# Why can a null measurement be more accurate than one using standard voltmeters and ammeters? What factors limit the accuracy of null measurements?

Question
Measurement
Why can a null measurement be more accurate than one using standard voltmeters and ammeters?
What factors limit the accuracy of null measurements?

2020-12-26
Step 1
A null measurement be more accurate than one using standard voltmeters and ammeters because standard measurements of voltage and current alter circuits. There are numerical uncertainties in standard measurement of voltage and current. Current flow reduced by an ammeters and voltmeter draws more current.
In null measurement circuit does not alter. In null measurement voltage balance and no current passing through the measuring device.
Step 2
factors which are affects the null measurement:
-Resistance of wiring'
-Change in resistance due to change in temperature
-current sensitivity of measuring device.

### Relevant Questions

A random sample of $$n_1 = 14$$ winter days in Denver gave a sample mean pollution index $$x_1 = 43$$.
Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
$$H_0:\mu_1=\mu_2.\mu_1>\mu_2$$
$$H_0:\mu_1<\mu_2.\mu_1=\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1<\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1\neq\mu_2$$
(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $$\mu_1 - \mu_2$$. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
$$\mu_1 - \mu_2$$.
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
(Measure the longest dimension of the room twice, using two different techniques. Do the measurement in feet and inches. Then convert to meters.) Below is what I got.
$$Tape Measure - 146 inches > 3.7084 meters / 12 feet and 2 inches = 12.1667 feet > 3.70840 meters$$
$$Ruler - 144.5 inches (144 + 1/2) = 3.6703 meters / 12.0416 feet = 3.67027 meters$$
Standard deviation is an indication of the...
a. precision of one measurement
b. accuracy of one measurement.
c. precision of repeated measurements.
Juan makes a measurement in a chemistry laboratory and records the result in his lab report. The stardard deviation of lab measurements made by students is $$\sigma=10$$ milligrams. Juan repeats the measurement 3 times and records the mean xbar of his 3 measurements.
(a) What is the standard deviation of Juan's mean result? (That is, if Juan kept making sets of 3 measurements and averaging them, what would be the standard deviation of all his xbar's?)
(b) How many times must juan repeat the measurement to reduce the standard deviation of xbar to5? Explain to someone who knows nothing about statistics the advantage of reporting the average of several measurements rather than the result of a single measurement.
We will now add support for register-memory ALU operations to the classic five-stage RISC pipeline. To offset this increase in complexity, all memory addressing will be restricted to register indirect (i.e., all addresses are simply a value held in a register; no offset or displacement may be added to the register value). For example, the register-memory instruction add x4, x5, (x1) means add the contents of register x5 to the contents of the memory location with address equal to the value in register x1 and put the sum in register x4. Register-register ALU operations are unchanged. The following items apply to the integer RISC pipeline:
a. List a rearranged order of the five traditional stages of the RISC pipeline that will support register-memory operations implemented exclusively by register indirect addressing.
b. Describe what new forwarding paths are needed for the rearranged pipeline by stating the source, destination, and information transferred on each needed new path.
c. For the reordered stages of the RISC pipeline, what new data hazards are created by this addressing mode? Give an instruction sequence illustrating each new hazard.
d. List all of the ways that the RISC pipeline with register-memory ALU operations can have a different instruction count for a given program than the original RISC pipeline. Give a pair of specific instruction sequences, one for the original pipeline and one for the rearranged pipeline, to illustrate each way.
Hint for (d): Give a pair of instruction sequences where the RISC pipeline has “more” instructions than the reg-mem architecture. Also give a pair of instruction sequences where the RISC pipeline has “fewer” instructions than the reg-mem architecture.
1. Who seems to have more variability in their shoe sizes, men or women?
a) Men
b) Women
c) Neither group show variability
d) Flag this Question
2. In general, why use the estimate of $$n-1$$ rather than n in the computation of the standard deviation and variance?
a) The estimate n-1 is better because it is used for calculating the population variance and standard deviation
b) The estimate n-1 is never used to calculate the sample variance and standard deviation
c) $$n-1$$ provides an unbiased estimate of the population and allows more variability when using a sample and gives a better mathematical estimate of the population
d) The estimate n-1 is better because it is use for calculation of both the population and sample variance as well as standard deviation.
$$\begin{array}{|c|c|}\hline \text{Shoe Size (in cm)} & \text{Gender (M of F)} \\ \hline 25.7 & M \\ \hline 25.4 & F \\ \hline 23.8 & F \\ \hline 25.4 & F \\ \hline 26.7 & M \\ \hline 23.8 & F \\ \hline 25.4 & F \\ \hline 25.4 & F \\ \hline 25.7 & M \\ \hline 25.7 & F \\ \hline 23.5 & F \\ \hline 23.1 & F \\ \hline 26 & M \\ \hline 23.5 & F \\ \hline 26.7 & F \\ \hline 26 & M \\ \hline 23.1 & F \\ \hline 25.1 & F \\ \hline 27 & M \\ \hline 25.4 & F \\ \hline 23.5 & F \\ \hline 23.8 & F \\ \hline 27 & M \\ \hline 25.7 & F \\ \hline \end{array}$$
$$\begin{array}{|c|c|}\hline \text{Shoe Size (in cm)} & \text{Gender (M of F)} \\ \hline 27.6 & M \\ \hline 26.9 & F \\ \hline 26 & F \\ \hline 28.4 & M \\ \hline 23.5 & F \\ \hline 27 & F \\ \hline 25.1 & F \\ \hline 28.4 & M \\ \hline 23.1 & F \\ \hline 23.8 & F \\ \hline 26 & F \\ \hline 25.4 & M \\ \hline 23.8 & F \\ \hline 24.8 & M \\ \hline 25.1 & F \\ \hline 24.8 & F \\ \hline 26 & M \\ \hline 25.4 & F \\ \hline 26 & M \\ \hline 27 & M \\ \hline 25.7 & F \\ \hline 27 & M \\ \hline 23.5 & F \\ \hline 29 & F \\ \hline \end{array}$$
Loretta, who turns eighty this year, has just learned about blood pressure problems in the elderly and is interested in how her blood pressure compares to those of her peers. Specifically, she is interested in her systolic blood pressure, which can be problematic among the elderly. She has uncovered an article in a scientific journal that reports that the mean systolic blood pressure measurement for women over seventy-five is 133.0 mmHg, with a standard deviation of 5.1 mmHg.
Assume that the article reported correct information. Complete the following statements about the distribution of systolic blood pressure measurements for women over seventy-five.
a) According to Chebyshev's theorem, at least $$?36\% 56\% 75\% 84\%\ or\ 89\%$$ of the measurements lie between 122.8 mmHg and 143.2 mmHg.
b) According to Chebyshev's theorem, at least $$8/9 (about\ 89\%)$$ of the measurements lie between mmHg and mmHg. (Round your answer to 1 decimal place.)