{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Stat 312 Homework Problems - Week 10

# Stat 312 Homework Problems - Week 10 - Homework Solutions...

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

Homework Solutions Chapter 9 Section 9.1 2. The data is summarized in the table. a. The critical value is z = 1.96. The standard error for a difference in means is: The confidence interval for X Y is b. The research question is; does the mean height of women aged 20-39 exceed the mean height of women 60 and older by at least one inch? The following hypotheses take opposite sides of the question. H 0 : - = 1 (it does not) versus H a : - > 1 (it does). The large sample test statistic is The rejection region for a one-sided test on the right is , into which the test statistic clearly falls. Therefore, we reject the null hypothesis. The data strongly indicate that the mean height of the younger women exceeds the mean height of the older women by more than one inch. (They haven’t had to raise teenage daughters yet!) c. Using the Excel NORMSDIST function, the p-value for this test is P(Z > 5.64) = 1  5.64) = 1 0.999 999 9915 = .000 000 0085. Unless one believes in 117,612,352 to 1 odds, one would reject the null hypothesis. d. H 0 : - = 1 versus H a : - < 1 5. The data is summarized in the table. Since the population data is normal with known standard deviations, a “large sample” procedure is appropriate. The respective standard errors are and . a. The hypotheses are H 0 : - = 1.0 (= 0 ) versus H a : - < 1.0. The alternate hypothesis says that the mean heat output for non- sufferers of Reynaud’s syndrome exceeds that of sufferers by more than one cal/cm 2 /min. The test statistic is . At the = .01 level of significance, the critical value for this one-sided test on the left is z .99 = -2.33. Since -2.90 < -2.33, we reject the null hypothesis and conclude that the mean heat output of Reynaud’s syndrome suffers is more than one cal/cm 2 /min than the mean for non-suffers. b. The p-value = P(Z < -2.90) = .0019. c. In this problem, ’ represents the difference in means under the alternate hypothesis. That is, - = 1.2. Now d. Using m = n, set So the goal will be accomplished if n = 66. 8. The data is summarized in the table. a. The hypotheses are H 0 : Y - X = 10 versus H a : Y - X > 10 The test statistic is . The p-value is essentially zero, so at any level of significance, we reject H 0 . The data overwhelmingly suggest that the mean tensile strength of the 1078 grade exceeds the mean of the 1064 grade by more than 10 kg/mm 2 . b. A 95% confidence interval for Y - X will provide information on precision and reliability. The confidence interval is given by . We are 95% confident that the mean tensile strength of all 1078 grade wire rods exceeds the mean tensile strength of the 1064 grade wire rods by 15.59 to 16.41 kg/mm 2 .

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 / 4

Stat 312 Homework Problems - Week 10 - Homework Solutions...

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

View Full Document
Ask a homework question - tutors are online