My teacher said that these questions would be similar to the questions on the fi

Question

My teacher said that these questions would be similar to the questions on the final and I was hoping that you could show me the work for this question so that i am not learning the incorrect methods.

Recall that 806 Stat 100 students responded to a survey in 2010 asking them, among other things, the number of drinks they had per week, the number of sex partners and the number of classes they skipped. Now a sample of 100 students is randomly chosen from all 806 students and the following multiple linear regression is fit for these 100 students: Slope SE Intercept 2.865 0.571 # of drinks 0.112 0.034 # of sex partners 0.087 0.050 Round all numbers to 2 decimal places. a. To see which slope is significant, compute the Z statistics for both slopes using the information in the table above (ignore intercept). Use this online calculator to find the p-value. The null hypothesis is that the population slope=0 and the alternative is that the population slope > 0. i) Z for slope of # of drinks = , p-value = Subm t Answer Tries 0/2 ii) Z for slope of # of sex partners = p-value= , Submt Answer Tries 0/2 b. Now do the same thing using the t statistics and the SE+ from the table below (ignore intercept). Use this online calculator to find the p-value Slope SE+ Intercept 2.865 0.580 # of drinks 0.112 0.035 # of sex partners 0.087 0.051 i) t for slope # of drinks = , p-value= Submt Answer Tries 0/3 ii) t for slope of # of sex partners = , p-value= Submt Answer Tries 0/3 The Z value for the 80% confidence interval is 1.28. Use the SE values in the first table to calculate the 80% confidence intervals for the slop i) 80% confidence interval for the slope of * of drinks = 0.112 ±

Explanation / Answer

Solutions:

Part a.i.

Z for slope of # of drinks = Slope coefficient / SE = 0.112/0.034 = 3.294118 = 3.29 approximately

P-value = 0.000494 = 0.00 approximately (by using z-table or excel)

Part a.ii.

Z for slope of # of sex partners = 0.087/0.050 = 1.74

P-value = 0.04093 = 0.04 approximately

Part b.i.

t for slope of # of drinks = 0.112/0.035 = 3.2

df = n – 1 = 100 – 1 = 99

p-value = 0.000924 = 0.00 approximately (by using t-table)

part b.ii.

t for slope of # of sex partners =0.087/0.051 = 1.705882 = 1.71 approximately

df = n – 1 = 100 – 1 = 99

p-value = 0.045583 = 0.05 approximately

part c.i.

80% confidence interval for the slope of # of drinks = 0.112 -/+ z*SE = 0.0112 -/+ 1.28*0.034

80% confidence interval for the slope of # of drinks = 0.112 -/+ 0.04352

Part c.ii.

80% confidence interval for the slope of # of sex partners = 0.087 -/+ z*SE = 0.087 -/+ 1.28*0.050

80% confidence interval for the slope of # of number of sex partners = 0.087 -/+ 0.064

Part d.i.

80% confidence interval for the slope of # of drinks = 0.112 -/+ t*SE = 0.0112 -/+ 1.29*0.035

80% confidence interval for the slope of # of drinks = 0.112 -/+ 0.04515

Part d.ii.

80% confidence interval for the slope of # of sex partners = 0.087 -/+ t*SE = 0.087 -/+ 1.29*0.051

80% confidence interval for the slope of # of number of sex partners = 0.087 -/+ 0.06579

Part e

Lower limit for confidence interval for the slope of # of drinks = 0.112 - 0.04515 = 0.06685

Upper limit for confidence interval for the slope of # of drinks = 0.112 + 0.04515 = 0.15715

Population slope of # of drinks lies within given confidence interval.

Lower limit for confidence interval for the slope of # of sex partners = 0.087 - 1.29*0.051 = 0.02121

Upper limit for confidence interval for the slope of # of sex partners =0.087 + 1.29*0.051 = 0.15279

Population slope of # of sex partners lies within given confidence interval.

So, both slopes fall within CI.

Part f

Correct answer = 80%

Explanation: As we know that the level of confidence interval denotes % of samples will have the sample statistic within given confidence interval.

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

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