Question

2. A new method is developed to determine the content of sodium in a biscuit sample. Student A gets the following results: 24.2, 23.9, 25.0, 23.7 and 25.2 mg, while student B obtains 22.6, 21.9, 23.2 and 23.1 mg.

(a) Report 95% confidence interval of their results individually. Why the standard deviations are non-zero?

(b) At the 95% confidence level, does one of them get more reproducible measurements?

(c) At the 95% confidence level, do they obtain different results? What about 90% confidence level?

(d) According to the manufacturer, the sodium content in the biscuit is 24.0 mg. Comment whether student B’s results are reliable at 90% confidence level.

(e) The results from student C are 25.3, 25.4, 24.9, 25.0, 23.2 and 24.5 mg. Determine whether there is an outlier at the confidence level of 95%? What about 90%?

Answer 1

(a) Report 95% confidence interval of their results individually. Why the standard deviations are non-zero?

24.4000	mean A
0.6671	std. dev.
0.2983	std. error
5	n
4	df
23.57	confidence interval 95.% lower
25.23	confidence interval 95.% upper
0.8283	margin of error

The 95% confidence interval for Student A is between 23.57 and 25.23.

22.7000	mean B
0.5944	std. dev.
0.2972	std. error
4	n
3	df
21.75	confidence interval 95.% lower
23.65	confidence interval 95.% upper
0.9459	margin of error

The 95% confidence interval for Student B is between 21.75 and 23.65.

The standard deviations are non-zero because the values are far from the mean.

(b) At the 95% confidence level, does one of them get more reproducible measurements?

Yes, Student A gets more reproducible measurements.

(c) At the 95% confidence level, do they obtain different results? What about 90% confidence level?

24.4000	mean A
0.6671	std. dev.
0.2983	std. error
5	n
4	df
23.76	confidence interval 90.% lower
25.04	confidence interval 90.% upper
0.6360	margin of error

The 90% confidence interval for Student A is between 23.76 and 25.04.

22.7000	mean B
0.5944	std. dev.
0.2972	std. error
4	n
3	df
22.00	confidence interval 90.% lower
23.40	confidence interval 90.% upper
0.6994	margin of error

The 90% confidence interval for Student B is between 22.00 and 23.40.

Yes, they do obtain different results at the 90% confidence level.

(d) According to the manufacturer, the sodium content in the biscuit is 24.0 mg. Comment whether student B’s results are reliable at 90% confidence level.

The 90% confidence interval for Student B is between 22.00 and 23.40.

Since the value 24 is not in the confidence interval, we cannot say that student B’s results are reliable at 90% confidence level.

(e) The results from student C are 25.3, 25.4, 24.9, 25.0, 23.2 and 24.5 mg. Determine whether there is an outlier at the confidence level of 95%? What about 90%?

24.7167	mean C
0.8085	std. dev.
0.3301	std. error
6	n
5	df
23.87	confidence interval 95.% lower
25.57	confidence interval 95.% upper
0.8485	margin of error

The 95% confidence interval for Student C is between 23.87 and 25.57.

24.7167	mean C
0.8085	std. dev.
0.3301	std. error
6	n
5	df
24.05	confidence interval 90.% lower
25.38	confidence interval 90.% upper
0.6651	margin of error

The 90% confidence interval for Student C is between 24.05 and 25.38.

There is no outlier in the data.