# Geographical Analysis (Oct. 2006) published a study of a new method for analyzing remote-sensing data from satellite pixels in order to identify urban land cover. The method uses a numerical measure of the distribution of gaps, or the sizes of holes, in the pixel, called lacunarity. Summary statistics for the lacunarity measurements in a sample of 100 grassland pixels are x¯=225 and s=20s=20. It is known that the mean lacunarity measurement for all grassland pixels is 220. The method will be effective in identifying land cover if the standard deviation of the measurements is 10% (or less) of the true mean (i.e., if the standard deviation is less than 22). a. Give the null and alternative hypotheses for a test to determine whether, in fact, the standard deviation of all grassland pixels is

Question
Analysis
Geographical Analysis (Oct. 2006) published a study of a new method for analyzing remote-sensing data from satellite pixels in order to identify urban land cover. The method uses a numerical measure of the distribution of gaps, or the sizes of holes, in the pixel, called lacunarity. Summary statistics for the lacunarity measurements in a sample of 100 grassland pixels are x¯=225 and s=20s=20. It is known that the mean lacunarity measurement for all grassland pixels is 220. The method will be effective in identifying land cover if the standard deviation of the measurements is 10% (or less) of the true mean (i.e., if the standard deviation is less than 22). a. Give the null and alternative hypotheses for a test to determine whether, in fact, the standard deviation of all grassland pixels is less than 22. b. A MINITAB analysis of the data is provided below. Locate and interpret the p-value of the test. Use α=.10α=.10. Test for One Standard Deviation Method Null hypothesisSigma = 22 Method Alternative hypothesisSigma = < 22 The standard method is only for the normal distribution. Statistics NStDevVariance 10020.0400 Tests

2020-11-24
(a)H0 :q = 22, Ha:q< 22 (b)P = 0.105, there is a 10.5% chance of obtaining « sample standard deviation of 20 or more extreme (among 100 observations), when the ture standard deviation is less than 22. There is not sufficient evidence to support the claim that the standard deviation of all grassland pixels is less than 22.

### Relevant Questions

True or False
1.The goal of descriptive statistics is to simplify, summarize, and organize data.
2.A summary value, usually numerical, that describes a sample is called a parameter.
3.A researcher records the average age for a group of 25 preschool children selected to participate in a research study. The average age is an example of a statistic.
4.The median is the most commonly used measure of central tendency.
5.The mode is the best way to measure central tendency for data from a nominal scale of measurement.
6.A distribution of scores and a mean of 55 and a standard deviation of 4. The variance for this distribution is 16.
7.In a distribution with a mean of M = 36 and a standard deviation of SD = 8, a score of 40 would be considered an extreme value.
8.In a distribution with a mean of M = 76 and a standard deviation of SD = 7, a score of 91 would be considered an extreme value.
9.A negative correlation means that as the X values decrease, the Y values also tend to decrease.
10.The goal of a hypothesis test is to demonstrate that the patterns observed in the sample data represent real patterns in the population and are not simply due to chance or sampling error.
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.
factor in determining the usefulness of an examination as a measure of demonstrated ability is the amount of spread that occurs in the grades. If the spread or variation of examination scores is very small, it usually means that the examination was either too hard or too easy. However, if the variance of scores is moderately large, then there is a definite difference in scores between "better," "average," and "poorer" students. A group of attorneys in a Midwest state has been given the task of making up this year's bar examination for the state. The examination has 500 total possible points, and from the history of past examinations, it is known that a standard deviation of around 60 points is desirable. Of course, too large or too small a standard deviation is not good. The attorneys want to test their examination to see how good it is. A preliminary version of the examination (with slight modifications to protect the integrity of the real examination) is given to a random sample of 20 newly graduated law students. Their scores give a sample standard deviation of 70 points. Using a 0.01 level of significance, test the claim that the population standard deviation for the new examination is 60 against the claim that the population standard deviation is different from 60.
(a) What is the level of significance?
State the null and alternate hypotheses.
$$H_{0}:\sigma=60,\ H_{1}:\sigma\ <\ 60H_{0}:\sigma\ >\ 60,\ H_{1}:\sigma=60H_{0}:\sigma=60,\ H_{1}:\sigma\ >\ 60H_{0}:\sigma=60,\ H_{1}:\sigma\ \neq\ 60$$
(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)
What are the degrees of freedom?
What assumptions are you making about the original distribution?
We assume a binomial population distribution.We assume a exponential population distribution. We assume a normal population distribution.We assume a uniform population distribution.
1. Find each of the requested values for a population with a mean of $$? = 40$$, and a standard deviation of $$? = 8$$ A. What is the z-score corresponding to $$X = 52?$$ B. What is the X value corresponding to $$z = - 0.50?$$ C. If all of the scores in the population are transformed into z-scores, what will be the values for the mean and standard deviation for the complete set of z-scores? D. What is the z-score corresponding to a sample mean of $$M=42$$ for a sample of $$n = 4$$ scores? E. What is the z-scores corresponding to a sample mean of $$M= 42$$ for a sample of $$n = 6$$ scores? 2. True or false: a. All normal distributions are symmetrical b. All normal distributions have a mean of 1.0 c. All normal distributions have a standard deviation of 1.0 d. The total area under the curve of all normal distributions is equal to 1 3. Interpret the location, direction, and distance (near or far) of the following zscores: $$a. -2.00 b. 1.25 c. 3.50 d. -0.34$$ 4. You are part of a trivia team and have tracked your team’s performance since you started playing, so you know that your scores are normally distributed with $$\mu = 78$$ and $$\sigma = 12$$. Recently, a new person joined the team, and you think the scores have gotten better. Use hypothesis testing to see if the average score has improved based on the following 8 weeks’ worth of score data: $$82, 74, 62, 68, 79, 94, 90, 81, 80$$. 5. You get hired as a server at a local restaurant, and the manager tells you that servers’ tips are $42 on average but vary about $$12 (\mu = 42, \sigma = 12)$$. You decide to track your tips to see if you make a different amount, but because this is your first job as a server, you don’t know if you will make more or less in tips. After working 16 shifts, you find that your average nightly amount is$44.50 from tips. Test for a difference between this value and the population mean at the $$\alpha = 0.05$$ level of significance.
Identify the appropriate hypothesis test for each of the following research situations using the options: The null hypothesis, The Test Statistics, The Sample Statistic, The Standard Error, and The Alpha Level.
A researcher conducts a cross-sectional developmental study to determine whether there is a significant difference in vocabulary skills between 8-year-old and 10-year-old children. A researcher determines that 8% of the males enrolled in Introductory Psychology have some form of color blindness, compared to only 2% of the females. Is there a significant relationship between color blindness and gender?
A researcher records the daily sugar consumption and the activity level for each of 20 children enrolled in a summer camp program. The researcher would like to determine whether there is a significant relationship between sugar consumption and activity level.
A researcher would like to determine whether a 4-week therapy program produces significant changes in behavior. A group of 25 participants is measured before therapy, at the end of therapy, and again 3 months after therapy.
A researcher would like to determine whether a new program for teaching mathematics is significantly better than the old program. It is suspected that the new program will be very effective for small-group instruction but probably will not work well with large classes. The research study involves comparing four groups of students: a small class taught by the new method, a large class taught by the new method, a small class taught by the old method, and a large class taught by the old method.
Hypothesis Testing Review
For each problem below, simply identify the null and alternative hypotheses. Use appropriate notation/symbols. You do not have to run any hypothesis tests, although it's good practice and I'll post answers for all of them.
1) A simple random sample of 44 men from a normally distributed population results in a standard deviation of 10.7 beats per minute. The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.10 significance level to test the claim that pulse rates of men have a standard deviation equal to 10 beats per minute.
2) In 1997, a survey of 880 households showed that 145 of them use e-mail. Use those sample results to test the claim that more than 15% of households use e-mail. Use a 0.05 significance level.
Hypothesis Testing Review
For each problem below, simply identify the null and alternative hypotheses. Use appropriate notation/symbols. You do not have to run any hypothesis tests, although it's good practice and I'll post answers for all of them.
1) A simple random sample of 44 men from a normally distributed population results in a standard deviation of 10.7 beats per minute. The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.10 significance level to test the claim that pulse rates of men have a standard deviation equal to 10 beats per minute.
2) In 1997, a survey of 880 households showed that 145 of them use e-mail. Use those sample results to test the claim that more than 15% of households use e-mail. Use a 0.05 significance level.
A random sample of $$\displaystyle{n}_{{1}}={16}$$ communities in western Kansas gave the following information for people under 25 years of age.
$$\displaystyle{X}_{{1}}:$$ Rate of hay fever per 1000 population for people under 25
$$\begin{array}{|c|c|} \hline 97 & 91 & 121 & 129 & 94 & 123 & 112 &93\\ \hline 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88 \\ \hline \end{array}$$
A random sample of $$\displaystyle{n}_{{2}}={14}$$ regions in western Kansas gave the following information for people over 50 years old.
$$\displaystyle{X}_{{2}}:$$ Rate of hay fever per 1000 population for people over 50
$$\begin{array}{|c|c|} \hline 94 & 109 & 99 & 95 & 113 & 88 & 110\\ \hline 79 & 115 & 100 & 89 & 114 & 85 & 96\\ \hline \end{array}$$
(i) Use a calculator to calculate $$\displaystyle\overline{{x}}_{{1}},{s}_{{1}},\overline{{x}}_{{2}},{\quad\text{and}\quad}{s}_{{2}}.$$ (Round your answers to two decimal places.)
(ii) Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use $$\displaystyle\alpha={0.05}.$$
(a) What is the level of significance?
State the null and alternate hypotheses.
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}<\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}\ne\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}>\mu_{{2}},{H}_{{1}}:\mu_{{1}}=\mu_{{12}}$$
(b) What sampling distribution will you use? What assumptions are you making?
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,
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
The Student's t. We assume that both population distributions are approximately normal with known standard deviations,
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimalplaces.)
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimal places.)
(c) Find (or estimate) the P-value.
P-value $$\displaystyle>{0.250}$$
$$\displaystyle{0.125}<{P}-\text{value}<{0},{250}$$
$$\displaystyle{0},{050}<{P}-\text{value}<{0},{125}$$
$$\displaystyle{0},{025}<{P}-\text{value}<{0},{050}$$
$$\displaystyle{0},{005}<{P}-\text{value}<{0},{025}$$
P-value $$\displaystyle<{0.005}$$
Sketch the sampling distribution and show the area corresponding to the P-value.
P.vaiue Pevgiue
P-value f P-value
Would you rather spend more federal taxes on art? Of a random sample of $$n_{1} = 86$$ politically conservative voters, $$r_{1} = 18$$ responded yes. Another random sample of $$n_{2} = 85$$ politically moderate voters showed that $$r_{2} = 21$$ responded yes. Does this information indicate that the population proportion of conservative voters inclined to spend more federal tax money on funding the arts is less than the proportion of moderate voters so inclined? Use $$\alpha = 0.05.$$ (a) State the null and alternate hypotheses. $$H_0:p_{1} = p_{2}, H_{1}:p_{1} > p_2$$
$$H_0:p_{1} = p_{2}, H_{1}:p_{1} < p_2$$
$$H_0:p_{1} = p_{2}, H_{1}:p_{1} \neq p_2$$
$$H_{0}:p_{1} < p_{2}, H_{1}:p_{1} = p_{2}$$ (b) What sampling distribution will you use? What assumptions are you making? The Student's t. The number of trials is sufficiently large. The standard normal. The number of trials is sufficiently large.The standard normal. We assume the population distributions are approximately normal. The Student's t. We assume the population distributions are approximately normal. (c)What is the value of the sample test statistic? (Test the difference $$p_{1} - p_{2}$$. Do not use rounded values. Round your final answer to two decimal places.) (d) Find (or estimate) the P-value. (Round your answer to four decimal places.) (e) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level alpha? At the $$\alpha = 0.05$$ level, we reject the null hypothesis and conclude the data are statistically significant. At the $$\alpha = 0.05$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the $$\alpha = 0.05$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the $$\alpha = 0.05$$ 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 sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters.
Using the Minitab statistical analysis program to enter the data and perform the analysis, complete the following: Construct a one-sided $$\displaystyle{95}\%$$ confidence interval for the true difference in population means. Test the null hypothesis that the population means are identical at the 0.05 level of significance.