Answer to Question #135360 in Statistics and Probability for Sana

Question

Eleven school boys were given a test in Drawing. They were given one
month's further tuition and a second test of equal difficulty was held at the end
of it. Do the marks give evidence that the students have benefited by the extra
coaching? Use α=0.05.
Roll
No
1 2 3 4 5 6 7 8 9 10 11
Marks
T1
28 20 19 21 18 20 18 17 23 16 19
Marks
T2
20 19 22 18 20 22 20 20 20 20 17
(b) The nine items of sample had the following values. 45, 47, 50, 52, 48, 47, 49,
53, 51. Does the mean of nine items differs significantly from an assumed
population mean of 47.5

Answer 1

solution

part a)

To compare means between 2 samples, we obtain the t value as:

$$t=\frac{\bar x_1-\bar x_²}{\sqrt{ \frac{\sigma_1^2}{n} + \frac{\sigma_2^2}{n} }}$$

Mean of test, T1:

$$\bar x_1 =\frac{\sum x_2}{n}=\frac{218}{11} =19.9091$$

Variance of test, T1:

$$\sigma_1^2=\frac{\sum (X_1-\bar x_1)^2}{n-1} =10.8909$$

Mean of test, T2:

$$\bar x_2 = \frac{\sum x_2}{n}=\frac{218}{11} =19.8182$$

Variance of test, T2:

$$\sigma_2^2=\frac{\sum (X_2-\bar x_2)^2}{n-1} =2.1636$$

Therefore, t value:

$$t=\frac{19.9091- 19.8182}{\sqrt{ \frac{10.8909}{11} + \frac{2.1636}{11} }}$$

$t\ value =0.083449$

The corresponding probability is $0.5324$ which is greater than the p value, 0.05

$Degrees\ of \ freedom, df = min(df_1,df_2) =10$

The critical value at $\alpha =0.05$

$$t_{\frac{\alpha}{2}, df} = t_{\frac{0.05}{2}, 10}=2.228$$

We conclude that there is no benefit from the coaching since the t value calculated (0.083449) is less than the critical value (2.228) and the probability (0.5324) is greater than 0.05

part b)

$$\sum x_i=442$$ $$Mean,\ \bar x= \frac{\sum x_i}{n} =49.11$$ $$Standard\ deviation, \sigma =\sqrt\frac{\sum (x_i-\bar x)^2}{n-1}=2.6194$$

The t statistic given $\mu=47.5$

$$t=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}} = \frac{49.11-47.5}{\frac{2.6194}{\sqrt{9}}}$$

$t \ value = 2.039974$

The corresponding probability is $0.102223$ which is greater than the p value 0.05

$Degrees\ of \ freedom, df=n-1=8$

The critical value at a 95% level of confidence

$$t_{\frac{\alpha}{2}, df} = t_{\frac{0.05}{2}, 8}=2.306$$

Since the t value (2.039974) is less than the critical value (2.306), and the p value (0.102223) is greater than 0.05, we conclude that the mean of the 9 items does not differ significantly from the assumed population with mean 47.5