Answer to Question #232912 in Statistics and Probability for Kudzie

Question

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? (12)

1.3 A random sample of 200 male shoppers in Pavilion shopping mall were interviewed to identify 

their reasons for coming to this particular mall. The factor of ‘I prefer Knox brand in the mall’

was the most important reason for 120 of those interviewed. Estimate the likely percentage 

of all male shoppers who frequent this shopping mall primarily because of the Knox brand in 

the mall, using 97.5% confidence limits

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 $(0.5224, 0.6776).$