decimal places as needed.
=18.2, =1.4 n= 144, = 17.5, = 0.05
Ho: = 18.2
Ha: 18.2
formula for test statistics is
z = -6.00
test statistics = -6.00
calculate p-Value
p-Value = 2 * P( z < -6.00)
find P( z < -6.00) using normal z table we get
P( z < -6.00) = 0
p-Value = 2 * 0
P-Value = 0
calculate critical value for two tailed test with = 0.05
using normal z table we get
Critical values are= ( -1.96 , 1.96 )
now rule to take decision is
Reject Ho if test statistics lies in rejection region.
from following graph we found that test statistics lies in rejection region
Hence, Null hypothesis is rejected.
Decision:
Therefore, there is enough evidence to support the alternative hypothesis.