Question

Consider the accompanying observations on bond strength. 9.8 9.1 7.9 5.9 6.3 5.6 5.5 11.3 12.3 13.5 17.3 4.7 10.8 3.7 3.2. 8.4 10.5 5.1 4.6 9.4 15.1 20.5 14.1 4.0 8.4 4.4 25.4 13.7 7.5 5.1 4.1 12.7 5.7 3.5 6.8 7.2 5.1 4.9 4.0 3.7 13.3 8.8 5.0 5.2 4.0 3.6 3.4 3.7 (a) Estimate true average bond strength in a way that conveys information about precision and reliability. (Hint: < x;= 387.8 and *? = 4,246.2.) (Round your answers to three decimal places.) A 95% CI for 4 - the true average strength is (b) Calculate a 95% CI for the proportion of all such bonds whose strength values would exceed 10. (Round your answers to three decimal places.)

Answer 1

The given data is,

11.3	12.3	9.8	9.1	7.9	6.3	6.8	7.2
13.5	17.3	9.4	5.5	5.9	5.6	5.1	4.9
4.7	10.8	15.1	8.4	4.4	4.1	4	3.7
3.7	3.2	20.5	25.4	13.7	12.7	13.3	8.8
8.4	10.5	14.1	7.5	5.1	5.7	5	5.2
5.1	4.6	4	4	3.6	3.5	3.4	3.7

The sample mean is,

= 387.8 48 = 8.07917

The sample standard deviation is,

$\small s =\sqrt \frac{\sum (x-\bar{x}) ^2}{n-1} = 4.8665$

Number of samples, n = 48

a)

we know that,

S Confidence Interval = 1 :* n

where z = 1.96 for 95% confidence level

$\small Confidence\; Interval = 8.07917 \pm 1.96 * \frac{4.8665} {\sqrt{48}}$

$\small Confidence\; Interval = [6.702,9.456]$

A 95% CI for the true average is (6.702 , 9.456)

b)

Proportion of all such bonds whose strnghth values would exceed 10 is,

$\small p = \frac{13}{48} = 0.27083$

The confidence interval is given by,

$\small C.I. =p\pm z* {\sqrt{\frac {p_{0}(1 - p_{0})} {n}}}$

$\small C.I. =0.27083\pm 1.96* {\sqrt{\frac {0.27083(1 - 0.27083)} {48}}}$

$\small C.I. =[0.145 , \; 0.397]$

(0.145 , 0.397)