Cayle the cat 3.1.4 Suppose I am conducting a test of significance where the null hypothesis is my cat Cayle will pick the correct cancer specimen 25% of the time and the alternative hypothesis is that she will pick the cancer specimen at a rate different than 25%. I end up with a p-value of 0.002. I also construct 95% and 99% confidence intervals from my data. What will be true about my confidence intervals

hello the options related to your question is missing attached below are the missing options

answer : Neither the 95% nor the 99% intervals will contain 0.25  ( B )

Step-by-step explanation:

Given that ;

H0 : = 0.25

Ha : ≠ 0.25

p-value = 0.002

also 95% and 99% confidence intervals are constructed

p value = 0.002        i.e. < 0.01  ∝ < 0.05 ∝

This means that we reject null hypothesis in both cases when  (∝ =0.05 and ∝ = 0.01 )

Hence The true statement about my confidence intervals is :

Neither the 95% nor the 99% intervals will contain 0.25

the answer is 375 miles

step-by-step explanation:

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40