{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework 6 Solutions

# Homework 6 Solutions - 10-20 a 1 The parameter of interest...

This preview shows pages 1–2. Sign up to view the full content.

10-20 a) 1) The parameter of interest is the difference in mean impact strength, , with 0 = 0 2) H 0 : or 3) H 1 : or 4) The test statistic is 5) Reject the null hypothesis if t 0 < where = 1.714 for α = 0.05 since (truncated) 6) 290 321 12 22 n 1 = 10 n 2 = 16 7) Conclusion: Because - 4.64 < - 1.714 reject the null hypothesis and conclude that supplier 2 provides gears with higher mean impact strength at the 0.05 level of significance. P -value = P( t < - 4.64): P -value < 0.0005 b) 1) The parameter of interest is the difference in mean impact strength, 2) H 0 : 3) H 1 : or 4) The test statistic is 5) Reject the null hypothesis if t 0 > = 1.714 for α = 0.05 where 6)290 321 =25 12 22 n 1 = 10 n 2 = 16 7) Conclusion: Because 0.898 < 1.714, fail to reject the null hypothesis. There is insufficient evidence to conclude that t the mean impact strength from supplier 2 is at least 25 ft-lb higher that supplier 1 using α = 0.05.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

Homework 6 Solutions - 10-20 a 1 The parameter of interest...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online