Question

Question 2. (10 points Use the following to answer questions a-d: A 2009 study to investigate the dominant paws in cats was described in Animal Behaviour Volume 78, Issue 2). The researchers used a random sample of 42 domestic cats. In this study, each cat was shown a treat (5 grams of tuna), and while the cat watched, the food was placed inside a jar. The opening of the jar was small enough that the cat could not stick its head inside to remove the treat. The researcher recorded the paw that was first used by the cat to try to retrieve the treat. This was repeated 100 times for each cat (over a span of several days). The paw used most often was deemed the dominant paw (note that one cat used both paws equally and was classified as "ambidextrous").Of the 42 cats studied, 20 were classified as "left-pawed" a. Verify that the sample is large enough to use the normal formula to find a confidence nterval for the proportion of domestic cats that are "left-pawed". (2 points) b. Construct a 95% confidence interval for the proportion of domestic cats that are-left. pawed". Use three decimal places in your margin of error. (2 points) c Provide an interpretation of your interval in the context of this data situation. (3 points) d. Another researcher wants to conduct a similar study to more precisely estimate the proportion of cats that are-left-pawed". They want to construct a 95% confidence interval that has a margin of error of 6%. How many cats does she need to use in her sample? (3 points

Answer 1

a) n = 42

    $\widehat p$ = 20/42 = 0.476

n$\widehat p$(1 - $\widehat p$) = 42 * 0.476 * (1 - 0.476) = 10.48

Since n$\widehat p$(1 - $\widehat p$) > 10, so we can use normal model.

b) At 95% confidence interval the critical value is z^* = 1.96

The 95% confidence interval for population proportion is

$\widehat p$ +/- z^* * sqrt($\widehat p$(1 - $\widehat p$)/n)

= 0.476 +/- 1.96 * sqrt(0.476 * (1 - 0.476)/42)

= 0.476 +/- 0.151

= 0.325, 0.627

c) We are 95% confident that the true proportion for the domestic cats that are "left pawed" lies between the confidence bounds 0.325 and 0.627.

d) Margin of error = 0.06

or, z^* * sqrt(p(1 - p)/n) = 0.06

or, 1.96 * sqrt(0.476 * (1 - 0.476)/n) = 0.06

or, n = (1.96 * sqrt(0.476 * (1 - 0.476))/0.06)^2

or, n = 267