In: Statistics and Probability
A technician at the National Bureau of Standards in Washington D.C. performs 16 measurements on a check-weight with a nominal (expected) weight of 1 KG. The average of the 16 measurements was LaTeX: \large \bar w = 1.00023\;w ¯ = 1.00023KG and the sample standard deviation (SD+) of the 16 measurements was LaTeX: \large s_w = 0.00043\;s w = 0.00043KG. Use this data to test the claim that the true weight of the check-weight is not 1 Kg. (a) State the formal hypotheses in terms of an appropriate parameter. ? Use a 5% significance level. (b) Find the test statistic and the p-value for the test and explain clearly how you found them (i.e., where do the EV and SE come from, what is the distribution of the test statistic, etc.) (c) Summarize the results of the test in terms of the original claim.
Here we have to test that
Null hypothesis:
Alternative hypothesis:
where
n = sample size = 16
Sample mean = = 1.00023
Sample standard deviation = = 0.00043
Here population standard deviation is not given so we use t test.
Test statistic:
t = 2.140 (Round to 3 decimal)
Test statistic = t = 2.140
Level of significance = = 0.05
Degrees of freedom = n - 1 = 16 - 1 = 15
Test is two tailed test.
P value from excel using function:
=T.DIST.2T(2.140,15)
= 0.0492 (Round to 4 decimal)
P value = 0.0492
Here p value <
So we reject the null hypothesis.
Conclusion: There is sufficient evidence to support the claim that the true weight of the check-weight is not 1 Kg.