Wildlife biologists inspect 153 deer taken by hunters and find 34 of them carrying ticks that test positive for Lyme disease. a) Create a 90% confidence interval for the percentage of deer that may carry such ticks. b) If the scientists want to cut the margin of error in half how many deer must they inspect? c) What concerns do you have about this sample?

(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,

The standard deviation is,

(a)

The confidence interval for proportions is:

For 90% confidence level the critical value of z is:

The confidence interval for the percentage of deer that may carry such ticks is:

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:

Half of this margin of error is:

The new sample size must be:

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.

(a) {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …}

(b) $59 398

step-by-step explanation:

(a) the annual sequence of profits

aₙ = a₁rⁿ⁻¹

the model for the sequence of profits is {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …} .

(b) profit after 15 yr

  a₁ = $ 30 000

  r = 1.05

a₁₅ = 30 000 × 1.05¹⁴

a₁₅ = 30 000 × 1.979 93

a₁₅ = $59 398

the profit in year 15 will be $59 398.