Respuesta :
Answer:
(a) The confidence interval for the percentage of deer that may carry such ticks is (16.6%, 27.8%).
(b) The new sample size is 150.
Step-by-step explanation:
Let X = number of deer that carry the ticks.
The sample size is, n = 153.
The random variable X follows a binomial distribution with parameters n and p.
As the sample size is large according to the central limit theorem the sampling distribution of sample proportion follows a normal distribution.
The mean is, [tex]\mu_{\hat p} =\hat p=\frac{34}{153}= 0.222[/tex]
The standard deviation is,
[tex]\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n} }\\=\sqrt{\frac{0.222(1-0.222)}{153} } \\=0.034[/tex]
(a)
The confidence interval for proportions is:
[tex]\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex]
For 90% confidence level the critical value of z is:
[tex]z_{\alpha /2}=z_{0.10/2}=z_{0.05}=1.645[/tex]
The confidence interval for the percentage of deer that may carry such ticks is:
[tex]\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }=0.222\pm 1.645\times 0.034\\=0.222\pm 0.056\\=(0.166, 0.278)\\\approx(16.6\%,27.8\%)[/tex]
Thus, the confidence interval for the percentage of deer that may carry such ticks is (16.6%, 27.8%).
(b)
The margin of error is: [tex]z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex]
Half of this margin of error is:
[tex]New\ MOE=\frac{1}{2} \times z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }\\=\frac{1}{2}\times 1.645\times0.034\\ =0.028[/tex]
The new sample size must be:
[tex]New\ MOE=\frac{1}{2}\times z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }\\0.028=\frac{1}{2}\times 1.645\times \sqrt{\frac{0.222(1-0.222)}{n} }\\n=(\frac{1.645}{2\times 0.028} )^{2}\times (0.222(1-0.222)) \\=149.035\approx150[/tex]
Thus, the new sample size is 150.
(c)
The sample size and margin of error are inversely proportional.
If the margin of error is cut in half the sample size becomes 4 times.
But the new sample size after cutting the margin of error in half is 150.
So this is not correct.
