Question

#15a

You are conducting a quality control test for bolts delivered by your supplier. The diameter of the bolt must be within 0.01 mm. of the size of pre-drilled holes in shaped metal parts for the machine that your company manufactures. Suppose that the true proportion of bolts that meet your specs is p = 0.7. What is the minimum sample size required for the sample proportion to have approximately a Normal distribution?

Round your answer to the nearest integer. (Note: in practice, you should round up. However, except for small minimum sample sizes, the two methods of rounding have about the same effect on the results.)

b.

A sample of 130 potential customers was asked to use your new product and the product of the leading competitor. After one week, they were asked to indicate which product they preferred. Let X = the number of customers who said that they preferred your product. In the sample, X = 57.

What is the lower endpoint of a 95% confidence interval for p, the proportion of potential customers who prefer your product to that of your competitor?

(Apply the large-sample confidence interval procedure. You will need to calculate z* in Excel, and do not round in your intermediate calculations.)

Express your answer in decimal form to two decimal places of accuracy.

Answer 1

Sol:

15.a).

Given

E = 0.01

p = 0.7

n = Z_α/2² ­*p*(1-p) / E²

= (2.58^2 * 0.7 * (1-0.7))/0.01^2

n =13978 .44

Sample size n = 13978

b).

Given,

To determine the lower bound

n = 130

x = 57

sample proportion = x/n

= 57/130

= 0.43846

p = 0.43846

Now here for the 95% confidence interval , z value is 1.96

So consider,

Margin of error E = z*sqrt(p(1-p)/n)

substitute the values

= 1.96*sqrt(0.43846(1-0.43846)/130)

E = 0.08529

Lower bound = p - E

= 0.43846 - 0.08529

= 0.34957

= 0.34

Lower bound = 0.34

Upper bound = 0.43846 + 0.08529

=0.52375

= 0.52

Upper bound = 0.52

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