Answer to Question #98087 in Statistics and Probability for denis

Question

Grouped data in a study to analyse performance in a 2012 examination series the following scores in percentage were compiled for a sample of 100 candidates.

42,51,52,53,61,62,63,64,71,72,81,46,53,54,55,63,64,65,66,73,74,82,48,57,58,59,67,68,69,77,78,87,43,54,56,64,65,66,67,74,75,84,47,52,53,54,62,63,64,65,72,73,82,45,56,57,58,66,67,68,69,76,77,86,44,55,56,57,65,66,67,75,76,85,49,50,51,52,60,61,62,70,71,51,52,53,61,62,63,71,72,58,59,57,68,69,67,78,79,80

a) Find an interval estimate of the actual average performance in the examination.

b) Suppose the expected average performance in the examination was 72 percent, would we be right to say that this group of candidates performed below the average?

c) A sample of 50 scores were obtained from 2011 series of a similar examination. The sample gave a mean score of 66.7 percent with a standard deviation of 3 percent.
Would we be right to conclude that there was no difference in performance between 2011 and 2012 series of the examination?

Answer 1

Arrange the data in equal classes of size 10 starting from 40 percent.

Frequency table

$$\begin{array}{c:c:c:c:c} \text{Class interval} &M & f &M\cdot f & M^2\cdot f\\ \hline 40-50 & 45 & 8 & 360 & 16200 \\ \hline 50-60 & 55 & 29 & 1595 & 87725 \\ \hline 60-70 & 65 & 35 & 2275 & 147875 \\ \hline 70-80 & 75 & 20 & 1500 & 112500 \\ \hline 80-90 & 85 & 8 & 680 & 57800 \\ \hdashline Sum & &100 & 6410 & 422100 \end{array}$$

From the table, we compute the sample mean $\bar{X}$ as follows

$$\bar{X}={1 \over N}\bigg(\displaystyle\sum_{i=1}^nM_i \cdot X_i\bigg)$$ $$\bar{X}={1 \over 100}\cdot 6410=64.1$$

Also, the sample variable is calculated as follows:

$$Var(X)={ 1\over N-1}\bigg(\displaystyle\sum_{i=1}^nM_i^2 \cdot X_i-{1 \over N}\big(\displaystyle\sum_{i=1}^nM_i \cdot X_i\big)^2\bigg)$$

$$Var(X)={ 1\over 100-1}\bigg(422100-{6410^2 \over 100}\bigg)\approx113.3232$$

Therefore, the sample sandartd deviation $s$ is directly computed by taking the square root from the sample variance.

$$s=\sqrt{Var(X)}\approx\sqrt{113.3232}\approx10.6453$$

a) The mean class is $60-70.$

b) Find 95 % confidence interval

$$z_c=1.96$$

$$CI=\bar{X}\pm z_c\cdot{s \over \sqrt{n}}$$ $$CI=64.1\pm 1.96\cdot{10.6453 \over \sqrt{100}}$$ $$CI=64.1\pm 2.086$$

$$(62.014, 66.186)$$

Find 99 % confidence interval

$$z_c=1.5758$$

$$CI=\bar{X}\pm z_c\cdot{s \over \sqrt{n}}$$

$$CI=64.1\pm 2.5758\cdot{10.6453 \over \sqrt{100}}$$

$$CI=64.1\pm 2.742$$

$$(61.358, 66.842)$$

We can be 99% confident that the population mean (μ) falls between 61.358 and 66.842.

Since $66.842<72$, then we would we be right to say that this group of candidates performed below the average.

c) A sample of 50 scores were obtained from 2011 series of a similar examination. The sample gave a mean score of 66.7 percent with a standard deviation of 3 percent.

Would we be right to conclude that there was no difference in performance between 2011 and 2012 series of the examination?

The provided sample means are shown below:

$\bar{X_1}=64.1, \bar{X_2}=66.7$

Also, the provided population standard deviations are:

$\sigma_1=10.6453, \sigma_2=3$

and the sample sizes are $n_1=100, n_2=50,$

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2$

$H_1: \mu_1\not=\mu_2$

This corresponds to a two-tailed test, for which a z-test for two population means, with known population standard deviations will be used.

Rejection Region

The significance level is $\alpha=0.05,$ and the critical value for a two-tailed test is $z_c=1.96$

The rejection region for this two-tailed test is $R=\{z:|z|>1.96\}$  

Test Statistics

The z-statistic is computed as follows:

$$z={\bar{X_1}-\bar{X_2} \over \sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}}$$

$$z={64.1-66.7 \over \sqrt{10.6453^2/100+3^2/50}}=-2.269$$

Since it is observed that $|z|=2.269>z_c=1.96,$  it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

$$P(Z<-2.269)=0.011634$$

$$p-value=2\cdot0.011634=0.023268$$

The p-value is $p=0.0233,$ and since $0.0233<0.05,$  it is concluded that the null hypothesis is rejected.

 We would not be right to conclude that there was no difference in performance between 2011 and 2012 series of the examination.