Answer to Question #211974 in Statistics and Probability for Divina

Question

A.Describe the population parameter of interest;
B.Formulate the appropriate null and alternative hypotheses;and
C.Compute for the test-statistic value for population proportion.

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 and social networking sites.At 95% confidence,is the claim true?

2.The SSG Presidential claims that more than 80% of the students loves the new uniform design.50 students were asked about the new uniform design and 42 said they love it.Using 0.01 level of significance,is the claim true?

3.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 $\alpha=0.05.$

The population parameter of interest is population proportion $p.$

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.

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 $R=\{z:|z|>1.96\}$

C. 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 $\alpha=0.01.$

The population parameter of interest is population proportion $p.$

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

$p\leq0.8$

$p>0.8$

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

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

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

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.84-0.8}{\sqrt{\dfrac{0.8(1-0.8)}{50}}}=0.7071$$

Since it is observed that $z=0.7071<2.3263=z_c,$ 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 greater than $0.8,$ at the $\alpha=0.01$ significance level.

Using the P-value approach: The p-value is $p=P(Z>0.7071)=0.239752,$ and since $p=0.239752>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is greater than $0.8,$ at the $\alpha=0.01$ significance level.

3.

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

The population parameter of interest is population proportion $p.$

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.