Prove - using hypothesis test - if this run is different from the desired weight

Question

Prove - using hypothesis test - if this run is different from the desired weight. (at 95% confidence level)
If the true weight differs from 100 by as much as 0.5, the test will detect this with a high probability (0.90).  Calculate the minimum recommended sample size.

97.71061 106.0732 108.5109 100.1829 97.00654 101.8695 98.6069 107.2078 97.3371 92.99752 102.3747 101.358 104.1138 99.18083 98.72277 102.9647 112.9711 99.54437 98.16246 99.2602 104.3621 101.3254 102.6288 109.371 106.003 102.1314 98.74391 117.051 106.9511 106.8762 108.0164 111.1756 100.8271 101.8082 95.95206 105.4606 104.7804 99.66627 106.1399 102.7639 107.0121 107.5361 115.7739 103.106 103.3009 105.4705 104.2064 98.26641 104.3478 110.8457

Explanation / Answer

mean=100

here we go for t-test and t=(x- -mean)/(S/sqrt(n))=(101.5938-100)/(4.8224/sqrt(18))=1.4022 with n-1=17 df

t-critical for 95%confidence=2.1

since t-calculate=1.4022 is less than t-critical for 95%confidence=2.1 , so we conclude that this run is not different from the desired weight.

answer of part 2) here margin of error=0.5

and 90 % margin of error=t(0.1/2,18)*(S/sqrt(n))=1.7341*4.8224/sqrt(n)

or 0.5=8.3624/sqrt(n)

or sqrt(n)=16.7248(next whole number is 17)

so minimum sample size should be 17

97.71061 106.0732 108.5109 100.1829 97.00654 101.8695 98.6069 107.2078 97.3371 92.99752 102.3747 101.358 104.1138 99.18083 98.72277 102.9647 112.9711 99.5 x-= 101.5938 S= 4.822384 n= 18

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.