Homework SourseThe National Institute of Standards and Technology NIST supp
The National Institute of Standards and Technology (NIST) supplies a “standard iron rod” whose electrical conductivity is supposed to be exactly 10.1. Is there reason to think that the true conductivity is not 10.1? To find out, NIST measures the conductivity of one rod 6 times. Repeated measurements of the same thing vary, which is why NIST makes 6 measurements. These measurements are an SRS from the population of all possible measurements. This population has a Normal distribution with mean equal to the true (given the population standard deviation is 0.1).
Step 1:
One set of measurements has mean conductivity x = 10.09. Find the value of the test statistic z and then use Table A to compute the p-value. (Hint: look at the alternative hypothesis. Is the p-value a left sided, right sided, or two sided area?)
What is the P-value?
Give your answer to 2 decimal places.
Fill in the blank:
Step 2:
Is this outcome statistically significant at the = 0.05 level? At the = 0.01 level?
Enter the number of the term that corresponds to each choice:
At the 0.05 level
At the 0.01 level
Step 3:
Another set of measurements has x = 9.95.
Find the P-value for this outcome.
Give your answer to 4 decimal places.
Fill in the blank:
Step 4:
Is it statistically significant at the = 0.05 level? At the = 0.01 level?
Enter the number of the term that corresponds to each choice:
At the 0.05 level
At the 0.01 level
| 1. Significant |
| 2. Not significant |
Solution
The National Institute of Standards and Technology (NIST) supplies a “standard iron rod” whose electrical conductivity is supposed to be exactly 10.1. Is there reason to think that the true conductivity is not 10.1? To find out, NIST measures the conductivity of one rod 6 times. Repeated measurements of the same thing vary, which is why NIST makes 6 measurements. These measurements are an SRS from the population of all possible measurements. This population has a Normal distribution with mean equal to the true (given the population standard deviation is 0.1).
Step 1:
One set of measurements has mean conductivity x = 10.09. Find the value of the test statistic z and then use Table A to compute the p-value. (Hint: look at the alternative hypothesis. Is the p-value a left sided, right sided, or two sided area?)
The test is two sided
Data
Null Hypothesis m=
10.1
Level of Significance
0.05
Population Standard Deviation
0.1
Sample Size
6
Sample Mean
10.09
Intermediate Calculations
Standard Error of the Mean
0.0408
Z Test Statistic
-0.2449
Two-Tail Test
Lower Critical Value
-1.9600
Upper Critical Value
1.9600
p-Value
0.8065
Do not reject the null hypothesis
What is the P-value?
Give your answer to 2 decimal places.
Fill in the blank: P=0.81
Step 2:
Is this outcome statistically significant at the = 0.05 level? At the = 0.01 level?
1. Significant
2. Not significant
Enter the number of the term that corresponds to each choice:
At the 0.05 level 2. Not significant
At the 0.01 level 2. Not significant
Step 3:
Another set of measurements has x = 9.95.
Z Test of Hypothesis for the Mean
Data
Null Hypothesis m=
10.1
Level of Significance
0.05
Population Standard Deviation
0.1
Sample Size
6
Sample Mean
9.95
Intermediate Calculations
Standard Error of the Mean
0.0408
Z Test Statistic
-3.6742
Two-Tail Test
Lower Critical Value
-1.9600
Upper Critical Value
1.9600
p-Value
0.0002
Reject the null hypothesis
Find the P-value for this outcome.
Give your answer to 4 decimal places.
Fill in the blank: P=0.002
Step 4:
Is it statistically significant at the = 0.05 level? At the = 0.01 level?
1. Significant
2. Not significant
Enter the number of the term that corresponds to each choice:
At the 0.05 level 1. Significant
At the 0.01 level 1. Significant
| Data | |
| Null Hypothesis m= | 10.1 |
| Level of Significance | 0.05 |
| Population Standard Deviation | 0.1 |
| Sample Size | 6 |
| Sample Mean | 10.09 |
| Intermediate Calculations | |
| Standard Error of the Mean | 0.0408 |
| Z Test Statistic | -0.2449 |
| Two-Tail Test | |
| Lower Critical Value | -1.9600 |
| Upper Critical Value | 1.9600 |
| p-Value | 0.8065 |
| Do not reject the null hypothesis |
