2. 7pts) As the population ages, there is increasing concern about accident-rela

Question

2. 7pts) As the population ages, there is increasing concern about accident-related injuries to the elderly. A research article reported on an experiment in which the maximum lean angle-the furthest a subject is able to lean and still recover in one ales and a sample of older females. The following observations are consistent with summary data given in the article: YF: 29,34, 33, 27, 28, 32. 31, 34. 32. 27 OF: 18, 15, 23, 13, 12 Does the data suggest that the true average maximum lean angle for older females is more than 10 degrees smaller than it is for younger females at a significant level of .10? (Show a hypotheses, test statistic value, rejection region, and conclusion in terms of problem context) (Assume that the population distributions are normal

Explanation / Answer

Using R we can have the results as,

CODE:

YF <- c(29,34,33,27,28,32,31,34,32,27)
OF <- c(18,15,23,13,12)
var.test(YF,OF)
t.test(YF,OF,paired = F,alternative = "greater",var.equal = T,mu=10)

OUTPUT:

> YF <- c(29,34,33,27,28,32,31,34,32,27)

> OF <- c(18,15,23,13,12)

> var.test(YF,OF)

F test to compare two

variances

data: YF and OF

F = 0.38409, num df = 9, denom

df = 4, p-value = 0.2147

alternative hypothesis: true ratio of variances is not equal to 1

95 percent confidence interval:

0.04313402 1.81218919

sample estimates:

ratio of variances

0.3840948

> t.test(YF,OF,paired = F,alternative = "greater",var.equal = T,mu=10)

Two Sample t-test

data: YF and OF

t = 2.4441, df = 13, p-value = 0.01477

alternative hypothesis: true difference in means is greater than 10

95 percent confidence interval:

11.23937 Inf

sample estimates:

mean of x mean of y

30.7 16.2

Here first we are checking if the variances of YF and OF are equals or not.

So, here we are testing the null hypothesis the ratio of the variances of YF and OF 1 against the alternative he ratio is not equal to one.

Here the value of the test statistic is 0.38409

The critical region is, less than 0.2752479 and more than 5.998779

Here the value of the test statistic lies in the interval [0.2752479, 5.998779]. So, we can't reject the null hypothesis at 10% level of significance. We can conclude that variances are equal.

So, the next step is two-sample t-test with equal variances.

The null hypothesis is, the mean difference of YF and OF is 10 against the alternative the mean difference of OF and YF is greater than 10.

Here the value of the test statistic is, 2.4441.

The critical region is, more than 1.350171.

Here the value of the test statistic belongs to the critical region. So, we can't reject the null at 10% level of significance and conclude that the mean difference is same (not greater than) 10 degrees.

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