Question

a) A random sample of 100 housewives shows that 60 prefer detergent A. Compute a 99% confidence interval for the fraction of total housewives favoring detergent A Give an interpretation of a confidence interval. b) A poll of 1000 randomly selected voters was taken to estimate the proportion of population that were in support of the president's current policy on foreign aid. If 555 were favorable, did we find sufficient evidence to support the president's claim that he has the support of the (simple) majority of all voters? Use a=0.05.

Answer 1

Answer:

n= 100, x= 60, c=99%

a)

$\hat p = \frac{x}{n}=\frac{60}{100}=0.60$

formula for confidence inteval is

p*(1-P) pZc* v

where Zc is the z critical value for c= 99%

Zc= 2.576

$0.6\pm 2.576*\sqrt{\frac{0.6*(1-0.6)}{100}}$

0.47381 < P < 0.72619

0.474 < P < 0.726

( 0.474 , 0.726 )

we are 99 % confident the proportion of housewife favouring detergent A is between 0.474 and  0.726

b)

n=1000, x= 555, $1589300202355_blob.png$ = 0.05

Ho: P $\leq$   0.50

H1: P >  0.50

$\hat p=\frac{x}{n}=\frac{555}{1000}=0.555$

formula for test statistics is

P-p p*(1-P)

$z = \frac{0.555 - 0.5}{\sqrt{\frac{0.5* ( 1-0.5)}{1000}}}$

test statistics: z = 3.48

calculate P-Value

P-Value = 1 - P(z < 3.48)

using z table we get

P(z < 3.48) = 0.9997

P-Value = 1 - 0.9997

P-Value = 0.0003

decision rule is

Reject Ho if ( P-value ) $1589300202194_blob.png$ ( $1589300202322_blob.png$ )

here, ( P-value= 0.0003 ) < ( $1589300202373_blob.png$ = 0.05)

Hence, we can say,

Null hypothesis is rejected.

therefore there is sufficient evidence to support the presidents claim that he has the majority of voters support.