ECON 321 Chapter 9 (with steps)
[Part of card #85] Compute the corresponding standardized sample test statistic. (Round your answer to two decimal places.)
-1.41 - Use SALT set to "One Sample Proportion" https://www.webassign.net/csalt/index.html#/toolset/inferential-statistics
[Part of card #65] What is the value of the sample test statistic? (Round your answer to three decimal places.)
-2.000 - Use SALT set to One Sample t https://www.webassign.net/csalt/index.html#/toolset/inferential-statistics (OR use the formula: t = (sample mean - population mean) / (standard deviation / SQRT(sample size)))
[Part of card # 21] (a) What is the level of significance?
0.01 - The value given for 𝛼.
[Part of card # 30] (a) What is the level of significance?
0.01 - The value given for 𝛼.
[Part of card #56] (a) What is the level of significance?
0.01 - The value given for 𝛼.
[Part of card #42] [Step 5] We have determined that the P-value falls between the corresponding right-tail areas 0.050 and 0.025. Therefore, we have the following interval. _____ < P-value < _____
0.025 < P-value < 0.050
[Part of card #65] (a) What is the level of significance?
0.05 Given value for 𝛼.
[Part of card #42] [Step 3] We have determined that the P-value falls between the corresponding two-tail areas 0.100 and 0.050. Therefore, we have the following interval. _____ < P-value < _____
0.050 < P-value < 0.100
[Part of card #65] (c) Estimate the P-value.
0.050 < P-value < 0.100
[Part of card #75] (b) Using methods of Chapter 8, find the P-value for the hypothesis test. (Round your answer to four decimal places.)
0.0778 - Find the z value using the formula: z = (x̄ - 𝜇) / (𝜎 / sqrt(n)) - You can use this calculator to find the P-value from the z value: https://www.socscistatistics.com/pvalues/normaldistribution.aspx
[Part of card #85] (d) Find the P-value of the test statistic. (Round your answer to four decimal places.)
0.1586 - Use SALT set to "One Sample Proportion" https://www.webassign.net/csalt/index.html#/toolset/inferential-statistics
[Part of card # 21] (c) Find (or estimate) the P-value. (Round your answer to four decimal places.)
0.1796 - Use a table (OR use this online calculator set to one-tailed: https://www.socscistatistics.com/pvalues/normaldistribution.aspx)
[Part of card #93] [Step 8] Calculate the P-value using the fact that P(z < −0.45) = 0.3274. P-value = 2P(z < −0.45) = 2(________) = ________
0.3274; 0.6548
[Part of card #85] (c) Compute p̂.
0.4 - Use SALT set to "One Sample Proportion" https://www.webassign.net/csalt/index.html#/toolset/inferential-statistics
[Part of card # 30] (c) Find (or estimate) the P-value. (Round your answer to four decimal places.)
0.4715 - Use a table (OR use this online calculator set to two-tailed: https://www.socscistatistics.com/pvalues/normaldistribution.aspx)
[Part of card # 30] Compute the z value of the sample test statistic. (Round your answer to two decimal places.)
0.72 z = (x̄-𝜇)/(𝜎/sqrt(n))
[Part of card # 21] Compute the z value of the sample test statistic. (Round your answer to two decimal places.)
0.92 Use the formula: z = (x̄-𝜇)/(𝜎/sqrt(n))
[Part of card #75] (a) What is the value of c = 1 − 𝛼?
0.95
[Part of card #48] (c) Compute the t value of the sample test statistic. (Round your answer to three decimal places.)
1.000 - Use SALT set to One Sample t: https://www.webassign.net/csalt/index.html#/toolset/inferential-statistics (OR use the formula: t = (sample mean - population mean) / (standard deviation / SQRT(sample size)))
[Part of card #42] [Step 1] (a) Find an interval containing the corresponding P-value for a two-tailed test. To determine where the given t-value would fall, we look for the row with heading d.f. = ___. In this row, the sample statistic t = 2.010 falls between t = 1.796 and t = ___.
11 (given); 2.201 (value to the right of 1.796)
[Part of card #48] How many degrees of freedom do we use?
15 - Formula: D.F. = n - 1
[Part of card #56] What is the value of the sample test statistic? (Round your answer to three decimal places.)
2.474 - Use SALT set to One Sample t https://www.webassign.net/csalt/index.html#/toolset/inferential-statistics (OR use the formula: t = (sample mean - population mean) / (standard deviation / SQRT(sample size)))
[Part of card#10] [Step 1] (a) State the null hypothesis. Our first step is to establish a working hypothesis about the population parameter in question. This hypothesis is called the null hypothesis, H0. Here the population parameter in question is the mean. The statement that is under investigation is that this mean is equal to ____. Therefore, we have the following null hypothesis. H0 : 𝜇 ___ 45
45 (given); =
[Part of card#10] [Step 2] (b) State the alternate hypothesis if you have no information regarding how the population mean might differ from 45. The alternate hypothesis is the statement we adopt in the situation where the evidence is so strong that we reject the null hypothesis. Here we have H0: 𝜇 = 45. However, we have no information about how the population might differ. So the mean could be higher or lower than 45. In this case, we would reject the null hypothesis if the population mean is any value EXCEPT ________. Therefore, we have the following alternate hypothesis. H1: 𝜇 ___ 45
45 (given); ≠
Briefly answer the following questions. (a) What is a null hypothesis H0?
A working hypothesis making a claim about the population parameter in question.
(b) What is an alternate hypothesis H1?
Any hypothesis that differs from the original claim being made.
[Part of card # 21] (d) 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 𝛼?
At the 𝛼 = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
[Part of card # 30] (d) 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 𝛼?
At the 𝛼 = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
[Part of card #56] (d) 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 𝛼?
At the 𝛼 = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant.
[Part of card #48] (e) Do we reject or fail to reject H0?
At the 𝛼 = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
[Part of card #65] (d) 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 𝛼?
At the 𝛼 = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
[Part of card #85] (e) Do you reject or fail to reject H0? Explain.
At the 𝛼 = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
[Part of card #75] Do we reject or fail to reject H0?
Fail to reject the null hypothesis, there is insufficient evidence that 𝜇 differs from 21.
[Part of card #75] Do we reject or fail to reject H0 based on this information?
Fail to reject, since 𝜇 = 21 is contained in this interval.
[Part of card #85] (b) State the hypotheses.
H0: p = 0.5; H1: p ≠ 0.5
[Part of card #56] State the null and alternate hypotheses.
H0: 𝜇 = 1.75 yr; H1: 𝜇 > 1.75 yr
[Part of card # 30] State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?
H0: 𝜇 = 11%; H1: 𝜇 ≠ 11%; two-tailed
[Part of card #48] (b) What are the hypotheses?
H0: 𝜇 = 12.5; H1: 𝜇 ≠ 12.5
[Part of card #65] State the null and alternate hypotheses.
H0: 𝜇 = 19.4; H1: 𝜇 ≠ 19.4
[Part of card # 21] State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?
H0: 𝜇 = 4.8%; H1: 𝜇 > 4.8%; right-tailed
[Part of card #93] [Step 3] (b) State the hypotheses. We now write the null hypothesis H0 and the alternate hypothesis H1. Here, we wish to test the claim that the population proportion of successes does not equal 0.50. Therefore, we have the following hypotheses.
H0:p = 50 H1:p ≠ 50
To test 𝜇 for an x distribution that is mound-shaped using sample size n ≥ 30, how do you decide whether to use the normal or Student's t distribution?
If 𝜎 is known, use the standard normal distribution. If 𝜎 is unknown, use the Student's t distribution with n - 1 degrees of freedom.
If we fail to reject (i.e., "accept") the null hypothesis, does this mean that we have proved it to be true beyond all doubt? Explain your answer.
No, it suggests that the evidence is not sufficient to merit rejecting the null hypothesis.
If we reject the null hypothesis, does this mean that we have proved it to be false beyond all doubt? Explain your answer.
No, the test was conducted with a risk of a type I error.
[Part of card #56] (c) Estimate the P-value.
P-value < 0.010
[Part of card #48] (d) Estimate the P-value for the test.
P-value > 0.250
Suppose you want to test the claim that a population mean equals 45. (a) State the null hypothesis. (b) State the alternate hypothesis if you have no information regarding how the population mean might differ from 45. (c) State the alternative hypothesis if you believe (based on experience or past studies) that the population mean may exceed 45. (d) State the alternative hypothesis if you believe (based on experience or past studies) that the population mean may be less than 45.
See cards #11-14.
The body weight of a healthy 3-month-old colt should be about 𝜇 = 76 kg.
See cards #16-20
Let x be a random variable representing dividend yield of bank stocks. We may assume that x has a normal distribution with 𝜎 = 2.0%. A random sample of 10 bank stocks gave the following yields (in percents). 5.7, 4.8, 6.0, 4.9, 4.0, 3.4, 6.5, 7.1, 5.3, 6.1 The sample mean is x̄ = 5.38%. Suppose that for the entire stock market, the mean dividend yield is 𝜇 = 4.8%. Do these data indicate that the dividend yield of all bank stocks is higher than 4.8%? Use 𝛼 = 0.01.
See cards #22-29
Nationally, about 11% of the total U.S. wheat crop is destroyed each year by hail.† An insurance company is studying wheat hail damage claims in a county in Colorado. A random sample of 16 claims in the county reported the percentage of their wheat lost to hail. Data: 14, 8, 10, 10, 11, 20, 15, 9, 7, 8, 22, 21, 12, 7, 13, 4 The sample mean is x = 11.9%. Let x be a random variable that represents the percentage of wheat crop in that county lost to hail. Assume that x has a normal distribution and 𝜎 = 5.0%. Do these data indicate that the percentage of wheat crop lost to hail in that county is different (either way) from the national mean of 11%? Use 𝛼 = 0.01.
See cards #31-38
For a Student's t distribution with d.f. = 11 and t = 2.010, consider the following. (a) Find an interval containing the corresponding P-value for a two-tailed test. (b) Find an interval containing the corresponding P-value for a right-tailed test.
See cards #43-47
A random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 13 and the sample standard deviation is 2. Use a level of significance of 0.05 to conduct a two-tailed test of the claim that the population mean is 12.5.
See cards #49-55
A random sample of 41 adult coyotes in a region of northern Minnesota showed the average age to be x = 2.09 years, with sample standard deviation s = 0.88 years. However, it is thought that the overall population mean age of coyotes is 𝜇 = 1.75. Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use 𝛼 = 0.01.
See cards #57-64
Socially conscious investors screen out stocks of alcohol and tobacco makers, firms with poor environmental records, and companies with poor labor practices. Some examples of "good," socially conscious companies are Johnson and Johnson, Dell Computers, Bank of America, and Home Depot. The question is, are such stocks overpriced? One measure of value is the P/E, or price-to-earnings ratio. High P/E ratios may indicate a stock is overpriced. For the S&P Stock Index of all major stocks, the mean P/E ratio is 𝜇 = 19.4. A random sample of 36 "socially conscious" stocks gave a P/E ratio sample mean of x = 17.6, with sample standard deviation s = 5.4. Does this indicate that the mean P/E ratio of all socially conscious stocks is different (either way) from the mean P/E ratio of the S&P Stock Index? Use 𝛼 = 0.05.
See cards #66-73
Is there a relationship between confidence intervals and two-tailed hypothesis tests? Let c be the level of confidence used to construct a confidence interval from sample data. Let 𝛼 be the level of significance for a two-tailed hypothesis test. The following statement applies to hypothesis tests of the mean. For a two-tailed hypothesis test with level of significance 𝛼 and null hypothesis H0: μ = k, we reject H0 whenever k falls outside the c = 1 - α confidence interval for 𝜇 based on the sample data. When k falls within the c = 1 - α confidence interval, we do not reject H0. Whenever the value of k given in the null hypothesis falls outside the c = 1 - α confidence interval for the parameter, we reject H0. For example, consider a two-tailed hypothesis test with 𝛼 = 0.05 and H0: 𝜇 = 21 H1: 𝜇 ≠ 21 A random sample of size 28 has a sample mean x = 23 from a population with standard deviat
See cards #75-81
A random sample of 50 binomial trials resulted in 20 successes. Test the claim that the population proportion of successes does not equal 0.50. Use a level of significance of 0.05.
See cards #86-92
A random sample of 20 binomial trials resulted in 9 successes. Test the claim that the population proportion of successes does not equal 0.50. Use a level of significance of 0.05. (a) Can a normal distribution be used for the p̂ distribution? Explain. (b) State the hypotheses. (c) Compute p̂ and the corresponding standardized sample test statistic. (d) Find the P-value of the test statistic. (e) Do you reject or fail to reject H0? Explain. (f) What do the results tell you?
See cards #94-103
For the same sample data and null hypothesis, how does the P-value for a two-tailed test of 𝜇 compare to that for a one-tailed test?
The P-value for a two-tailed test is twice the P-value for a one-tailed test.
[Part of card #56] (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.
The Student's t, since the sample size is large and 𝜎 is unknown.
[Part of card #65] (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.
The Student's t, since the sample size is large and 𝜎 is unknown.
In a statistical test, we have a choice of a left-tailed test, a right-tailed test, or a two-tailed test. Is it the null hypothesis or the alternate hypothesis that determines which type of test is used? Explain your answer.
The alternative hypothesis because it specifies the region of interest for the parameter in question.
[Part of card # 30] Sketch the sampling distribution and show the area corresponding to the P-value.
The only graph with two tails.
[Part of card #65] Sketch the sampling distribution and show the area corresponding to the P-value.
The only two-tailed graph.
(d) What is the level of significance of a test?
The probability of a type I error.
[Part of card #56] Sketch the sampling distribution and show the area corresponding to the P-value.
The right-tailed graph with the shaded area that starts around the value of the sample test statistic.
[Part of card # 21] Sketch the sampling distribution and show the area corresponding to the P-value.
The right-tailed graph with the shaded area that starts from the z value.
[Part of card #85] (f) What do the results tell you?
The sample p̂ value based on 50 trials is not sufficiently different from 0.50 to justify rejecting H0 for 𝛼 = 0.05.
[Part of card # 21] (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.
The standard normal, since we assume that x has a normal distribution with known 𝜎.
[Part of card # 30] (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.
The standard normal, since we assume that x has a normal distribution with known 𝜎.
To use the normal distribution to test a proportion p, the conditions np > 5 and nq > 5 must be satisfied. Does the value of p come from H0, or is it estimated by using p̂ from the sample?
The value of p comes from H0.
[Part of card # 30] (e) State your conclusion in the context of the application.
There is insufficient evidence at the 0.01 level to conclude that the average hail damage to wheat crops in the county in Colorado differs from the national average.
[Part of card # 21] (e) State your conclusion in the context of the application.
There is insufficient evidence at the 0.01 level to conclude that the average yield for bank stocks is higher than that of the entire stock market.
[Part of card #65] (e) Interpret your conclusion in the context of the application.
There is insufficient evidence at the 0.05 level to conclude that the mean P/E ratio of all socially conscious stocks differs from the mean P/E ratio of the S&P Stock Index.
[Part of card #48] (f) Interpret the results.
There is insufficient evidence at the 0.05 level to reject the null hypothesis.
[Part of card #56] (e) Interpret your conclusion in the context of the application.
There is sufficient evidence at the 0.01 level to conclude that coyotes in the specified region tend to live longer than 1.75 years.
[Part of card #75] Compare your result to that of part (a).
These results are the same.
(c) What is a type I error?
Type I error is rejecting the null hypothesis when it is true.
What is a type II error?
Type II error is failing to reject the null hypothesis when it is false.
[Part of card #48] (a) Is it appropriate to use a Student's t distribution? Explain.
Yes, because the x distribution is mound-shaped and symmetric and 𝜎 is unknown.
[Part of card #85] (a) Can a normal distribution be used for the p̂ distribution? Explain.
Yes, np and nq are both greater than 5.
In general, if sample data are such that the null hypothesis is rejected at the 𝛼 = 1% level of significance based on a two-tailed test, is H0 also rejected at the 𝛼 = 1% level of significance for a corresponding one-tailed test? Explain your answer.
Yes. If the two-tailed P-value is smaller than 𝛼, the one-tailed area is also smaller than 𝛼.
[Part of card#10] [Step 3] (c) State the alternative hypothesis if you believe (based on experience or past studies) that the population mean may exceed 45. Here we have H0: 𝜇 = 45. However, testing might show that the mean may be more than 45. In this case, rejecting the null hypothesis would mean claiming that the population mean is ________. Therefore, we have the following alternate hypothesis. H1: 𝜇 ___ 45
greater than 45; >
[Part of card #75] What is the value of 𝜇 given in the null hypothesis (i.e., what is k)? Is this value in the confidence interval?
k = 21; Yes - Given: H0: 𝜇 = 21, H1: 𝜇 ≠ 21
[Info on card #15] (e) For each of the tests in parts (b), (c), and (d), respectively, would the area corresponding to the P-value be on the left, on the right, or on both sides of the mean?
left; right; both
[Part of card#10] [Step 4] (d) State the alternative hypothesis if you believe (based on experience or past studies) that the population mean may be less than 45. Here we have H0: 𝜇 = 45. However, testing might show that the mean may be less than 45. In this case, rejecting the null hypothesis would mean claiming that the population mean is ________. Therefore, we have the following alternate hypothesis. H1: 𝜇 ___ 45
less than 45; <
[Part of card #75] Using the methods of Chapter 7, construct a 1 − 𝛼 confidence interval for 𝜇 from the sample data. (Round your answers to two decimal places.)
lower limit = 20.78 upper limit = 25.22 - Use the formula x̄ ± (z-score of c) * 𝜎 / sqrt(n) (OR use this online calculator https://www.omnicalculator.com/statistics/confidence-interval)
When using the Student's t distribution to test 𝜇, what value do you use for the degrees of freedom?
n - 1
[Part of card #93] [Step 1] (a) Can a normal distribution be used for the p̂ distribution? Explain... We have a random sample of 20 binomial trials resulting in 9 successes and we wish to test the claim that the population proportion of successes does not equal 0.50 using a significance level of 0.05. Therefore, we can define n, p, and q as follows.
n = 20 - n = the random sample p = 0.5 - population proportion of success does not equal __ q = 1-p = 0.5
[Part of card #93] [Step 2] Now calculate np and nq for n = 20, p = 0.50, and q = 0.50. We conclude that since _____ both np and nq are greater than 5 both np and nq are less than 5 , the requirements have been met to use the normal distribution for the p̂ distribution.
np = 20(0.5) = 10 nq = 20(0.5) = 10 both np and nq are greater than 5
[Part of card #42] [Step 4] (b) Find an interval containing the corresponding P-value for a right-tailed test. We now need to determine the corresponding areas. We look directly above the t values t = 1.796 and t = 2.201 to find the corresponding values in one of the first rows of the table. Since this is a right-tailed test, we use entries from the row labeled _____. Therefore, the P-value falls between the corresponding areas 0.050 and _____.
one-tail area; 0.025 (one-tail area value associated with the second t value)
[Part of card #93] [Step 4] (c) Compute p̂ and the corresponding standardized sample test statistic. First, calculate p̂. We have a random sample of 20 binomial trials resulting in 9 successes. So, we have r = ___ and n = ___. Use these values to calculate p̂, writing the result as an exact decimal value. p̂ = r/n
r = 9 - # of successes n = 20 - # of trials p̂ = 0.45
Consider a binomial experiment with n trials and r successes. To construct a test for a proportion p, what value do we use for the z value of the sample test statistic?
r/n
[Part of card #42] [Step 2] We now look directly above these values to find the corresponding values in one of the first rows of the table. Since this is a two-tailed test, we use entries from the row labeled ___. Therefore, the P-value falls between the corresponding areas 0.100 and ___.
two-tail area; 0.050 (two-tail area value associated with the t value you found in step 1)
[Part of card #93] [Step 6] (d) Find the P-value of the test statistic. To find the P-value of the test statistic, z = −0.45, we will use the Standard Normal Distribution Table keeping in mind that this is a _______, therefore we use the following formula. P-value = ___ P(z < −0.45)
two-tailed test; 2
[Part of card #93] [Step 5] Now, calculate the corresponding standardized sample test statistic, the z value, using the formula z = (p̂ − p)/SQRT(pq/n). We previously determined that p̂ = 0.45, p = 0.50, q = 0.50, and n = 20. Use these values to calculate z, rounding the final result to two decimal places.
z = -0.45
What is the probability of a type II error?
𝛽
[Info on card #15] (b) In Nevada, there are many herds of wild horses. Suppose you want to test the claim that the average weight of a wild Nevada colt (3 months old) is less than 76 kg. What would you use for the alternate hypothesis H1?
𝜇 < 76 kg
[Info on card #15] (a) If you want to set up a statistical test to challenge the claim that 𝜇 = 76 kg, what would you use for the null hypothesis H0?
𝜇 = 76 kg
[Info on card #15] (c) Suppose you want to test the claim that the average weight of such a wild colt is greater than 76 kg. What would you use for the alternate hypothesis?
𝜇 > 76 kg
[Info on card #15] (d) Suppose you want to test the claim that the average weight of such a wild colt is different from 76 kg. What would you use for the alternate hypothesis?
𝜇 ≠ 76 kg
