# Twenty-one independent measurements were taken of the hardness (on the Rockwell C scale) of HSLA-100 steel base metal, and another 21 independent measurements were made of the hardness of a weld produced on this base metal. The standard deviation of the measurements made on the base metal was 3.06, and the standard deviation of the measurements made on the weld was 1.41. Assume that the measurements are independent random samples from normal populations. Need to conclude that measurements made on the base metal are more variable than measurements made on the weld?

Question
Measurement
Twenty-one independent measurements were taken of the hardness (on the Rockwell C scale) of HSLA-100 steel base metal, and another 21 independent measurements were made of the hardness of a weld produced on this base metal.
The standard deviation of the measurements made on the base metal was 3.06, and the standard deviation of the measurements made on the weld was 1.41.
Assume that the measurements are independent random samples from normal populations.
Need to conclude that measurements made on the base metal are more variable than measurements made on the weld?

2021-02-06
Step 1
According to the given question, twenty-one independent measurements were taken of the hardness (on the Rockwell C scale) of HSLA-100 steel base metal, and another 21 independent measurements were made of the hardness of a weld produced on this base metal.
The standard deviation of the measurements made on the base metal was 3.06, and the standard deviation of the measurements made on the weld was 1.41.
Assume that the measurements are independent random samples from normal populations.
Therefore the given data are as follows:
$$n_{1}=21, s_{1}=3.06\ and\ n_{2}=21, s_{2}=1.41$$
In order to test whether the measurements made on the base metal are move variable than measurements made on the weld, we define the test hypothesis as:
$$H_{0}:\sum_{1}^{2}=\sigma_{2}^{2}$$
Against the alternative hypotesis as
$$H_{0}:\sum_{1}^{2}\neq \sigma_{2}^{2}$$
This hypothesis follows a F statistics with $$df_{1}=n_{1}-1=20,df_{2}=n_{2}-1=20$$
and the confidence level at $$\alpha = 0.05$$:
Step 2
$$F=\frac{S_{1}^{2}}{S_{2}^{2}}=\frac{9.832}{2.088}=4.709$$
Where
$$S_{1}^{2}=\frac{n_{1}}{n_{1}-1}s_{1}^{2}=9.832\ and\ S_{2}^{2}=\frac{n_{2}}{n_{2}-1}s_{2}^{2}=2.088$$
Tabulated F statistics value is determined as:
$$F_{20,20,0.05}=2.124$$
As the calculated value is more that the tabulated value
$$F_{observed}=4.709>F_{20,20,0.05}=2.124$$
We reject the null hypothesis at 55 level of significance and we have sufficient evidence to conclude that measurements made on the base metal are more variable than measurements made on the weld.

### 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.
A new thermostat has been engineered for the frozen food cases in large supermarkets. Both the old and new thermostats hold temperatures at an average of $$25^{\circ}F$$. However, it is hoped that the new thermostat might be more dependable in the sense that it will hold temperatures closer to $$25^{\circ}F$$. One frozen food case was equipped with the new thermostat, and a random sample of 21 temperature readings gave a sample variance of 5.1. Another similar frozen food case was equipped with the old thermostat, and a random sample of 19 temperature readings gave a sample variance of 12.8. Test the claim that the population variance of the old thermostat temperature readings is larger than that for the new thermostat. Use a $$5\%$$ level of significance. How could your test conclusion relate to the question regarding the dependability of the temperature readings? (Let population 1 refer to data from the old thermostat.)
(a) What is the level of significance?
State the null and alternate hypotheses.
$$H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}>?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}\neq?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}?_{2}^{2},H1:?_{1}^{2}=?_{2}^{2}$$
(b) Find the value of the sample F statistic. (Round your answer to two decimal places.)
What are the degrees of freedom?
$$df_{N} = ?$$
$$df_{D} = ?$$
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. We have random samples from each population.The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings. Fail to reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.Reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.
$$\begin{array}{cc}\hline & \text{Afraid to walk at night?} \\ \hline & \text{Yes} & \text{No} & \text{Total} \\ \hline \text{Male} & 173 & 598 & 771 \\ \hline \text{Female} & 393 & 540 & 933 \\ \hline \text{Total} & 566 & 1138 & 1704 \\ \hline \text{Source:}2014\ GSS \end{array}$$
If the chi-square $$(\chi 2)$$ test statistic $$=73.7$$ what is the p-value you would report? Use Table C.Remember to calculate the df first.
$$P>0.250$$
$$P=0.01$$
$$P<0.001$$
$$P<0.002$$

$$\begin{array}{|c|cc|}\hline & \text{Right-Tail Probability} \\ \hline df & 0.250 & 0.100 & 0.050 & 0.025 & 0.010 & 0.005 & 0.001 \\ \hline 1 & 1.32 & 2.71 & 3.84 & 5.02 & 6.63 & 7.88 & 10.83 \\ 2 & 2.77 & 4.61 & 5.99 & 7.38 & 9.21 & 10.60 & 13.82 \\ 3 & 4.11 & 6.25 & 7.81 & 9.35 & 11.34 & 12.84 & 16.27 \\ 4 & 5.39 & 7.78 & 9.49 & 11.14 & 13.28 & 14.86 & 18.47 \\ 5 & 6.63 & 9.24 & 11.07 & 12.83 & 15.09 & 16.75 & 20.52 \\ 6&7.84&10.64&12.59&14.45&16.81&18.55&22.46 \\ 7&9.04&12.02&14.07&16.01&18.48&20.28&24.32\\ 8&10.22&13.36&15.51&17.53&20.09&21.96&26.12 \\ 9&11.39&14.68&16.92&19.02&21.67&23.59&27.88 \\ 10&12.55&15.99&18.31&20.48&23.21&25.19&29.59 \\ 11&13.70&17.28&19.68&21.92&24.72&26.76&31.26 \\ 12&14.85&18.55&21.03&23.34&26.22&28.30&32.91 \\ 13&15.98&19.81&22.36 & 24.74 & 27.69 & 29.82 & 34.53 \\ 14 & 17.12 & 21.06 & 23.68 & 26.12 & 29.14 & 31.32 & 36.12 \\15 & 18.25 & 22.31 & 25.00 & 27.49 & 30.58 & 32.80 & 37.70 \\ 16 & 19.37 & 32.54 & 26.30 & 28.85 & 32.00 & 34.27 & 39.25 \\ 17 & 20.49 & 24.77 & 27.59 & 30.19 & 33.41 & 35.72 & 40.79 \\ 18 & 21.60 & 25.99 & 28.87 & 31.53 & 34.81 & 37.16 & 42.31 \\ 19 & 22.72 & 27.20 & 30.14 & 32.85 & 36.19 & 38.58 & 43.82 \\ 20 & 23.83 & 28.41 & 31.41 & 34.17 & 37.57 & 40.00 & 45.32 \\ \hline \end{array}$$
A certain scale has an uncertainty of 3 g and a bias of 2 g. a) A single measurement is made on this scale. What are the bias and uncertainty in this measurement? b) Four independent measurements are made on this scale. What are the bias and uncertainty in the average of these measurements? c) Four hundred independent measurements are made on this scale. What are the bias and uncertainty in the average of these measurements? d) As more measurements are made, does the uncertainty get smaller, get larger, or stay the same? e) As more measurements are made, does the bias get smaller, get larger, or stay the same?
A certain scale has an uncertainty of 3 g and a bias of 2 g.
a) A single measurement is made on this scale. What are the bias and uncertainty in this measurement?
b) Four independent measurements are made on this scale. What are the bias and uncertainty in the average of these measurements? c) Four hundred independent measurements are made on this scale. What are the bias and uncertainty in the average of these measurements?
d) As more measurements are made, does the uncertainty get smaller, get larger, or stay the same?
e) As more measurements are made, does the bias get smaller, get larger, or stay the same?

Assume that the random variable Z follows standard normal distribution, calculate the following probabilities (Round to two decimal places)
a)$$P(z>1.9)$$
b)$$\displaystyle{P}{\left(−{2}\le{z}\le{1.2}\right)}$$
c)$$P(z\geq0.2)$$

1. The standard error of the estimate is the same at all points along the regression line because we assumed that A. The observed values of y are normally distributed around each estimated value of y-hat. B. The variance of the distributions around each possible value of y-hat is the same. C. All available data were taken into account when the regression line was calculated. D. The regression line minimized the sum of the squared errors. E. None of the above.
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.
$$\begin{array}{|c|c|}\hline \text{Shoe Size (in cm)} & \text{Gender (M of F)} \\ \hline 25.7 & M \\ \hline 25.4 & F \\ \hline 23.8 & F \\ \hline 25.4 & F \\ \hline 26.7 & M \\ \hline 23.8 & F \\ \hline 25.4 & F \\ \hline 25.4 & F \\ \hline 25.7 & M \\ \hline 25.7 & F \\ \hline 23.5 & F \\ \hline 23.1 & F \\ \hline 26 & M \\ \hline 23.5 & F \\ \hline 26.7 & F \\ \hline 26 & M \\ \hline 23.1 & F \\ \hline 25.1 & F \\ \hline 27 & M \\ \hline 25.4 & F \\ \hline 23.5 & F \\ \hline 23.8 & F \\ \hline 27 & M \\ \hline 25.7 & F \\ \hline \end{array}$$
$$\begin{array}{|c|c|}\hline \text{Shoe Size (in cm)} & \text{Gender (M of F)} \\ \hline 27.6 & M \\ \hline 26.9 & F \\ \hline 26 & F \\ \hline 28.4 & M \\ \hline 23.5 & F \\ \hline 27 & F \\ \hline 25.1 & F \\ \hline 28.4 & M \\ \hline 23.1 & F \\ \hline 23.8 & F \\ \hline 26 & F \\ \hline 25.4 & M \\ \hline 23.8 & F \\ \hline 24.8 & M \\ \hline 25.1 & F \\ \hline 24.8 & F \\ \hline 26 & M \\ \hline 25.4 & F \\ \hline 26 & M \\ \hline 27 & M \\ \hline 25.7 & F \\ \hline 27 & M \\ \hline 23.5 & F \\ \hline 29 & F \\ \hline \end{array}$$
A log 10 m long is cut at 1 meter intervals and itscross-sectional areas A (at a distance x from theend of the log) are listed in the table. Use the Midpoint Rule withn = 5 to estimate the volume of the log. (in $$\displaystyle{m}^{{{3}}}$$ )
Show transcribed image text A log 10 m long is cut at 1 meter intervals and itscross-sectional areas A (at a distance x from theend of the log) are listed in the table. Use the Midpoint Rule withn = 5 to estimate the volume of the log. (in $$\displaystyle{m}^{{{3}}}$$)