Question

A. Choose the significance level a.
B. Identify if it is two-tailed or one-tailed
C. Get the critical values from the test statistic table.
D. Sketch the critical regions.

1. The principal claims that 6 of every 10 learners have access to internet and social networking sites. Upon verification, only 40 out of 70 learners have access to internet & social networking sites. At 95% confidence,is the claim true?

2. The health worker claims that less than 30% of covid-19 cases are asymptomatic. Upon looking at 100 cases, it was found out that 19 of which are asymptomatic. At 99% confidence level,is the claim true?

Answer 1

1.

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

B. The following null and alternative hypotheses for the population proportion needs to be tested:

$p=0.6$

$p\not=0.6$

This corresponds to a two-tailed test, for which a z-test for one population proportion will be used.

C. The critical value for a two-tailed test with $\alpha=0.05$ is $z_c=1.96.$

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

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{\dfrac{40}{70}-0.6}{\sqrt{\dfrac{0.6(1-0.6)}{70}}}=-0.488$$

Since it is observed that $|z|=0.488<1.96=z_c,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is different than 0.6, at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value is $p=2P(Z<-0.488)=0.62555,$ and since $p=0.62555>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is different than 0.6, at the $\alpha=0.05$ significance level.

2.

A. Based on the information provided, the significance level is $\alpha=0.01.$

B. The following null and alternative hypotheses for the population proportion needs to be tested:

$p\geq0.3$

$p<0.3$

This corresponds to a left-tailed test, for which a z-test for one population proportion will be used.

C. The critical value for a left-tailed test with $\alpha=0.01$ is $z_c=-2.3263.$

The rejection region for this left-tailed test is $R=\{z:z<-2.3263\}$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.19-0.3}{\sqrt{\dfrac{0.3(1-0.3)}{100}}}=-2.4004$$

Since it is observed that $z=-2.4004<-2.3263=z_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is less than $0.3,$ at the $\alpha=0.01$ significance level.

Using the P-value approach: The p-value is $p=P(Z<-2.4004)=0.008189,$ and since $p=0.008189<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is less than $0.3,$ at the $\alpha=0.01$ significance level.