In: Statistics and Probability
A machine is producing metal pieces that are cylindrical in shape. A sample of 12 pieces is taken and the diameters are: 1.01, 0.97, 1.03, 1.04, 0.8, 0.78, 1.06, 0.95, 0.87,1.0,0.88 and 0.90 centimeters.
a. Find 98% confidence interval for the true average diameter of metal pieces produced by the machine.
We would like to test whether the true average diameter of metal pieces produced by the machine is less than 1 centimeter.
b. Assuming the population distribution is approximately normal and using level of significance ? = 0.05, carry out a hypothesis test.
For the given sample 1.01, 0.97, 1.03, 1.04, 0.8, 0.78, 1.06, 0.95, 0.87,1.0,0.88 and 0.90.
The mean is calculated as:
Mean = (1.01 + 0.97 + 1.03 + 1.04 + 0.8 + 0.78 + 1.06 + 0.95 +
0.87 + 1.0 + 0.88 + 0.90)/12
= 11.29/12
=Mean = 0.9408
and the sample standard deviation as:
Standard Deviation s = √(1/12 - 1) x ((1.01 - 0.9408)2 + ( 0.97 - 0.9408)2 + ( 1.03 - 0.9408)2 + ( 1.04 - 0.9408)2 + ( 0.8 - 0.9408)2 + ( 0.78 - 0.9408)2 + ( 1.06 - 0.9408)2 + ( 0.95 - 0.9408)2 + ( 0.87 - 0.9408)2 + ( 1.0 - 0.9408)2 + ( 0.88 - 0.9408)2 + ( 0.90 - 0.9408)2)
= √(1/11) x ((0.0692)2 + (0.0292)2 + (0.0892)2 + (0.0992)2 + (-0.1408)2 + (-0.1608)2 + (0.1192)2 + (0.0092)2 + (-0.0708)2 + (0.0592)2 + (-0.0608)2 + (-0.0408)2)
= √(0.0909) x ((0.00478864) + (0.00085264) + (0.00795664) + (0.00984064) + (0.01982464) + (0.02585664) + (0.01420864) + (8.464E-5) + (0.00501264) + (0.00350464) + (0.00369664) + (0.00166464))
= √(0.0909) x (0.09729168)
= √(0.008843813712)
= 0.094
a) Now the confidence interval at 98% confidence level is calculated as:
Where the tc( critical t ) is calculated using excel formula for t-distribution at given confidence level and calculated degree of freedom (n-1) =12-1=11 as =T.INV.2T(0.02,11):
tc computed by formula as tc = 2.718
so, now the confidence interval is calculated as:
so, the 98% confidence interval for the true average diameter of metal pieces produced by the machine is [0.867, 1.015}
b) Assuming that the distribution is normal but the population standard deviation is unknown hence we will use t-distribution to conduct the hypothesis test.
Based on the claim that the true average diameter of metal pieces produced by the machine is less than 1 centimeter.
The hypotheses are:
Based on the hypothesis it will be a left-tailed test.
Rejection region:
At 98% confidence level and degree of freedom =n-1=12-1=11 for the left-tailed test, the critical value is calculated using excel formula for t-distribution which is =T.INV(0.02,11), the tc value calculated is -2.328.
So, reject Ho if t-calculated is less than tc =-2.328.
Test Statistic:
The test statistic is calculated as:
P-value:
The P-value is also computed using excel formula for t-distribution using the calculated t-value and the degree of freedom as =T.DIST(-2.182,11,TRUE).
The p-value computed is 0.0259.
Conclusion:
Based on the P-value calculation and the t-value since P-value is greater than 0.02 and 't' is greater than tc hence we cannot reject the null hypothesis and conclude that there is insufficient evidence to support the claim.