Question

Please show work

2.) Which is better for student learning - online HW or handwritten HW? Suppose students who do HW online are equally successful as students who complete handwritten HW, so we take a nationwide sample of college classes & measure success based on the average score on exams. Below is the data: Online HW Handwritten HW Conduct a hypothesis test at the 5% level to determine if the average success using online HW is different a) State the null & alternative hypotheses, in the context of the problem. b) Describe the type I & type II errors in the context of the problem. c) Describe the sample statistic in the context of the problem. Find its distribution & explain why d) Use your TI to find the p-value, sketch the region whose area the probability represents (label the appropriate values along the horizontal axis), and define the p-value in the context of the problem. e) Make your conclusion in the context of the problem. Explain why. we are interested in whether sample average score 77.5, sample standard deviation 12, sample size 60 sample average score 75.8, sample standard deviation = 12, sample size = 45

Answer 1

a) The null hypothesis is  
Ho1-20
   and the alternative hypothesis is  
Ha 12 O
.

(where   $\mu_1$    population average using online HW and      population average using handwritten HW.)

2

b) Type-I error:

In this problem type-I error is the error generated due to rejecting the null hypothesis based on the sample information i.e. concluding that the average success using online HW is different when actually the average success using online HW is not significantly different for the population.

Type-II error:

In this problem type-II error is the error generated due to accepting the null hypothesis based on the sample information i.e. concluding that the average success using online HW is not different when actually the average success using online HW is significantly different for the population.

c) (Assuming the two population standard deviations are the same because the two given sample standard deviations are same and we know that, sample standard deviations are the good unbiased estimators of population standard deviations)

The sample statistic for this test is   ,

X T = 2 1 1 s S n1 n2 E

where  
(n,-1)s (n2 -1) s2 308.038835 n1n2 2

X, Х
   are the sample means   and   $n_1,n_2$    are the sample sizes of the respective samples.

Let the population standard deviation for both the samples is   $\sigma$ .

(Assuming the samples are from  
2 W(11,0
   and   $N(\mu_2,\sigma^2)$    respectively.)

We know  
X~ N(1,
   and  
X2~ N(42 n2

$\therefore \bar{X_1}-\bar{X_2} \sim N\left ( \mu_1-\mu_2, \frac{\sigma^2}{n_1}+\frac{\sigma^2}{n_2} \right )$

i.e.   $\frac{(\bar{X_1}-\bar{X_2})-(\mu_1-\mu_2)}{\sigma\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \sim N(0,1)$

Now under   $H_0$ ,

$\frac{\bar{X_1}-\bar{X_2}}{\sigma\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \sim N(0,1)$

And we know   $\frac{(n_1-1)s_1^2}{\sigma^2} \sim \chi^2_{n_1-1}$    and   $\frac{(n_2-1)s_2^2}{\sigma^2} \sim \chi^2_{n_2-1}$

So,   $\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{\sigma^2} \sim \chi^2_{n_1+n_2-2} \Rightarrow \frac{(n_1+n_2-2)s^2}{\sigma^2}\sim \chi^2_{n_1+n_2-2}$

So under   $H_0$ ,   $\frac{\frac{\bar{X_1}-\bar{X_2}}{\sigma\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}}{ \sqrt{\frac{(n_1+n_2-2)s^2}{\sigma^2}/(n_1+n_2-2)}} \sim t_{n_1+n_2-2} \Rightarrow T \sim t_{n_1+n_2-2}$ .

So the sample statistic follows t distribution with degrees of freedom $(n_1+n_2-2)$ .

Here the value of the sample statistic is   $T=\frac{77.5-75.8}{\sqrt{308.038835}\times\sqrt{\frac{1}{60}+\frac{1}{45}}}=0.491172(approx.)$

d) Here p-value is the probability of the sample statistic being more extreme than the calculated value of it, i.e. the probability of      being more extreme than the calculated value 0.491172.

The p-value of the test is,

$2Min \left \{ P(T\leq0.491172),P(T\geq0.491172) \right \} =2P(T\geq0.491172)$

$=2\times 0.3121749=0.6243498(approx.)$

T049 T O491172 CS Scanned with CamScanner

The above probability represents the shaded area is the picture.

e) In this test, we failed to reject the null hypothesis based on the sample information since the p-value of the test is greater than the given significance level   $\alpha=0.05$ .