Question

The following data represent the number of cycles to failure of aluminum test coupons subjected to repeated alternating stress at 21,000 psi, 18 cycles per second: 115 2130 1421 1674 3010 1260 1730 1535 1310 1016 1910 1888 1102 1781 1540 1109 1102 1018 1782 1452 1522 758 1792 1000 1578 1750 1502 1481 1605 1501 1258 1567 706 1890 1416 1238 1315 1883 2215 2100 1560 990 1085 798 1203 785 885 1594 1820 1055 1764 1468 1270 2023 1940 1512 1020 1015 845 1223 475 1315 1269 1120 1330 1750 865 910 1608 1642 Assuming that any outliers detected in your first assignment are the result of recording error and can be excluded: 1. Is it reasonable to assume sampling from a normal distribution? Explain 2. Would you accept the claim that the average number of cycles to failure is less than 1500? State the hypotheses and test using a- 2 %. 3. Construct a 99% confidence interval for estimating the standard deviation of the number of cycles to failure. Interpret your results.

Answer 1

Given that,
population mean(u)=1500
sample mean, x =1415.728
standard deviation, s =427.694
number (n)=70
null, Ho: μ=1500
alternate, H1: μ<1500
level of significance, α = 0.02
from standard normal table,left tailed t α/2 =2.093
since our test is left-tailed
reject Ho, if to < -2.093
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =1415.728-1500/(427.694/sqrt(70))
to =-1.649
| to | =1.649
critical value
the value of |t α| with n-1 = 69 d.f is 2.093
we got |to| =1.649 & | t α | =2.093
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :left tail - Ha : ( p < -1.6485 ) = 0.05189
hence value of p0.02 < 0.05189,here we do not reject Ho
ANSWERS
---------------
1.
normal distribution
2.
null, Ho: μ=1500
alternate, H1: μ<1500
test statistic: -1.649
critical value: -2.093
decision: do not reject Ho
p-value: 0.05189
we do not have enough evidence to support the claim that average number of cycles to failure
is less than 1500.
3.
CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ᴪ^2 right < σ^2 < (n-1) s^2 / ᴪ^2 left
where,
s = standard deviation
ᴪ^2 right = (1 - confidence level)/2
ᴪ^2 left = 1 - ᴪ^2 right
n = sample size
since alpha =0.01
ᴪ^2 right = (1 - confidence level)/2 = (1 - 0.99)/2 = 0.01/2 = 0.005
ᴪ^2 left = 1 - ᴪ^2 right = 1 - 0.005 = 0.995
the two critical values ᴪ^2 left, ᴪ^2 right at 69 df are 102.9962 , 42.494
s.d( s )=427.694
sample size(n)=70
confidence interval for σ^2= [ 69 * 182922.1576/102.9962 < σ^2 < 69 * 182922.1576/42.494 ]
= [ 12621628.8769/102.9962 < σ^2 < 12621628.8769/42.4935 ]
[ 122544.6072 < σ^2 < 297024.9303 ]
and confidence interval for σ = sqrt(lower) < σ < sqrt(upper)
= [ sqrt (122544.6072) < σ < sqrt(297024.9303), ]
= [ 350.0637 < σ < 544.9999 ]