Answer:
Kindly check explanation
Step-by-step explanation:
H0 : μ = 5500
H1 : μ > 5500
The test statistic assume normal distribution :
Test statistic :
(Xbar - μ) ÷ s/sqrt(n)
(5625.1 - 5500) ÷ 226.1/sqrt(15) = 2.1429 = 2.143
Pvalue from test statistic :
The Pvalue obtained using the calculator at df = 15 - 1 = 14 is 0.025083
α = 0.05
Since ;
Pvalue < α
0.025083 < 0.05 ; Reject H0
The confidence interval :
Xbar ± Tcritical * s/sqrt(n)
Tcritical at 95% = 1.761 ;
margin of error = 1.761 * 226.1/sqrt(15) = 102.805
Lower boundary : (5625.1 - 102.805) = 5522.295
(5522.295 ; ∞)
The hypothesized mean does not occur within the constructed confidence boundary. HENCE, there is significant eveidbce to support the claim that the true mean life of a biomedical device is greater than 5500
Answer =pi² Quotient[m, n]
Step-by-step explanation:
3x³+2ײ-39×-26
the answer is still the same
Answer:
2:1
Step-by-step explanation:
I hope it's helped.......
Add every decimal, then divide by 14 (two weeks, 7 days, total of 14 values), and the closest you get is 0.42.
