Respuesta :
Answer:
a) Population proportion = 24%
b) Standard deviation of the sample proportion = 0.0302
c) Yes. The proportion sampling distribution can be modeled with a binomial distribution, and satisfies the conditions to be approximated to a normal distribution.
d) P(p<0.25)=0.6927
e) Based on the sampling distribution model for this scenario, we should expect about 95% of the samples to show that between 18.1% and 29.9% celebrate their pet’s birthday.
Step-by-step explanation:
a) The expected proportion of the population that celebrate their pet's birthday can be estimated without bias from the sample proportion. Then, we can say that the expected population proportion is 24%.
b) The standard deviation of the sample proportion is:
[tex]\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.24*0.76}{200}}\\\\\\ \sigma_p=\sqrt{0.00091}=0.0302[/tex]
c) It is expected to be normal, as it can be modeled with a binomial distribution (constant probability of success, repeated Bernoulli experiments).
This binomial distribution can be approximated to a normal distribution is np and (1-n)p are greater than 10.
[tex]np=200*0.24=48\\\\n(1-p)=200*0.76=152[/tex]
d) To calculate this probability we use the z-score for 25%.
[tex]z=\dfrac{p-\pi}{\sigma_p}=\dfrac{0.25-0.24}{0.0302}=\dfrac{0.01}{0.0302}=0.3311\\\\\\P(p<0.25)=P(z<0.3311)=0.6297[/tex]
The probability that the proportion of people in the sample who celebrate their pet’s birthday is less than 25% is P=0.6297.
e) The critical z-value for a 95% confidence interval is z=1.96.
The margin of error (MOE) can be calculated as:
[tex]MOE=z\cdot \sigma_p=1.96 \cdot 0.0302=0.059[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=p-z \cdot \sigma_p = 0.24-0.059=0.181\\\\UL=p+z \cdot \sigma_p = 0.24+0.059=0.299[/tex]
The 95% confidence interval for the population proportion is (0.181, 0.299).
