Respuesta :
Answer:
a) The probability is P=0.3982.
b) No. There is no enough evidence to say that the proportion of the population is not 0.6.
Step-by-step explanation:
a) In this question, we have a sample of the population, of size n=1000. To know what is the probability of observing a sample proportion that is at least 0.64 we have to know the parameters of the sampling distribution.
The parameters of the sampling distribution of the proportion will be:
[tex]\mu=\pi=0.6\\\\ \sigma=\sqrt{\frac{\pi(1-\pi)}{N} } =\sqrt{\frac{0.6(1-0.6)}{1000} }= 0.0155[/tex]
With these parameters, we can calculate the z-value for p=0.64 as:
[tex]z=\frac{p-\pi}{\sigma}=\frac{0.64-0.6}{0.0155}= 0.258[/tex]
Then, we can calculate the probability of having a sample with p≥0.64:
[tex]P(p\geq0.64)=P(z\geq0.258)=0.3982[/tex]
b) To know if the proportion of the population is greater than 0.6 the rigth thing to do is perform a hypothesis test, in which we test the following hypothesis:
[tex]H_0: \pi\leq0.6\\\\H_1:\pi>0.6[/tex]
If the null hypothesis is rejected, we can conclude that there is evidence that the proportion of the population is greater than 0.6.
First, we assume a significance level of 0.05.
Second, we calculate the z value:
[tex]z=\frac{p-\pi-0.5/N}{\sigma} =\frac{0.64-0.6-0.5/1000}{0.0155} =\frac{0.0395}{0.0155} =2.54[/tex]
The P-value for this z is P=0.4. The P-value is greater than the significance level, what means that there is no evidence to reject the null hypothesis.
In other words, a sample mean of 0.64 is a quite probable value even if the proportion of the population is 0.6.
The probability of observing a sample proportion that is at least 0.64 is 0.3892.
How to calculate probability?
From the information given the z value for p will be:
= (0.64 - 0.60)/0.0155
= 0.258
The probability using the normal table will be 0.3892. Also, it can be deduced that there's no evidence to reject the null hypothesis.
Learn more about probability on:
https://brainly.com/question/24756209
