Answer to Question #347108 in Statistics and Probability for erika miguel

Question

A study believes that 70% of adults in the Philippines own a cellphone.

A cellphone manufacturer believes that the actual number is much

less than 70%. 100 Filipino adults were surveyed, of which 74 have

cellphones. Using a 5% level of significance, is the cellphone

manufacturer’s claim valid or not?

Answer 1

Claim: the actual number is much less than 70%.

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

$H_0:p\ge0.70$

$H_a:p<0.70$

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

Evidence:

Based on the information provided, the significance level is $\alpha = 0.05 ,$ and the critical value for a left-tailed test is $z_c = -1.6449.$

The rejection region for this left-tailed test is $R = \{z: z < -1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{74/100-0.7}{\sqrt{\dfrac{0.7(1-0.7)}{100}}}=2.1822$$

Since it is observed that $z = 2.1822 \ge-1.6449= z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=P(Z<2.1822)= 0.985453,$ and since $p= 0.985453>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is less than 0.70, at the $\alpha = 0.05$ significance level.