A researcher claims that 90% of people trust DNA testing. In a survey of 100 people, 91 of them said that they trusted DNA testing. Is the actual proportion of people who trust DNA testing larger than 90%? Test the researcher’s claim at the 1% level of significance.

Respuesta :

Keisho

You must know that % is a number out of 100.

Therefore given 91/100 trusted DNA surveying, we know as a percentage this is 91% from the equation below:

(91/100)*100=91

We can conclude that 1%=1 person.

Although the difference is only 1 (91-90=1), the actual proportion of people who trust DNA testing is larger than the 90% by 1%.

Testing the hypothesis, it is found that since the p-value of the test is of 0.37 > 0.01, there is not enough evidence to conclude that the actual proportion of people who trust DNA testing larger than 90%.

At the null hypothesis, we test if the proportion is of 90%, that is:

[tex]H_0: p = 0.9[/tex]

At the alternative hypothesis, it is tested if the proportion is larger than 90%, that is:

[tex]H_1: p > 0.9[/tex]

The test statistic is given by:

[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

In which:

[tex]\overline{p}[/tex] is the sample proportion.

p is the proportion tested at the null hypothesis.

n is the sample size.

For this problem, the parameters are: [tex]n = 100, \overline{p} = \frac{91}{100} = 0.91, p = 0.9[/tex]

Hence, the value of the test statistic is:

[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

[tex]z = \frac{0.91 - 0.9}{\sqrt{\frac{0.9(0.1)}{100}}}[/tex]

[tex]z = 0.33[/tex]

The p-value of the test is the probability of finding a sample proportion above 0.91, which is 1 subtracted by the p-value of z = 0.33.

Looking at the z-table, z = 0.33 has a p-value of 0.63.

1 - 0.63 = 0.37.

Since the p-value of the test is of 0.37 > 0.01, there is not enough evidence to conclude that the actual proportion of people who trust DNA testing larger than 90%.

A similar problem is given at https://brainly.com/question/24166849