A physicist is trying to determine whether doping material A with a certain type
ID: 3200935 • Letter: A
Question
A physicist is trying to determine whether doping material A with a certain type of chemical will increase its conductivity under room temperature. 8 bulks of material A was prepared. She rst measures each of their conductivity, then dope them with the chemical, and then measures their conductivity again. The results of the experiment are as in the table below. You may assume that all usual assumptions for condence interval and signicant test are met.
Table 1: Conductivity of material A
(S/m) Sheet 1 2 3 4 5 6 7 8
Before Doping 262.2 262.5 262.7 261.3 263.2 261.8 263.5 261.9
After Doping 262.3 262.1 262.8 262.2 262.9 262.1 263.1 262.9
(a) State the null and alternative hypotheses for this experiment’s signicance test. (b) Compute the test statistic for this signicance test. (c) Using the t-table, nd a range of values that the p-value falls in, and state the conclusions of this test, using = 0.05. (d) Compute the appropriate 95% condence interval for this experiment and give the interpretation. (e) Is there agreement between the condence interval and the signicance test? Explain.
Explanation / Answer
(a) State the null and alternative hypotheses for this experiment’s signicance test.
Answer:
The null and alternative hypothesis for this test is given as below:
Null hypothesis: H0: Doping material A with a certain type of chemical will not change its conductivity under room temperature.
Alternative hypothesis: Ha: Doping material A with a certain type of chemical will increase its conductivity under room temperature.
(b) Compute the test statistic for this signicance test.
Answer:
Here, we have to use paired t test. The test statistic formula is given as below:
Test statistic = t = dbar / [Sd/sqrt(n)]
Where, dbar is the mean of differences between before and after scores, Sd is the standard deviation of these differences and n is the sample size.
From the given data we have
Before
After
Di
(Di - DBar)^2
262.2
262.3
-0.1
0.00390625
262.5
262.1
0.4
0.31640625
262.7
262.8
-0.1
0.00390625
261.3
262.2
-0.9
0.54390625
263.2
262.9
0.3
0.21390625
261.8
262.1
-0.3
0.01890625
263.5
263.1
0.4
0.31640625
261.9
262.9
-1
0.70140625
Dbar = -0.1625
Sd = 0.5502
Sample size = n = 8
Degrees of freedom = n – 1 = 8 – 1 = 7
Test statistic = t = -0.1625 / [0.5502/sqrt(8)]
Test statistic = -0.8354
(c) Using the t-table, nd a range of values that the p-value falls in, and state the conclusions of this test, using = 0.05.
Answer:
Here, we have to find the critical values for alpha = 0.05
By using t table
Critical value = -1.8946
P-value = 0.2155
Alpha value = 0.05
P-value > Alpha value
So, we do not reject the null hypothesis that Doping material A with a certain type of chemical will not change its conductivity under room temperature.
(d) Compute the appropriate 95% condence interval for this experiment and give the interpretation.
Answer:
Confidence interval = Dbar -/+ t*SE
SE = Sd/sqrt(n) = 0.5502/sqrt(8) = 0.1945
Critical t value = 1.8946
Confidence interval = -0.1625 -/+ 0.3685
Lower limit = -0.1625 – 0.3685 = -0.5310
Upper limit = -0.1625 + 0.3685 = 0.2060
Confidence interval = (-0.5310, 0.2060)
(e) Is there agreement between the condence interval and the signicance test? Explain.
Answer:
Yes, there is an agreement between confidence interval and significance test. In the significance test we do not reject the null hypothesis however in confidence interval it is observed that the value 0 lies between the confidence interval.
Before
After
Di
(Di - DBar)^2
262.2
262.3
-0.1
0.00390625
262.5
262.1
0.4
0.31640625
262.7
262.8
-0.1
0.00390625
261.3
262.2
-0.9
0.54390625
263.2
262.9
0.3
0.21390625
261.8
262.1
-0.3
0.01890625
263.5
263.1
0.4
0.31640625
261.9
262.9
-1
0.70140625
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.