According to the Fiji Diabetes Association, 23.1% of Fijians aged 60 years or older had diabetes in 2017. A recent random sample of 200 Fijians aged 60 years or older showed that 52 of them have diabetes. Using a 5% significance level, perform a test of hypothesis to determine if the current percentage of Fijians aged 60 years or older who have diabetes is higher than that in 2017. Use the P-value method

Respuesta :

Answer:

The point estimated is 0.260

Step-by-step explanation:

The computation is shown below:

Sample in which the people have diabetes = 52

Random sample = 200

Significance level = 5%

Based on this, the, estimation point is

= Sample in which the people have diabetes ÷ Random sample

= 52 ÷ 200

= 0.260

hence, the point estimated is 0.260

We simply applied the above formula

And, the same is to be considered

The Null Hypothesis is accepted because the value of z is 0.973 and this can be determined by performing the test of the Hypothesis.

Given :

  • Association, 23.1% of Fijians aged 60 years or older had diabetes in 2017.
  • A recent random sample of 200 Fijians aged 60 years or older showed that 52 of them have diabetes.
  • Use a 5% significance level.

Hypothesis are as follows:

Null hypothesis,  [tex]\rm H_0: p\leq 0.231[/tex]

Alternate hypothesis,  [tex]\rm H_a : p>0.231[/tex]

Now, performing the test of the hypothesis. The critical value z is given by:

[tex]z = \dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}[/tex]    ----- (1)

Now, the value of [tex]\hat {p}[/tex] is given by:

[tex]\hat{p}=\dfrac{52}{200} = 0.26[/tex]

Now, put the values of known terms in equation (1).

[tex]z = \dfrac{0.26-0.231}{\sqrt{\dfrac{0.231\times 0.769}{200}}}[/tex]

[tex]z = \dfrac{0.029}{0.0298}[/tex]

z = 0.973

Therefore, the null hypothesis is accepted.

For more information, refer to the link given below:

https://brainly.com/question/19613147