In: Statistics and Probability
The weights of 40 bags of sugar from a certain producer, labeled as containing 1 kg each, were measured. The sample mean weight of the bags of sugar was 1.0042 kg, and the sample standard deviation was 0.042 kg.
(a) Calculate the value of the estimated standard error of the sample mean
(b) Calculate a 99% confidence interval for the population mean weight of bags of sugar labeled 1 kg from this producer. Interpret the confidence interval in terms of all possible random samples of bags of sugar labeled 1 kg from this producer.
(c) On the basis of the confidence interval from part (b), what would have been the outcome of a z-test of the null hypothesis that the population mean of bags of sugar labeled 1 kg from this producer is in fact 1.03 kg? Interpret the result of the test. (Note that you should use the confidence interval from part (b) to answer this question, and you should not perform a z-test.)
X̅ = 1.0042
kg
Sample Mean
s = 0.042
kg
Standard error of the mean
n =
40
Sample Size
a) Standard Error of the sample mean is given by
Standard error of the sample mean is 0.0066
b) Confidence interval for mean is given
by
99% confidence interval for population
mean
For 99%, α = 0.01, α/2 = 0.005
From the z-tables, or Excel function
NORM.S.INV(α/2)
z = NORM.S.INV(0.005) = 2.576 (We take the
positive value for calculations)
= (0.9872, 1.0212)
99% confidence interval is (0.9872,
1.0212)
Interpretation
:
If repeated samples of sugar bags are taken and the confidence
interval
was computed for each sample, 99% of the intervals would contain
the population
mean weight of the bags of
sugar.
c) The z-test would be to test if the bags of sugar have weight
equal to 1.03 kg
Since 1.03 is not included in the interval (0.9872,
1.0212)
the z-test would be statistically significant
.
That is the null hypothesis would be
rejected
Conclusion :
The weight of the bags of sugar labeled 1 kg does not equal 1.03
kg