Question

4. One defence that organisms have against predators is having a disagreeable taste, odour or spray. One such insect is the walking stick. Of 50 randomly sampled blue jays that had been sprayed by a walking stick, 42 remained aloof from the insect for at least 2 weeks. Let p be the true proportion of blue jays that would avoid further contact with a walking stick for at least 2 weeks. (a) Use this observation to find a 90% confidence interval for p. (b) Use these data as a pilot study to determine the sample size needed to estimate p within 2 percentage points with 98% confidence. (c) Does this observation provide evidence that p is greater than 0.80? i. Using standard notation, define the population parameter(s) being tested. ii. Specify the null and alternative hypotheses. iii. Compute the observed value of the test statistic. iv. Specify the distribution to be used for computing the p-value and compute the p- value within table accuracy. v. State your conclusion and report the estimated value of the parameter being tested and the estimated) standard error. vi. What assumptions have you made to carry out this test?

Answer 1

p: be the true proportion of blue jays that would avoid further contact with a walking stick for at least 2 weeks.

x = number of blue jays that remained aloof from the insect from at least 2 weeks = 42

n = number of blue jays randomly selected = 50

42 sample proportion = 50 - 0.84

a) 90% confidence interval for p

Formula of confidence interval for populaton proportion p is

p(1-7) Ê – ZcV !<p<ộ+ Ze p(1-P) n n

Where Z is the critical value at given confidence level, and the confidence interval is two-sided

c = 90% = 0.90

alpha = 1 - c = 0.10

1 - (alpha/2) = 1 - (0.10/2) = 1 - 0.05 = 0.95

Using the z table the criitcal value for the area 0.95 is 1.645

Plug the values in the formula

0.84 – 1.645 1.645V 0.84(1 – 0.84) 50 <p < 0.84 + 1.645V 0.84(1 – 0.84) 50

0.75 < p < 0.93

(0.75, 0.93) is the 90% confidence interval for population proportion p.

b)

Margin of error = E = 2% = 0.02

phat = p^ = 0.84 as the pilot study estimate

c = 98% = 0.98

The formula of sample size is,

$n = \hat p (1 - \hat p)\left ( \frac{Z_c}{E} \right )^2$

Where Zc is the critical value for the given confidence level.

Alpha = 1 - c = 1 - 0.98 = 0.02

1 - (alpha/2) = 1 - 0.01 = 0.99

Using the z table the z critical value for area 0.99 is 2.33

Plug the values in the formula of n,

$n = \hat p (1 - \hat p)\left ( \frac{Z_c}{E} \right )^2 = 0.84(1-0.84)\left ( \frac{2.33}{0.02} \right ) =1824.11\approx 1824$

Sample size needed = 1824

c) Claim p is greater than 0.80 that is p > 0.80

i. Population parameter is p

p: be the true proportion of blue jays that would avoid further contact with a walking stick for at least 2 weeks.

ii. Null and alternative hypotheses

$H_0: p\leq 0.80 \ and \ H_1: p > 0.80$

iii. Observed value of test statistics

The formula of test statistics is,

$z = \frac{\hat p - p}{\sqrt{p(1-p)/n}} = \frac{0.84 - 0.80}{\sqrt{0.80(1-0.80)/50}} =0.71$

Observed value of tets statistics = Z = 0.71

iv) The distribution used to compute the p-value is normal distribution.

Here the normal approximation is used for sampling distribution.

The alternative hypothesis contains greater than sign so the test is right tailed test.

P-value = P(Z > test statistics) = P(Z > 0.71)

Using the normal table the probability for z = 0.71 is 0.7611, but table provides the less than probability. To find the greater than probability just subtract the less than from 1

P-value = 1 - 0.7611 = 0.2389

v) If P-value > alpha then fail to reject the null hypothesis otherwise reject the null hypothesis.

Alpha is not given so take 0.05

p-value (0.2389) is more than 0.05 so fail to reject the null hypothesis

If alpha is 0.10 or 0.02, then that are also less than p-value so fail to reject the nul hypothesis.

Fail to reject the null hypothesis means there is no sufficient evidence to support the claim that the proportion p is greater than 0.80

Estimated value of paramter

$\mu_{\hat p} = p = 0.80$

Estimated standard error

$\sigma_{\hat p} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.80(1-0.80)}{50}} = 0.051845926$

vi) Assumptions.

-The sample should be a random sample.

-The sample size is no more than 5% of the population size.

- n*p and n * (1-p) both are greater than or equal to 5.