Answer:
a; H0: u= 140 Ha: u ≠ 140 two tailed test.
b. Therefore reject H0 as t= 3 ≠ t≤ t ( ∝/2) (n-1) =2.353 or
3 > t ( ∝/2) (n-1) =2.353
c. If we check from the table the p- value is 0.6 which lies between 0.1 and 0.05 therefore reject H0.
d. Again reject H0 as 140 < 150.163
e. A type II error has not been committed as H0 is rejected.
Step-by-step explanation:
We formulate the null and alternative hypotheses as
a; H0: u= 140 Ha: u ≠ 140 two tailed test.
For a two tailed test the significance level ∝= 0.1 the critical region is given by
t ≤ t ( ∝/2) (n-1) and t > t ( ∝/2) (n-1)
So the critical region will be
t≤ t ( ∝/2) (n-1) =2.353
where
t= x` - u / s/ √n
Sr. No X X²
1 150 22500
2 150 22500
3 180 32400
4 170 28900
∑ 650 106,300
X`= ∑x/n = 650/4= 162.5
s²= 1/n-1 (x-x`)²= 1/n-1 [ ∑x² -(∑x)²/n ]
= 1/3[106,300 -650²/4] = 225
s= 15
Putting the values in the above equation
t= 162.5- 140/ 15/ √4
t= 3
So calculated value of t= 3
b. Therefore reject H0 as t= 3 ≠ t≤ t ( ∝/2) (n-1) =2.353 or
3 > t ( ∝/2) (n-1) =2.353
c. If we check from the table the p- value is 0.6 which lies between 0.1 and 0.05 therefore reject H0.
d. a 90% confidence interval based on the calculated values will be
x`± 1.645 (s)/ √n
Putting the values
162.5 ±1.645 ( 15/2)
162.5 ±12.3375
174.84 , 150.163
d. Again reject H0 as 140 < 150.163
e. A type II error has not been committed as H0 is rejected.
