Question

You are working on a manufacturing process that should produce products that have a diameter of 3/4 of an inch. You take a random sample of 35 products and find the diameter to be 13/16 of an inch with a standard deviation of 5/32 of an inch. 3. Calculate a 95% confidence interval around the sample mean. What do you notice about this interval relative to the desired diameter and how does this relate to your answers in questions 1 and 2?

Answer 1

Sample Mean Diameter = 13/16 = 0.8125

Standard Deviation = 5/32 = 0.15625

Confidence Interval Calculator

Step 1: Find ?/2
Level of Confidence = 95%
? = 100% - (Level of Confidence) = 5%
?/2 = 2.5% = 0.025

Step 2: Find t_?/2
Calculate t_?/2 by using t-distribution with degrees of freedom (DF) as n - 1 = 35 - 1 = 34 and ?/2 = 0.025 as right-tailed area and left-tailed area.

t_?/2 = 2.032244 (Obtained using online t value calculator screenshot attached)

Step 3: Calculate Confidence Interval

Lower Bound = x? - t_?/2•(s/?n) = 0.8125 - (2.032244)(0.15625/?35) = 0.7588
Upper Bound = x? + t_?/2•(s/?n) = 0.8125 + (2.032244)(0.15625/?35) = 0.8662

Confidence Interval = (0.7588, 0.8662)

The confidence interval does not contains desired diameter 0.75. If the confidence interval does not contain the null hypothesis value, the results are statistically significant. We reject null hypothesis. This means that there is sufficient evidence to support the claim that mean diameter of the products produced is not equal to 0.75.