Question

A lecturer wants to know if his statistics class has a good grasp of basic maths from matric level. Eleven students are chosen at random from the class and given a maths proficiency test. The lecturer wants the class to be able to score above 75 on the test to show that they had a good grasp of basic maths. The eleven students get scores shown below: 62 88 71 50 67 70 92 75 68 83 95 Can the lecturer be 95 percent confident that the students had a good grasp of basic maths?

Answer 1

$$\bar{x}=\dfrac{\sum_ix_i}{n}=\dfrac{1}{11}(62+88+71+50+67+70$$ $$+92+ 75 +68+ 83+ 95)=\dfrac{821}{11}=74.6364$$

$$s^2=\dfrac{\sum_i(x_i-\bar{x})^2}{n-1}=\dfrac{1}{11-1}((62-\dfrac{821}{11})^2$$

$$+(88-\dfrac{821}{11})^2+(71-\dfrac{821}{11})^2+(50-\dfrac{821}{11})^2$$

$$+(67-\dfrac{821}{11})^2+(70-\dfrac{821}{11})^2+(92-\dfrac{821}{11})^2$$

$$+(75-\dfrac{821}{11})^2+(68-\dfrac{821}{11})^2+(83-\dfrac{821}{11})^2$$

$$+(95-\dfrac{821}{11})^2=186.854545$$

$$s=\sqrt{s^2}=13.6695$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu\leq75$

$H_1:\mu>75$

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha=0.05,$

$df=n-1=11-1=10$ ​degrees of freedom, and the critical value for a right-tailed test is$t_c= 1.812461.$

The rejection region for this right-tailed test is $R=\{t:t>1.812461\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{74.6364-75}{13.6695/\sqrt{11}}=-0.088223$$

Since it is observed that $t=-0.088223<1.812461=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for right-tailed, $\alpha=0.05, df=10,$

$t=-0.088223$ ​is$p=0.534279,$ and since $p=0.534279>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected

Therefore, there is not enough evidence to claim that the population mean $\mu$ is greater than 75, at the $\alpha=0.05$ significance level.

Therefore, there is not enough evidence to claim that the students had a good grasp of basic maths, at the $\alpha=0.05$ significance level.

1.3 The sample proportion is computed as follows, based on the sample size $n=200$ and the number of favorable cases $X=120$

$$\hat{p}=\dfrac{X}{n}=\dfrac{120}{200}=0.6$$

The critical value for $\alpha=0.025$ is $z_c=z_{1-\alpha/2}=2.2414.$

The corresponding confidence interval is computed as shown below:

$$CI(proportion)=\bigg(\hat{p}-z_c\times\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}},$$

$$\hat{p}+z_c\times\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\bigg)$$

$$=\bigg(0.6-2.2414\times\sqrt{\dfrac{0.6(1-0.6)}{200}},$$

$$0.6+2.2414\times\sqrt{\dfrac{0.6(1-0.6)}{200}}\bigg)$$

$$=(0.5224, 0.6776)$$

Therefore, based on the data provided, the 97.5% confidence interval for the population proportion is $0.5224<p<0.6776,$ which indicates that we are 97.5% confident that the true population proportion $p$ is contained by the interval $(0.5224, 0.6776).$