Question

a)A university planner wants to determine the proportion of spring semester students who will attend summer school. Suppose the university would like a 0.90 probability that the sample proportion is within 0.281 or less of the population proportion.What is the smallest sample size to meet the required precision? (There is no estimation for the sample proportion.)  (Enter an integer number.)

b)A university planner wants to determine the proportion of fall semester students who will attend summer school. She surveys 30 current students discovering that 20 will return for summer school.At 90% confidence, compute the margin of error for the estimation of this proportion.

c)For the t distribution with 14 degrees of freedom, calculate P(T < 2.624)!

Answer 1

Solution,

Given that,

a) =  1 - = 0.5

margin of error = E = 0.281

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05  = 1.645

sample size = n = (Z / 2 / E )2 * * (1 - )

= (1.645 / 0.281)2 * 0.5 * 0.5

= 8.56

sample size = n = 9

b) Given that,

n = 30

x = 20

Point estimate = sample proportion = = x / n = 20 / 30 = 0.67

1 - = 1 - 0.67 = 0.33

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

Margin of error = E = Z / 2 * (( * (1 - )) / n)

E = 1.645 (((0.67 * 0.33) / 30)

E = 0.141

c) degrees of freedom = 14

P(T < 2.624 )

= 0.9900