Estimation

Estimation

Answers

Group 1

The mean S. indirect bilirubin level of 16 four days old infants was found to be 5.98 mg/dl. The population SD (σ)=3.5 mg/dl. Assuming normality , find 90,95, 99% CI for µ:
_ _
{X- Z1-α /2 x σ /√ n < µ < X + Z1-α /2 x σ /√ n}

_ _

CI{X- Z1-α /2 x σ /√ n < µ < X + Z1-α /2 x σ /√ n}=1-α

95%CI {5.98- 1.96 * 3.5 /√ 16 < µ < 5.98 + 1.96 * 3.5 /√ 16}

95%CI {5.98- 1.715 < µ < 5.98 + 1.715}

95%CI {4.265 < µ < 7.695}

_ _
CI{X- Z1-α /2 x σ /√ n < µ < X + Z1-α /2 x σ /√ n}=1-α


99%CI{5.98- 2.58 * 3.5 /√ 16 < µ < 5.98 + 2.58 * 3.5 /√ 16}

99%CI{5.98- 2.258 < µ < 5.98 + 2.258}

99%CI={ 3.72 < µ < 8.24}

Group 2

A sample of 10 twelve years old boys and a sample of 10 twelve years old girls yielded mean height of 59.8 inches (boys), and 58.5 inches (girls). Assuming normality and σ1=2 inches, and σ2= 3 inches . Find 90% CI for the difference in means of height between girls and boys at this age.

Group 2

_ _ _ _
CI{( X1-X2) -Z √ (σ21/n1)+ (σ22/n2)< (µ1-µ2)< ( X1-X2)+
Z√ (σ21/n1)+ (σ22/n2)}

90%CI{( 59.8-58.5) -1.645 √ (2)2/10)+ (3)2/10)< (µ1-µ2)< ( 59.8-58.5)+1.645√ (2)2/10)+ (3)2/10)}

90%CI{1.3 -1.88< (µ1-µ2)< 1.3+ 1.88}

90%CI{ -0.58< (µ1-µ2)< 3.18}
If 0 is included not significant


Group 3
In a survey 300 adults were interviewed , 123 said they had yearly medical checkup. Find the 95% for the true proportion of adults having yearly medical checkup.
~ 123
P=-------=0.41
300

Group 3

~ ~ ~ ~ ~ ~
CI{P-Z √ p(1-p)/n<P<P+Z √ p(1-p)/n}=1-α

95%CI{0.41-1.96 √ 0.41(1-0.41)/300<P<0.41+1.96

√ 0.41(1-0.41)/300}

95%CI{0.41- 0.06<P<0.41+0.06}

95%CI{0.35<P<0.47}

95%CI= 35-47%

Group 4
200 patients suffering from a certain disease were randomly divided into two equal groups. The first group received NEW treatment, 90 recovered in three days. Out of the other 100 who received the STANDARD treatment 78 recovered within three days. Find the 95% CI for the difference between the proportion of recovery among the populations receiving the two treatments


Answer
~ ~ 90 78
P1-P2=------- - ---------=0.12
100 100

Answer

~ ~ ~ ~ ~ ~ ~ ~
CI ( P1-P2 )-Z √ p1(1-p1)/n1 + p2(1-p2)/n2 < P1-P2 < ( P1-P2 )+Z

~ ~ ~ ~

√ p1(1-p1)/n1 + p2(1-p2)/n2

95% CI=0.12± 1.96 √ 0.9(1-0.9)/100 + 0.78(1-0.78)/100

95%CI=0.12 ± 0.1

95%CI =0.02-0.22 ( 2-22%)

Group 5
A random sample of 400 passengers of an airline is polled after their flights. Of the passengers, 300 say they will fly again with the same airline. Which of the following is a 90% confidence interval for the proportion of passengers that will fly again with the same airline?
• 0.75 +/- 0.066
• 0.75 +/- 0.005
• 0.75 +/- 0.045
• 0.75 +/- 0.036
• 0.75 +/- 0.15