Question

4. A survey of 35 individuals reveals that the proportion of people in a residential area who patronize a particular supermarket is 0.42; a repeat of the survey in the following year, using 52 individuals finds that the proportion is 0.56. Find a 95% confidence interval for the difference in sample proportions. (3)

5. A researcher suspects that the mean pollutant level in a region is higher than the state average of 4.3 grams/cubic liter. Data are collected; from 56 samples, the mean is found to be 5.4, with a standard deviation of 5.0. Test the null hypothesis that the true mean in the region is 4.3, against the alternative that it is higher. State the null and alternative hypotheses, give the critical value, find the test statistic, make a decision, and give the p-value. (5)

PLEASE HELP!!!!!!

Answer 1

4)

p̂1=x1/n1 =	0.4200	p̂2=x2/n2 =	0.5600
n1                       =	35	n2                       =	52
estimated difference in proportion   =p̂1-p̂2   =			-0.1400
std error Se =√(p̂1*(1-p̂1)/n1+p̂2*(1-p̂2)/n2) =			0.1082
for 95 % CI value of z=		1.960
margin of error E=z*std error =		0.2120
lower bound=(p̂1-p̂2)-E=		-0.3520
Upper bound=(p̂1-p̂2)+E=		0.0720
from above 95% confidence interval for difference in population proportion =(-0.352,0.072)

5)

null hypothesis: HO: μ		=	4.3
Alternate Hypothesis: Ha: μ		>	4.3

population mean μ=	4.3
sample mean 'x̄=	5.400
sample size   n=	56.00
sample std deviation s=	5.000
std error 'sx=s/√n=	0.668
test stat t ='(x-μ)*√n/sx=	1.646

p value      =

0.0527 (from excel: function tdist(1.646,55,1)

since p value <0.05 ; we can not reject null hypothesis that  true mean in the region is higher than 4.3