Question

Homework 5, STAT 33001 Due April 9, 2019 For problems with R, include the R-code and the output from R with answers For test of hypothesis problems, if it is not specified, include the null- and alternative hy- pothesis, test statistic, critical value or P-value, and conclusion Learning Outcomes from this assignment: .Students will be able to perform hypothesis testing for one and two samples. Students will be able to perform hypothesis testing for paired samples. . Students will be able to caleulate and interpret p-values Students will be able to perform hypothesis testing in R and interpret the R output 1. The composition of the earths atmosphere may have changed over time. To try to discover the nature of the atmosphere long ago, we can examine the gas in bubbles inside ancient amber. Amber is tree resin that has hardened and been trapped in rocks. The gas in bubbles within amber should be a sample of the atmosphere at the time the amber was formed. Measurements on specimens of amber from the late Cretaceous era (75 to 95 million years ago) give these percents of nitrogen 63.4 65 64.4 63.3 54.8 64.5 60.8 49.1 51 Assume (this is not yet agreed on by experts) that these observations are an SRS from the late Cretaceous atmosphere. Test the hypothesis that the mean percent of nitrogen in ancient air is different than 57 percent at the 1% significance level. (A) Set up the null- and alternative hypothesis. (B) Determine the test statistic and its value. (C) Determine the critical value. (D) Determine the P-value. (E) Test this hypothesis using R (F) State your conclusion in practical term. (not just reject or fail to reject).

Answer 1

x<- c(63.4,65,64.4,63.3,54.8,64.5,60.8,49.1,51)
t.test(x,mu = 57,conf.level=0.99)

> t.test(x,mu = 57,conf.level=0.99)

        One Sample t-test

data:  x
t = 1.2416, df = 8, p-value = 0.2495
alternative hypothesis: true mean is not equal to 57
99 percent confidence interval:
 52.59259 66.58519
sample estimates:
mean of x 
 59.58889 

a)

$H_o : \mu = 57$

$H_a : \mu \neq 57$

b)

TS = (Xbar - mu)/(sd/sqrt(n))

TS = 1.2416

c)

df = n-1 = 8

critical value =

qt(0.995,8)

= 3.355387

d)

p-value = 0.2495

e) t.test(x,mu = 57,conf.level=0.99)

F)

since p-value > alpha

we fail to reject the null hypothesis

we conclude that there is not enough evidence to conclude that mean in different from 57% at 1 % level

Please rate