Question

Suppose a purchasing manager wants to determine the average cost of machined metal parts in the US. How many machined metal parts would they have to sample? The manager wants to be 90% confident with a margin of error of 6.74. Suppose that the costs of the metal parts ranges from 93 to 220. When entering the values into JMP, round the planning value for the standard deviation to four values after the decimal. Round final answer as you normally would.

Answer 1

Answer:

they will have to sample 60 machined metal parts

Step-by-step explanation:

Given that;

Margin of Error E = 6.74

at 90% Confidence Interval, z = 1.645  { from table }

Range R = maximum range - minimum range = 220 - 93 = 127

standard deviation σ = R / 4 = 127/4= 31.7500

How many machined metal parts would they have to sample n = ?

to get our sample size, we use the following formula

n = [ zσ / E ]²

we substitute in our values

n = [ 1.645 × 31.7500 / 6.74 ]²  

n = [ 52.2287 / 6.74 ]²  

n = [ 7.7490 ]²          

n = 60.047 ≈ 60  

Therefore, they will have to sample 60 machined metal parts

Answer 2

You can use the margin of error with the z value and standard deviation to get the result.

The number of metal machine parts would be 60.

Given that:

• The confidence level needed is 90%
• The margin of error is 6.74
• The cost of metal parts  ranges from 93 to 200

To find:

The number of machined metal parts that have to be sampled.

Finding the range of the distribution:

Range = Max. value - min. value = 220 - 93 = 127

Finding the standard deviation:

The standard deviation is approximately the quarter of the range of the distribution.

Thus, $\sigma = 127 \div 4 = 31.75$

For 90% confidence, the z value is 1.645 (from z tables)

Thus, from the formula for size of sample, we have:

$n = (\dfrac{z\sigma}{E})^2 = (\dfrac{1.645 \times 31.75}{6.74})^2 = (7.749)^2 \approx 60$

The number of metal machine parts would be 60.

Learn more about sample size here:

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