Question

You no longer believe that the population standard deviation in the amount of financial aid awarded to this population of students is a known quantity. You therefore will use the standard deviation in the amount of financial aid awarded for a representative sample of students as an estimate of the unknown population standard deviation. You collect award data from a random sample of students once again. This data is shown in appendix one below. Once again, at each of the 5% and 10% levels of significance, is the financial aid award different from $15,000? Again, in your discussion, comment upon the effect of the change in the level of significance on your decision, if necessary. Also, compare, at each level of significance, the results of this portion of the problem to those of the previous part. Account for any difference in your decisions at each level of significance between the two parts of the problem. Make this accounting based not only on a mathematical approach but rather on a conceptual justification.

(Amount of Financial Aid Awarded)

$14,200         $16,100         $18,400         $22,000         $15,200   $17,800         $19,300         $19,400         $13,500         $15,500 $21,100         $14,900         $12,000         $ 9,500           $14,000 $18,600         $13,800         $17,600         $16,000         $18,500 $19,300         $25,400         $11,500         $ 8,000           $ 6,000 $17,400         $14,400         $17,500         $12,900         $10,200 $20,400         $13,300         $18,000         $19,800         $12,600 $ 7,500           $15,900         $21,800

Answer 1

solution : The population standard deviation is unkonwn. Therefore we use the standard deviation in the amount of financial aid awarded for a representative sample of students as an estimate of the unknown population standard deviation.

The data given is,

(Amount of Financial Aid Awarded)

$14,200         $16,100         $18,400         $22,000         $15,200   $17,800         $19,300         $19,400         $13,500         $15,500 $21,100         $14,900         $12,000         $ 9,500           $14,000 $18,600         $13,800         $17,600         $16,000         $18,500 $19,300         $25,400         $11,500         $ 8,000           $ 6,000 $17,400         $14,400         $17,500         $12,900         $10,200 $20,400         $13,300         $18,000         $19,800         $12,600 $ 7,500           $15,900         $21,800

Here the sample mean is

> x=c(14200,16100,18400,22000 ,15200, 17800, 19300 ,19400 ,13500,15500,21100,
+ 14900 ,12000, 9500,14000,18600 ,13800,17600 ,16000 ,18500,19300,25400,
+ 11500 ,8000,6000, 17400 ,14400,17500 ,12900 ,10200,20400,13300,18000,
+ 19800,12600,7500, 15900 ,21800)
> x
[1] 14200 16100 18400 22000 15200 17800 19300 19400 13500 15500 21100 14900
[13] 12000 9500 14000 18600 13800 17600 16000 18500 19300 25400 11500 8000
[25] 6000 17400 14400 17500 12900 10200 20400 13300 18000 19800 12600 7500
[37] 15900 21800
> mean(x) # gives the sample mean
[1] 15771.05
> v=var(x)

> v
[1] 18426977
# gives vaience with (n-1) as denominator
> n=length(x)
> n
[1] 38

sd = square root of var

> sd= sqrt(v)
> sd
[1] 4292.666

thus s=4235.807

Here the estimated varience is, 18426977 as sd is 4292.666

We have to test, if the financial aid award different from $15,000

To test:  Test of means of one sample mean when population varience is not known

To test :

H0: µ =$15000

Vs H1: µ≠ $15000

Test statistic : t cal =  

T s/n

= $\frac{15771.05-15000 }{4292.666/\sqrt{38}}$

  

t cal = 1.1072

Decision : Here we reject H0 if table value |tcal | > t _{38, $\alpha$}    else accept it.

For α=0.05, from t distribution table we get,

Here, tcal = 1.1073 < t _{37 , 0.025 = 2.042}

Hence we accept H0

Now , For α=0.01, from t distribution table we get,

Here, tcal = 1.1073 < t _{37 , 0.005 = 2.524}

Hence we accept H0

Conclusion :

For both the critical regions we accept H0. hence at both los we can say that the financial aid award is not different from $15,000

The R commands are as follows

> t.test(x,mu=15000,conf.level=0.99) # at 1% los

One Sample t-test

data: x
t = 1.1073, df = 37, p-value = 0.2753
alternative hypothesis: true mean is not equal to 15000
99 percent confidence interval:
13880.14 17661.96
sample estimates:
mean of x
15771.05

> t.test(x,mu=15000,conf.level=0.95) # at 5% los

One Sample t-test

data: x
t = 1.1073, df = 37, p-value = 0.2753
alternative hypothesis: true mean is not equal to 15000
95 percent confidence interval:
14360.09 17182.02
sample estimates:
mean of x
15771.05

Now , we have to comment upon the effect of the change in the level of significance on decision

Here we can see that the CI for 99% is (13880.14, 17661.96)

Here we can see that the CI for 95% is ( 14360.09 ,17182.02)

The confidence length for 99% CI is larger.

Also, compare, at each level of significance,

When we compare the level of significance, we surely can say that if the hypothesis is accepted at 99% confidence then it will also accepted at 95% confidence. inverse is not true.

the hypothesis rejected at 99% confidence can be or can not be accepted at 95% confidence level.