Question

2. Let u be the population mean of Tasmanian devil head-to-tail lengths (in centimetres) in all of Tasmania now (2020). The research team decides to compare the current Tasmanian devil head-to-tail lengths mean with the mean from 20 years ago. The known mean Tasmanian devil head-to-tail lengths from 20 years ago is 65 cm. (a) Test the hypothesis whether y has changed from 65 cm. You must summarize all steps: state the null (H) and alternative hypotheses (Ha) relevant to the research objectives stated in this scenario, a suitable test statistic, its observed value in the sample, the sampling distribution for this statistic, the P-value, your summary of significance and your conclusion in plain language. (b) Some assumptions need to be made for the sampling distribution of the test statistic (as given in Part 2a) to be valid. State these assumptions, and briefly discuss whether these assumptions are satisfied. (c) Produce a 95% confidence interval for y, the true mean head-to-tail length. For this question you may assume that it is appropriate to use a t-distribution. Make sure you write down all the required steps to calculate this interval. (d) Does your confidence interval (constructed in Part 2c) include the value 65 cm? (e) Explain whether your confidence interval (constructed in Part 2c) is consistent conclusions from the hypothesis test in Part 2a. 1 your

The diagram above is data of current Tasmanian devils HTL length.

Answer 1

Solution : Given that the population mean is the mean of Tasmanian devils HTL length

let the given data of tasmanian devils HTL length is x. Here we know the population mean as 65cm but we do not know about the sd

thus we have to estimate the sd as below, R commands as follows

the estimate of population sd is sample varience.

> x
[1] 50.73 60.71 62.15 57.23 53.87 66.34 55.58 60.82 58.91 49.16 59.80 53.76
[13] 61.93 59.48 56.62 61.38 62.90 64.53 62.50 61.72 48.31 61.17 50.95 60.35
[25] 56.15 52.70 57.87 53.86 61.32 62.23 50.70 66.15
> mean(x)
[1] 58.18375
> v=var(x) # var of X
> v
[1] 24.70048
> n=length(x)
> n
[1]32
> sd= sqrt(v) # sd of X
> sd
[1] 4.969957

a) We have to test the mean is different from 65.

The population sd is unkonwn thus we have to use one sample t test.

To test

H0: The mean is not different from 65

vs H1 : The mean is different from 65

Test statistics :

t cal=  $\frac{\bar{x}-\mu }{s/\sqrt{n}}$

= $\frac{ 58.18375- 65 }{ 4.969957/\sqrt{32}}$

= -7.75832

Thus ,

decision : We reject H0 if |t cal | > t _{n-1 , $\alpha$ /2}

Here the |t cal|= 7.75832 >  t table value = 2.042 , thus we reject H0 at 5% los

Here we take los as 5% .

Conclusion : hence the value mean of Tasmanian devils HTL length has changed from 65 cm.

R commands to perform test are as,

> t.test(x,mu=65,conf.level=0.95)

One Sample t-test

data: x
t = -7.7583, df = 31, p-value = 9.402e-09
alternative hypothesis: true mean is not equal to 65
95 percent confidence interval:
56.39189 59.97561
sample estimates:
mean of x
58.18375

Here p value is 9.402e-09 < 0.05

thus at 5% los we reject the null hypothesis that the mean population THT has not changed from 65 cm

Therefore, the value mean of Tasmanian devils HTL length has changed from 65 cm.

b) The assumption that need to be satisfied are

i) The sample size must be less than 30

This assumption need to satisfy becouse t distribution is exct test and not approximation test as z test.

ii) The popultaion mean should be known and the population sd has to be estimated

we estimate it by using the sample varience

The Z test is general test but t test is exact test thus it gives rigid results about the sample

c)

Confidence interval for population mean

(
T
- t _n-1,∝/2 * s/$\sqrt{n}$ ,  
T
+ t_n-1∝/2 * s/ $\sqrt{n}$ )

Lower confidence bound =
T
- t_n-1∝/2 * s /  $\sqrt{n}$

=  58.18375 -   2.042*  4.969957/ 5.65685
= 56.3997

And upper confidence bound =  
T
+ t_n-1,∝/2 * s/ $\sqrt{n}$

= 58.18375+ 2.042 * 4.969957 / 5.65685

= 59.9777952

Thus the CI is given by ( 56.3997 , 59.9777952)

d) Here the population mean 65 does not lie in the confidence interval

e) This supports the decision of the rejection of the null hypothesis. Since the population mean is 65 which does not lie in the confidence indicates that the data does does not support the assumption that sample mean is 65 cm

Hence the test done in part 2a is correct.