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. 2. Considering a 5% level of significance, what can you state statistically about the process? Use the same null and alternative hypotheses as you did in question 1 but this time use the p-Value approach. Note: you will reach the same conclusion whether you interpolate in the tables in the appendices or use the appropriate distribution in Excel although your numerical values will be slightly different.

Answer 1

Product should have a Diameter = 3/4 of an inch. = 3/4 times 2.54 cm [Therefore inch =] mu = 1.095 Sample information sample size n = 35 sample mean X^- = 13/16 of an inch X^- = 13/16 times 2.54 = 2.0657 sample deviation = 5/12 of an inch s = 5/12 times 2.4 = 0.3656 For solving information about statistical process we have Null Hypothesis H_0: = 1.905 H_1: notequalto 1.905 test t = z - h_0/S/squareroot n t_ = 2.0658 - 1.905/0.3650 /squareroot 35 = 2.5183 Using function in we (0.025 24) = 2.8450 Decision Rule Since we reject as null hypothesis, \r\n Using P - Value: P - Value is the level of significance within a hypothesis test Representing The Prob of occur of agent event . P - value is need as an the Rejection points to provide The smallest level of significance of with the Null hypothesis would be Rejected. Decision RULE: A small P value (lessthanorequalto 0.05) indicates Rejected of the Null hypothesis.