Answer:
(69.708 ; 74.292) ; 78% ; 350 ;
H0 : μ = 72
H0 : μ > 72
Pvalue = 0.8897
there is not enough evidence to support the claim that the mean time needed to complete the step is greater than 73.
Explanation:
99.5% confidence level, = 2.807
Sample size, n = 150
xbar = 72
Standard deviation, s = 10
Xbar ± Margin of error
Margin of Error = Zcritical * s/sqrt(n)
Margin of Error = 2.807 * 10/sqrt(150)
Margin of Error = 2.292
Lower boundary = 72 - 2.292 = 69.708
Upper boundary = 72 + 2.292 = 74.292
(69.708 ; 74.292)
B.)
Confidence level of the interval (71, 73)
(72 - 71 ) or (73 - 72)
Margin of Error = 1
Margin of Error = Zcritical * s/sqrt(n)
1 = Zcritical * 10/sqrt(150)
1 = Zcritical * 0.8164965
Zcritical = 1 / 0.8164965
Zcritical = 1.2247
α = 0.22 = 1 - 0.22 = 0.78 = 78%
C.)
Margin of Error = ±1.5
1.5 = Zcritical * s/sqrt(n)
99.5% confidence level, = 2.807
1.5 = 2.807 * 10/sqrt(n)
1.5 = 28.07 / sqrt(n)
Square both sides
1.5² = 28.08² / n
2.25n = 788.4864
n = 788.4864 / 2.25
n = 350.4384
n = 350 samples
D.)
H0 : μ = 72
H0 : μ > 72
Test statistic :
(xbar - μ) ÷ s/sqrt(n)
(73 - 72) ÷ 10/sqrt(150)
Test statistic = 1.2247
Pvalue :
P(Z < 1.2247) = 0.8897
Pvalue > α
0.8897 > 0.005
We fail to reject the Null ; Hence there is not enough evidence to support the claim that the mean time needed to complete the step is greater than 73.