Question

The diameter of bushings turned out by a manufacturing process is a normally distributed random variable with a mean of 4.035 mm and a standard deviation of 0.005 mm. A sample of 25 bushings is taken once an hour. (a) Within what interval should 95 percent of the bushing diameters fall? (Round your answers to 4 decimal places.) The 95% confidence interval is from to (b) Within what interval should 95 percent of the sample means fall? (Round your answers to 5 decimal places.) The 95% confidence interval is from to (c-1) What conclusion would you reach if you saw a sample mean of 4.020? The sample came from a population that a population mean equal to 4.035. (c-2) What conclusion would you reach if you saw a sample mean of 4.055? The sample came from a population that a population mean equal to 4.035.

Answer 1

A) the z score for 95% is 1.96

Multiply by the deviation:

1.96 x 0.005 = 0.0098

Now add and then subtract that from the mean:

4.035 - 0.0098 = 4.0252

4.035 + 0.0098 = 4.0448

The interval is (4.0252, 4.0448)

B) 4.035 +/- 1.96 x sqrt( 0.005/sqrt(25))

= 4.035 +/-0.00196

Answer: ( 4.033, 4.037)

C) the conclusion is that both 4.020 and 4.055 are out of the range.

4.020 is below the lowest range and 5.055 is higher than highest range.

Answer 2

The 95% confidence interval for the sample means is; CI = (4.033, 4.037)

What is the Confidence Interval?

A) We are given;

Mean; x' = 4.035 mm

standard deviation; σ = 0.005 mm

sample size; n = 25

Formula for confidence interval is;

CI = x' ± z(σ/√n)

The z score for 95% confidence level is 1.96. Thus;

CI = 4.035 ± 1.96(0.005)

CI = 4.035 ± 0.0098

CI = (4.025, 4.045)

B) From the formula for confidence interval earlier stated, we have;

CI = 4.035 ± 1.96(0.005/√(25))

CI = 4.035 ± 0.00196

CI = (4.033, 4.037)

C)i) The conclusion is that both 4.020 is out of the range of the confidence interval.

ii) The conclusion is that 4.055 is higher than the confidence interval

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