Question

commercial perceptions is developed, and we want to estimate the mean score for all pilots. How many randomly selected pilots must be tested if we want the sample mean to be in error by no more than 3 points, with a 98% confidence? A preliminary study suggests that the population standard deviation is 24 points. 24. A college wishes to know the proportion of American adults who speak two or more languages. The survey includes asking 565 randomly selected American adults and finds that 226 speak two or more languages. a) Find the point estimate of the percentage of adults who speak two or more languages. b) Find a 96% confidence interval estimate of the percentage of Americans who speak two or more languages. Does the majority of Americans speak two or more languages? Explain.

Answer 1

We need to construct the 96% confidence interval for the population proportion. We have been provided with the following information about the number of favorable cases:

Favorable Cases X =	226
Sample Size N=	565

Part a

The sample proportion is computed as follows, based on the sample size N=565 and the number of favorable cases X=226:

X 226 Ô = Ñ = 565 = 0.4

part b

The critical value for α=0.04 is zc​=z1−α/2​=2.054. The corresponding confidence interval is computed as shown below:

CI(Proportion) = (0--p00,- 2) 2 + 36p .-)) 0.4(1 – 0.4) 0.4 +2.054 x V - 10.4 – 2.054 x 1 0.4(1 – 0.4) 565 565 = (0.358, 0.442)

Therefore, based on the data provided, the 96% confidence interval for the population proportion is 0.358<p<0.442, which indicates that we are 96\%96% confident that the true population proportion p is contained by the interval (0.358, 0.442).

since the 96 confidence interval interval points are less than 50% hence we can say that the majority of american do not speak two or more languages