Question

20. A film distribution manager calculates that 8% of the films released are flops. If the manager is right, what is the probability that the proportion of flops in a sample of 409 released films would differ from the population proportion by more than 3%? Round your answer to four decimal places.

1. A soft drink manufacturer wishes to know how many soft drinks adults drink each week. They want to construct a 98% confidence interval with an error of no more than 0.07. A consultant has informed them that a previous study found the mean to be 6.4 soft drinks per week and found the variance to be 0.81. What is the minimum sample size required to create the specified confidence interval? Round your answer up to the next integer.

Answer 1

Solution:

20)

Solution:

Given ,

p = 8% = 0.08(population proportion)

n = 409  (sample size)

Let be the sample proportion.

The sampling distribution of is approximately normal with

mean = =  p = 0.08

SD =   =     

=    \sqrt{0.08(1-0.08)/409}

=  0.01341458535

Now ,

P( will differ from p by more than 3%)

=  P( will differ from p by more than 0.03)

= 1 - P( will differ from μ by less than 0.03)

= 1 - P(p - 0.03 <   < p + 0.03 )

= 1 - P( 0.08 - 0.03 < < 0.08 + 0.03)

= 1 - P(0.05 < < 0.11)

= 1 - { P( < 0.11) - P( < 0.05) }

= 1 - { P(Z <(0.11 - 0.08)/0.01341458535) - P(Z <(0.05 - 0.08)/0.01341458535) }

= 1 - { P(Z < 2.24) - P(Z < -2.24) }

= 1 - { 0.9875 - 0.0125} .. (use z table)

= 1 - 0.975

= 0.0250

P( will differ from p by more than 3%) = 0.0250

1)

c = 98% = 0.98

E = 0.07

² = 0.81

So , = 0.9

c = 0.98

= 1- c = 1- 0.98 = 0.02

  /2 = 0.01

Using Z table ,

= 2.326

Now, sample size (n) is given by,

=  {(2.326* 0.9)/ 0.07 }²

= 894.351746939

= 895 ..(round to the next whole number)

Answer : Required Sample size is n = 895