Question

Test the hypothesis from an article in a newspaper, magazine or on the internet that makes a claim about a population mean or a population proportion. Select a variable (mean or proportion) to test. Define the population. Select a sample of at least 10 data values for mean testing or 30 to 50 data values for proportion testing. Construct a 90% interval confidence interval. and state the statistical decision based on the P-value with 5% of level significance.

Answer 1

we might be interested in studying the proportion of children living near by a factory who have colic. The prevalence of colic in the general public is estimated to be as low as 6%. let out of surveyed 48 children 5 were found with symptom of colic .

here we want to test the

null hypothesis H0:P=0.06 ( children and general public have same )

alternate hypothesis Ha:P0.06 ( ( children and general public have different )

p=x/n=/48=0.1042,

here we use z-test and statistic z=(p-P)/SE(p)=(p-P)/sqrt(P(1-P)/n)=(0.1042-0.06)/sqrt(0.06*(1-0.06)/48))=1.2994

the two tailed p-value=0.1972 is more than alpha=0.05, so we accept the null hypothesis and conclude that proportion of colic incidence in children and general public are same/equivalent.

(1-alpha)*100% confidence interval for population proportion (P)=sample proportion (p) ±z(alpha/2)*SE(p)

90% confidence interval for P=0.1042±z(0.1/2)*SE(p)=0.1042±1.645*0.0343=0.1042±0.0564=(0.0036,0.1164)

sample proportion
	sample	n	x	p	var(p)=p(1-p)/n
	sample	48	5	0.0600	0.001175
	SE(p)=	SQRT(VAR(p))		0.0343
	z-value	margin of error		lower limit	upper limit
90% confidence interval	1.645	0.0564		0.0036	0.1164