question archive (a) Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the proportions of boys and girls under 10 years old who are afraid of spiders

(a) Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the proportions of boys and girls under 10 years old who are afraid of spiders

Subject:StatisticsPrice:4.89 Bought3

(a) Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the proportions of boys and girls under 10 years old who are afraid of spiders. Assume the "worst case" scenario for the value of both sample proportions. We want a 98 % confidence level and for the error to be smaller than 0.08' Answer: (b) Again find the sample size required, as in part (a), but with the knowledge that a similar student last year found that the proportion of boys afraid of spiders is 0.64 and the proportion of girls afraid of spiders was 0.87. Answer:

pur-new-sol

Purchase A New Answer

Custom new solution created by our subject matter experts

GET A QUOTE

Answer Preview

We will use maximum error formula =za S.D/SQRT(n)

but we will find sample size "n"

sqrt(n)=za. S.D/maximum error

Squaring on both sides , we get

(n)=za. {S.D/maximum error}2

Given 98% confidence interval (z value) = 2.3263

given maximum error = 0.02

(n)=za. {p(1-p)/maximum error}2

n≤ ( 2.3263*0.5/0.02)2   ( here S.D = p(1-p) ≤ 1/2

on simplification , we get n = 3381

Conclusion:

The sample size of two samples is  n = 3381

verification:-

We will use maximum error formula =za S.D/SQRT(n)  = 2.3263*0.5/sqrt(3381)= 0.02

substitute all values and simplify we get maximum error is 0.02

 

 

 

LetP1 and P2 denotes the proportion of boys and girls under 10 years old who are afraid of spiders respectively.

 Let required sample size for both is n that is n1 =n2 = n Given, margin of error MOE < 0.02 with confidence level of 98%.

 Given: p? 1 = 0.64   p? 2  =0.87

 

 

2.3263 * sqrt {0.64*0.36)/n   +   0.87*0.13/n}

=2.3263 * sqrt {0.2304)/n   +   0.1131/n}

=2.3263 * sqrt {0.3435/n}

please see the attached file for the complete solution.

Where (1 -α) x 100 = confidence level 

Z is standard normal distribution so Z0.01 is given by

 P(Z>Z0.01) = 0.01 ----P(Z≤ Z 0.01) =0.99

 And from Z table we know that P(Z≤ 2.3263) = 0.99 so Z0.01 =2.3263

 Now as MOE < 0.02 hence

 

=2.32632 *  {0.3435/n} <0.02

n> {2.3263/0.02)2 * 0.3435

n>4647

Therefore, the sample sizes are more than 4647

 

Related Questions