Solved: Xavier Jiménez Albán
Last Update: 2024-11-19 13:15:06
The general population scores an average of \(\mu_0=80\) with a standard deviation of \(\sigma = 4.5\) on a memory test. A researcher conducts an experiment with 40 alcoholics. She believes the alcoholics will score lower on the average on the memory test. Is this test a one-sided or two-sided test?
Solution
This is a one-sided test.
A test to measure anxiety and depression has a mean score of \(\mu_0 = 55\) with a standard deviation equal to 3.5 for the general population. A group of 45 persons known to be suffering from anxiety and depression is given the test. It is of interest to know if the group will score differently than the general population on the test. Is this a one-sided or two-sided test?
Solution
This is a two sided test.
Rats are known to take a mean of 10 minutes to find their way through a maze to reach a food source with a standard deviation equal to 2.5 minutes. A researcher wishes to determine if taking a certain drug will change the time it takes to run the maze. Fifty rats are given the drug and timed on their run times through the maze. Is this a one-sided or two-sided test?
Solution
This is a two sided test.
A psychological researcher would like to evaluate a new approach to teaching statistics for the behavioral sciences. She knows that the average score that has been attained in this course over the past several years is 75 with \(\sigma = 7\). She uses the new approach on a sample of 75 to see if they score differently. Should she conduct a one-side or two-sided test to find the answer?
Solution
She should conduct a two-sided test.
State the null hypothesis and the alternative hypothesis in Problem 1.
Solution
Let \(\mu\) represent the average score on the memory tests for alcoholics.
\[H_0: \mu = \mu_0 \qquad H_1: \mu < \mu_0\] \[H_0: \mu = 80 \qquad H_1: \mu < 80\]
State the null hypothesis and the alternative hypothesis in Problem 2.
Solution
Let \(\mu\) represent the average score on anxiety and depression test for the group known to be suffering from depression anxiety
\[H_0: \mu = \mu_0 \qquad H_1: \mu \neq \mu_0\] \[H_0: \mu = 55 \qquad H_1: \mu \neq 55\]
State the null hypothesis and the alternative hypothesis in Problem 3.
Solution
\[H_0: \mu = \mu_0 \qquad H_1: \mu \neq \mu_0\] \[H_0: \mu = 10 \qquad H_1: \mu \neq 10\]
State the null hypothesis and the alternative hypothesis in Problem 4.
Solution
\[H_0: \mu = \mu_0 \qquad H_1: \mu \neq \mu_0\] \[H_0: \mu = 75 \qquad H_1: \mu \neq 75\]
Refer to Problems 1 and 5. Suppose the mean of the sample of 40 memory test scores for the 40 alcoholics is \(M = 78.4\). Give the computed z-test statistic, compute the p-value, and give your conclusion for \(\alpha = 0.05\).
Solution
\[H_0: \mu = 80 \qquad H_1: \mu < 80 \qquad Z = \frac{M-\mu_0}{\frac{\sigma}{\sqrt{n}}} =\frac{78.4-80}{\frac{4.5}{\sqrt{40}}} = -2.25\]
The p-value is the area under the standard normal curve to the left of -2.25. Using R, we find \(z\):
(z <- qnorm(pnorm(q = 78.4, mean = 80, sd = 4.5 / sqrt(40))))## [1] -2.248731Therefore,
pnorm(q = z, mean = 0, sd = 1) # p-value## [1] 0.01226481The line qnorm(pnorm(...)) illustrates the process of
converting a value to its cumulative probability and then converting
that probability back to the original value. This demonstrates the
relationship between values and probabilities in a normal distribution,
enhancing the understanding of these fundamental statistical
concepts.
The above is equivalent to:
pnorm(q = 78.4, mean = 80, sd = 4.5 / sqrt(40)) # p-value## [1] 0.01226481The p-value for this hypothesis test is \(0.0122\), which is less than \(\alpha = 0.05\) and we therefore reject the null hypothesis.
Refer to Problems 2 and 6. Suppose the mean of the sample of 45 depression and anxiety test scores for the 45 known anxiety and depression sufferers is \(M = 56.1\). Give the computed z-test statistic, compute the p-value, and give your conclusion for \(\alpha = 0.05\).
Solution
\[H_0: \mu = 55 \qquad H_1: \mu \neq 55 \qquad Z = \frac{M-\mu_0}{\frac{\sigma}{\sqrt{n}}} =\frac{56.1-55}{\frac{3.5}{\sqrt{45}}} = 2.11\]
The p-value is the area under the standard normal curve to the left of -2.11 plus the area of the right of 2.11. We may find the area to the left of -2.11 and double it to find the p-value.
(z <- (56.1 - 55) / (3.5 / sqrt(45)))## [1] 2.108293Therefore,
2 * pnorm(q = -z, mean = 0, sd = 1)## [1] 0.03500568This is equivalent to:
2 * pnorm(q = 56.1, mean = 55, sd = 3.5 / sqrt(45), lower.tail = FALSE)## [1] 0.03500568Remember, we find the area in two tails to find the p-value because it is a two-tailed test. The p-value for this hypothesis test is 0.0349, which is less than \(\alpha= 0.05\), and we therefore reject the null hypothesis.
Refer to Problems 3 and 7. Suppose the mean of the sample of run times for the 50 rats taking the drug is \(M = 10.5\) minutes. Give the computed z-test statistic, compute the p-value, and give your conclusion for \(\alpha = 0.05\).
Solution
\[H_0: \mu = 10 \qquad H_1: \mu \neq 10 \qquad Z = \frac{M -\mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{10.5 -10}{\frac{2.5}{\sqrt{50}}} = 1.41\]
The p-value is the area to the left of -1.41 plus the area to the right of 1.41
z <- (10.5 -10) / (2.5 / sqrt(50))
z## [1] 1.414214\(Z\)-value is equivalent to
z <- qnorm(pnorm(q = 10.5, mean = 10, sd = 2.5 / sqrt(50)))
z## [1] 1.414214Therefore,
2 * pnorm(q = -z, mean = 0, sd = 1)## [1] 0.1572992Or directly,
2 * pnorm(q = 10.5, mean = 10, sd = 2.5 / sqrt(50), lower.tail = FALSE)## [1] 0.1572992Because the p-value is greater than 0.05 we do not have enough evidence to reject the null hypothesis.
Refer to Problems 4 and 8.Suppose the average score on the test in behavioral statistics for the sample is 77.5. Give the computed z-test statistic, compute the p-value, and give your conclusion for \(\alpha = 0.05\).
Solution
\[H_0: \mu = 75 \qquad H_1: \mu \neq 75 \qquad Z = \frac{M-\mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{77.5-75}{\frac{7}{\sqrt{75}}} = 3.09\]
The p-value is equal to area to the left of -3.09 plus the area to the right of 3.09.
In R this is equivalent to:
z <- qnorm(pnorm(q = 77.5, mean = 75, sd = 7 / sqrt(75)))
z## [1] 3.092948Therefore,
pnorm(q = -z, mean = 0, sd = 1) +
pnorm(q = z, mean = 0, sd = 1, lower.tail = F)## [1] 0.001981789Directly:
2 * pnorm(q = 77.5, mean = 75, sd = 7 / sqrt(75), lower.tail = FALSE)## [1] 0.001981789The p-value is \(2(0.001) = 0.002\). The new method gives better scores on the average.
In Problem 9, describe a type 1 and a type 2 error.
Solution
A type 1 error occurs if it is concluded that \(\mu < 80\) when \(\mu = 80\).
A type 2 error occurs if it is concluded that \(\mu = 80\) when \(\mu < 80\).
In problem 10, describe a type 1 and a type 2 error.
Solution
A type 1 error occurs if the sample evidence leads us to conclude that \(\mu \neq 55\), when \(\mu = 55\).
A type 2 error occurs if the sample evidence leads us to conclude that \(\mu = 55\), when \(\mu \neq 55\).
In problem 11, describe a type 1 and a type 2 error.
Solution
A type 1 error occurs if it is concluded that \(\mu \neq 10\) when \(\mu = 10\).
A type 2 error occurs if it is concluded that \(\mu = 10\) when \(\mu \neq 10\).
In problem 12, describe a type 1 and a type 2 error.
Solution
A type 1 error occurs if it is concluded that \(\mu \neq 75\) when \(\mu = 75\).
A type 2 error occurs if it is concluded that \(\mu = 75\) when \(\mu \neq 75\).
After working your way through Problems 1, 5, 9, and 13, put your z-test together as a complete test of hypotheses.
Solution
Let \(\mu\) represent the average score on the memory test for alcoholics.
\(H_0: \mu = 80\) The average score on the memory test for alcoholics equals 80, the same as the general population.
\(H_1: \mu < 80\) The average score on the memory test for alcoholics is less than 80, the average score for the general population.
\(\alpha\) is set equal to 0.05.
Forty alcoholics are administered the memory test and the sample mean, \(M\), is found to equal 78.4.
The value of the test statisticis is \(Z = \frac{78.4-80}{\frac{4.5}{\sqrt{40}}} = -2.25\)
We reject null hypothesis because \(Z=-2.25\) is less than \(-1.65\):
qnorm(0.05)## [1] -1.644854pnorm(q = 78.4, mean = 80, sd = 4.5 / sqrt(40)) # p-value## [1] 0.01226481After working your way through Problems 1 to 16, put your z-test together as a complete test of hypotheses for Problems 1 through 4.
Solution
Let \(\mu\) represent the average score on the anxiety and depression test for the group known to be suffering from depression and axiety.
\(H_0: \mu = 55 \quad\) The average made on the test given to the group known to be suffering from depression and anxiety is 55, the same as it is for the general population.
\(H_0: \mu \neq 55 \quad\) The average made on the test given to the group known to be suffering from depression and anxiety is not 55, the value it is for the general population.
\(\alpha\) is set equal to 0.05.
Forty-five subjects known to be suffering from anxiety and depression are administered the test for anxiety and depression and the sample mean for that group is \(M = 56.1\).
The value of the test statisticis is \(Z = \frac{56.1-55}{\frac{3.5}{\sqrt{45}}} = 2.11\)
At this point we reject null hypothesis if \(z\) is not between than
c(qnorm(0.05/2), qnorm((1 - 0.95) /2, lower.tail = FALSE))## [1] -1.959964 1.9599642 * pnorm(q = 56.1, mean = 55, sd = 3.5 / sqrt(45), lower.tail = FALSE)## [1] 0.03500568After working your way through Problems 1 to 16, put your z-test together as a complete test of hypotheses for Problems 1 through 4.
Solution
Let \(\mu\) represent the average time to run the maze when under the influence of the drug.
\(H_0: \mu = 10\) The average time to run the maze when under the influence of the drug is the 10 minutes, the same as when not under the influence of the drug.
\(H1: \mu \neq 10\) The average time to run the maze when under the influence of the drug is not 10 minutes, which is the time when not under the influence of the drug.
\(\alpha\) is set equal to 0.05.
Fifty rats are given the drug and timed when running the maze. The average time to run the maze when under the influence is 10.5 minutes.
The value of the test statisticis is \(Z = \frac{10.5-10}{\frac{2.5}{\sqrt{50}}} = 1.41\)
At this point we cannot reject null hypothesis, because \(z\) is between:
c(qnorm(0.05/2), qnorm(1 - 0.05/2))## [1] -1.959964 1.9599642 * pnorm(q = 10.5, mean = 10, sd = 2.5 / sqrt(50), lower.tail = FALSE)## [1] 0.1572992After working your way through Problems 1 to 16, put your z-test together as a complete test of hypotheses for Problems 1 through 4.
Solution
Let \(\mu\) represent the average grade made when teaching statistics for the behavioral sciences using the new approach.
\(H_0: \mu = 75\) The new approach produces the same average results as the old technique.
\(H_1: \mu \neq 75\) The new approach produces the different average results as the old technique.
\(\alpha\) is set equal to 0.05.
Seventy-five students are taught using the new approach. The average score made by this sample of 75 using using the new approach is \(M=77.5\)
The value of test statistic is \(Z = \frac{M - \mu_0}{\frac{\sigma}{\sqrt{n}}}=\frac{77.5 - 75}{\frac{7.0}{\sqrt{75}}} = 3.09\)
At this point we reject null hypothesis if \(z\) is not between than
c(qnorm(0.05/2), qnorm(1-0.05/2))## [1] -1.959964 1.9599642 * pnorm(q = 77.5, mean = 75, sd = 7 / sqrt(75), lower.tail = FALSE)## [1] 0.001981789In Problems 21 through 24, the hypotheses in Problems 1 through 4 are tested when the data are in raw form rather than summarized form. This is the situation most often encountered in actual research studies.
Table below contains the test scores made on the memory test for 40 alcoholics. Test that the mean of these scores are less than 80, the average score for the general population.
| 66 | 79 | 76 | 77 |
| 75 | 70 | 80 | 69 |
| 67 | 67 | 67 | 70 |
| 85 | 75 | 75 | 87 |
| 81 | 72 | 79 | 75 |
| 81 | 75 | 76 | 75 |
| 72 | 74 | 69 | 70 |
| 73 | 80 | 76 | 70 |
| 72 | 78 | 75 | 81 |
| 75 | 77 | 80 | 71 |
Solution
The R output follows. The standard deviation is assumed to be 4.5, the value of \(M\) computed from the data is 74.8, the computed test statistic is z = -7.31, and the p-value is 0.000. The null hypothesis is rejected.
BSDA::z.test(x = sample, sigma.x = 4.5, mu=80, alternative = "less")##
## One-sample z-Test
##
## data: sample
## z = -7.3084, p-value = 1.352e-13
## alternative hypothesis: true mean is less than 80
## 95 percent confidence interval:
## NA 75.97033
## sample estimates:
## mean of x
## 74.8The BSDA::z.test function in R conducts a one-sample z-test assuming
the population standard deviation (\(\sigma\)) is known. The function takes
several arguments: x = sample specifies the sample data,
sigma.x = 4.5 is the known population standard deviation,
mu = 80 is the population mean under the null hypothesis,
and alternative = "less" indicates a one-sided test where
the alternative hypothesis is that the sample mean is less than the
population mean (\(\mu < 80\)). The
function calculates the z-statistic and the p-value, determining whether
the observed sample mean is significantly lower than the population
mean.
The result of the one-sample z-test provides several key pieces of information. The calculated z-statistic is -7.3084, indicating that the sample mean is significantly lower than the population mean of 80 under the null hypothesis (The sample mean (74.8) is 7.3084 standard deviations below the expected population mean (80).) The p-value is extremely small (1.352e-13), meaning the probability of observing a sample mean as low as 74.8 or lower, if the true population mean were 80, is virtually zero. The alternative hypothesis states that the true mean is less than 80. The 95 percent confidence interval for the population mean is [NA, 75.97033], indicating that the lower bound cannot be calculated due to the one-sided nature of the test. Finally, the sample mean is 74.8. These results provide strong evidence to reject the null hypothesis and accept that the true mean is indeed less than 80.
Table below contains the test scores made on the anxiety and depression test given to 45 individuals suffering from anxiety and depression. Test that the average of these scores is not equal to 55, the average score for the general population. (See problem 2)
| 55 | 55 | 55 | 60 | 50 |
| 50 | 54 | 57 | 51 | 62 |
| 57 | 59 | 57 | 60 | 59 |
| 59 | 49 | 63 | 50 | 58 |
| 58 | 54 | 56 | 56 | 54 |
| 61 | 57 | 58 | 52 | 54 |
| 60 | 55 | 55 | 59 | 60 |
| 50 | 57 | 54 | 61 | 51 |
| 59 | 55 | 56 | 59 | 54 |
Solution
The R output follows. The standard deviation is assumed to be 3.5, the value of M computed from the data is 56.111, the computed test statistic is z = 2.13, and the p-value is 0.033. The null hypothesis is rejected.
BSDA::z.test(x = p_8_22, sigma.x = 3.5, mu = 55, alternative = "two.sided")##
## One-sample z-Test
##
## data: p_8_22
## z = 2.1296, p-value = 0.03321
## alternative hypothesis: true mean is not equal to 55
## 95 percent confidence interval:
## 55.08850 57.13372
## sample estimates:
## mean of x
## 56.11111Table below contains the run times for 50 rats when under the influence of a drug. Test that the average run time is different than 10 minutes for the 50 rats. See problem 3.
| 16.2 | 13.6 | 6.1 | 9.9 | 6.8 |
| 9.5 | 11.8 | 15.6 | 9.4 | 14.2 |
| 8.3 | 9.9 | 15.3 | 11.3 | 3.6 |
| 11.1 | 8.8 | 10.1 | 12.1 | 10.1 |
| 10.6 | 9.2 | 6.8 | 8.8 | 12.0 |
| 11.0 | 9.1 | 11.6 | 12.7 | 10.9 |
| 8.7 | 10.7 | 14.7 | 9.0 | 13.2 |
| 11.5 | 8.5 | 9.5 | 9.9 | 9.2 |
| 2.3 | 11.4 | 10.0 | 14.5 | 12.9 |
| 11.1 | 7.7 | 14.4 | 9.4 | 10.3 |
Solution
The R output follows. The standard deviation is assumed to be 2.5, the value of M computed from the data is 10.506, the computed test statistic is z = 1.43, and the p-value is 0.152. The null hypothesis is not rejected.
BSDA::z.test(x = p_8_23, sigma.x = 2.5, mu = 10, alternative = "two.sided" )##
## One-sample z-Test
##
## data: p_8_23
## z = 1.4312, p-value = 0.1524
## alternative hypothesis: true mean is not equal to 10
## 95 percent confidence interval:
## 9.813048 11.198952
## sample estimates:
## mean of x
## 10.506Table below contains the test scores for 75 students in statistics for the behavioral sciences when taught using the new technique. Test that the average test score is different than 75, the average when the course is taught using the traditional method. See problem 4
| 67 | 74 | 67 | 77 | 76 |
| 73 | 72 | 80 | 77 | 73 |
| 67 | 73 | 73 | 75 | 85 |
| 77 | 79 | 80 | 79 | 74 |
| 75 | 79 | 72 | 78 | 82 |
| 74 | 72 | 80 | 64 | 69 |
| 84 | 72 | 70 | 79 | 81 |
| 75 | 81 | 79 | 68 | 79 |
| 71 | 71 | 75 | 70 | 65 |
| 68 | 62 | 69 | 85 | 77 |
| 74 | 82 | 84 | 77 | 71 |
| 72 | 63 | 73 | 75 | 71 |
| 82 | 61 | 66 | 69 | 85 |
| 61 | 79 | 68 | 85 | 65 |
| 67 | 75 | 73 | 72 | 73 |
Solution
The R output follows. The standard deviation is assumed to be 7, the value of \(M\) computed from the data is 73.96, the computed test statistic is z = -1.29, and the p-value is 0.198. The null hypothesis is not rejected.
BSDA::z.test(x = p_8_24, sigma.x = 7, mu = 75, alternative = "two.sided")##
## One-sample z-Test
##
## data: p_8_24
## z = -1.2867, p-value = 0.1982
## alternative hypothesis: true mean is not equal to 75
## 95 percent confidence interval:
## 72.37578 75.54422
## sample estimates:
## mean of x
## 73.96A psychological researcher believes that extra playtime with infants will result in a higher average weight at 2 years age. The average weight of humans at 2 years of age is 26 pounds with a standard deviation of 3.5 pounds for the general population. The researcher plans an experiment with 36 infants. She instructs the parents of the 36 infants to spend extra time playing with the infants and then determines their weights at 2 years of age. Would this be a one-sided or two-sided test?
Solution
This is a one-sided test.
A psychological researcher wishes to test whether a drug reduces appetite and, as a result,food consumption. Suppose rats are known to consume 12 grams of food per day with a standard deviation of 3 grams. Fifty of these rats are given the drug over a period of time and their average food consumption is determined. Would this be a one-sided or two-sided test?
Solution
This is a one-sided test.
An educational psychologist wishes to increase the usage of statistical software in her Statistics for Psychology course. She wants to determine if it will change the average performance of her students. The average previously on a final exam in the course has been 70 with a standard deviation of 5. Fifty students are taught the course with the increased use of statistical software. Would this be a one-sided or two-sided test?
Solution
This is a two-sided test.
Rats are known to reach an average weight at maturity of 930 grams with a standard deviation equal to 25 grams. Researchers believe that if the rats are fed a growth hormone, they will reach an average weight that exceeds 930 grams. Would this be a one-sided or two-sided test?
Solution
This is a one-sided test.
State the null hypothesis and the alternative hypothesis in Problem 25.
Solution
Let \(\mu\) represent the average weight for the infants who have increased playtime before age 2.
\[H_0: \mu = 26 \qquad H_1: \mu > 26\]
State the null hypothesis and the alternative hypothesis in Problem 26.
Solution
Let \(\mu\) represent the average food consumption of the rats given the weight-reducing drug.
\[H_0: \mu = 12 \qquad H_1: \mu < 12\]
State the null hypothesis and the alternative hypothesis in Problem 27.
Solution
Let \(\mu\) represent the average test score for the students who have their course taught with increased use of the statistical software
\[H_0: \mu = 70 \qquad \mu \neq 70\]
State the null hypothesis and the alternative hypothesis in Problem 28.
Solution
Let \(\mu\) represent the average weight of the rats given the growth hormone.
\(H_0: \mu = 930 \qquad H_1: \mu > 930\)
Refer to Problems 25 and 29. Suppose the mean weight of the sample of 36 two-year-olds who had longer playtimes as infants is 27.1 pounds. Give the computed z-test statistic, compute the p-value, and give your conclusion for \(\alpha = 0.05\).
Solution
\[H_0: \mu = 26 \qquad H_1: \mu > 26 \qquad Z = \frac{M-26}{\frac{\sigma}{\sqrt{n}}} = \frac{27.1-26}{\frac{3.5}{\sqrt{36}}} \approx 1.89\]
pnorm(1.885714286, lower.tail = FALSE)## [1] 0.02966673The p-value is the area under the standard normal curve to the right of 1.89. Using R, we find this area equals 0.029. The p-value for this hypothesis test is 0.029, which is less than \(\alpha = 0.05\), and therefore we reject the null hypothesis.
Alternatively, we have the same result with:
pnorm(q = 27.1, mean = 26, sd = 3.5 / sqrt(36), lower.tail = FALSE)## [1] 0.02966673Refer to Problems 26 and 30. The average weight of the sample of 50 rats when given the appetite suppressant is 11.2 grams. Give the computed z-test statistic, compute the p-value, and give your conclusion for \(\alpha = 0.05.\)
Solution
\[H_0: \mu = 12 \qquad H_1: \mu <12 \qquad Z = \frac{M-\mu}{\frac{\sigma}{\sqrt{n}}} = \frac{11.2-12}{\frac{3}{\sqrt{50}}} \approx -1.89\]
pnorm(-1.89)## [1] 0.02937898The p-value is the area under the standard normal curve to the left of -1.89. Using R, we find is 0.029. The p-value for this hypothesis test is 0.029, which is less than \(\alpha = 0.05\), and therefore we reject the null hypothesis.
Alternatively, we have the same result with:
pnorm(q = 11.2, mean = 12, sd = 3 / sqrt(50), lower.tail = TRUE)## [1] 0.02967322Refer to Problems 27 and 31. Suppose the average score on the test when computer software is used for a sample of 50 students is 69.3. Give the computed z-test statistic, compute the p-value, and give your conclusion for \(\alpha = 0.05\).
Solution
\[H_0: \mu = 70 \qquad \mu \neq 70 \qquad Z = \frac{M - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{69.3 - 70}{\frac{5}{\sqrt{50}}} \approx -0.99\]
pnorm(-0.99, lower.tail = TRUE) + pnorm(0.99, lower.tail = FALSE)## [1] 0.3221741The p-value is the area under the standard normal curve to the left
of -0.99 plus the area to the right of 0.99, or simply double the area
to the left of -0.99. Using R, we find
= 2 * pnorm(-0.99), which equals 0.322. The p-value is
\(0.322\), which is greater than \(\alpha = 0.05\) and we are unable to reject
the null hypothesis.
Refer to Problems 28 and 32. Suppose the average weight of 50 rats, when given the growth hormone, is 945 grams. Give the computed z-test statistic, compute the p-value, and give your conclusion for \(\alpha = 0.05\).
Solution
\[H_0: \mu = 930 \qquad \mu > 930 \qquad Z = \frac{M - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{945 - 930}{\frac{25}{\sqrt{50}}} \approx 4.24\]
qnorm(pnorm(q = 945, mean = 930, sd = 25 / sqrt(50)))## [1] 4.242641pnorm(4.242641, mean = 0, sd = 1, lower.tail = FALSE)## [1] 1.104523e-05Equivalent to:
pnorm(q = 945, mean = 930, sd = 25 / sqrt(50), lower.tail = FALSE)## [1] 1.104525e-05Equivalent to:
1 - pnorm(q = 945, mean = 930, sd = 25 / sqrt(50))## [1] 1.104525e-05The p-value is the area under the standard normal curve to the right
of 4.24. This area is given by 1 - pnorm(q = 4.242641).
Because the p-value \(< \alpha\) ,
reject the null hypothesis.
In Problem 33, describe a type 1 and a type 2 error.
Solution
A type 1 error occurs if the sample evidence leads us to conclude that \(\mu > 26\) when \(\mu=26\)
A type 2 error occurs if sample evidence does not lead us to reject that \(\mu = 26\) when \(\mu > 26\)
In Problem 34, describe a type 1 and a type 2 error
Solution
A type 1 error occurs if the sample evidence leads us to conclude that \(\mu < 12\) when \(\mu = 12\).
A type 2 error occurs if the sample evidence leads us to conclude that \(\mu = 12\) when \(\mu < 12\).
In Problem 35, describe a type 1 and a type 2 error.
Solution
A type 1 occurs if the sample evidence lead us to conclude that \(\mu \neq 70\) when \(\mu = 70\).
A type 2 occurs if the sample evidence does not lead us to reject that \(\mu = 70\) when \(\mu \neq 70\).
In Problem 36, describe a type 1 and a type 2 error.
Solution
A type 1 occurs if the sample evidence lead us to conclude that \(\mu > 930\) when \(\mu = 930\).
A type 2 occurs if the sample evidence lead us to conclude that \(\mu = 930\) when \(\mu > 930\).
After working your way through Problems 25, 29, 33, and 37, put your z-test together as a complete test of hypotheses.
Solution
\(H_0: \mu = 26\) The average weight of 2-years-olds who had increased playtime as infants is equal to 26 pounds, the same as the general population.
\(H_1: \mu > 26\) The average weight of 2-years-olds who had increased playtime as infants is greater than 26 pounds, the general population mean for 2-years olds.
\(\alpha\) is set equal to 0.05.
Thirty-six children had increased playtime as infants and the average weight is determined (at age 2) to be 27.1 pounds.
The value of the test statistic is \(Z = \frac{M - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{27.1 - 26}{\frac{3.5}{\sqrt{36}}} = 1.89\)
We can reject \(H_0\) because \(Z\) is greater than
qnorm(p = 0.95)## [1] 1.644854pnorm(1.885714286, lower.tail = FALSE)## [1] 0.02966673After working your way through Problems 26, 30, 34, and 38, put your z-test together as a complete test of hypotheses.
Solution
Let \(\mu\) represent the average food consumption of the rats given the appetite-reducing drug.
\(H_0: \mu = 12\) The average food consumption of the rats given the appetite suppressant is the same as the general population.
\(H_1: \mu < 12\) The average food consumption of the rats given the appetite suppresant is less than the food consumption of the general population.
\(\alpha\) is set to \(0.05\).
Fifty rats are given the food suppressant and their average food consumption is 11.2 grams.
The value of the test statistic is \[Z = \frac{M-\mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{11.2-12}{\frac{3}{\sqrt{50}}} \approx -1.89\]
We can reject \(H_0\) because \(Z\) is less than:
qnorm(p = 0.05)## [1] -1.644854pnorm(q = -1.89)## [1] 0.02937898After working your way through Problems 27, 31, 35, and 39, put your z-test together as a complete test of hypotheses.
Solution
\(H_0: \mu = 70\) The average test scores are the same as when the course is taught in the usual manner.
\(H_1: \mu \neq 70\) The average tests scores are not the same as when the course is taught in the usual manner.
\(\alpha\) is set equal to \(0.05\).
Fifty students are taught the course using statistical software. The average test score for the sample is \(69.93\).
The value of the test statistic is \(Z = \frac{M - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{69.3 - 70}{\frac{5.0}{\sqrt{50}}} \approx - 0.99\)
c(qnorm(0.05/2), qnorm((1-0.95) / 2, lower.tail = FALSE))## [1] -1.959964 1.959964We cannot reject \(H_0\) because \(Z\) is between \(-1.96\) and \(1.96\)
(z <- qnorm(pnorm(q = 69.3, mean = 70, sd = 5 / sqrt(50))))## [1] -0.9899495pnorm(z, lower.tail = TRUE) + pnorm(-z, lower.tail = FALSE)## [1] 0.3221988The above code is equivalent to:
2 * pnorm(q = 69.3, mean = 70, sd = 5 / sqrt(50))## [1] 0.3221988After working your way through Problems 28, 32, 36, and 40, put your z-test together as a complete test of hypotheses.
Solution
\(H_0: \mu = 930\) The average weight of the rats given the growth hormone is the same as the general population.
\(H_1: \mu > 930\) The average weight of the rats given the growth hormone is greater than the general population.
\(\alpha\) is set equal to \(0.05\).
The average weight of the \(50\) rats given the growth hormone is 945 grams.
The value of the test statistic is \(Z = \frac{M - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{945 - 930}{\frac{25}{\sqrt{50}}} \approx 4.24\)
qnorm(p = 0.95)## [1] 1.644854We can reject \(H_0\) because \(Z = 4.24\) is greater than \(1.65\)
Table 8.7 gives the weights of \(n = 36\) two-year-olds who received extra playtime as infants. Use R to test \(H_0 : \mu = 26 \quad H_1 : \mu > 26\) with \(\alpha = 0.05\). The standard deviation is assumed to be \(3.5\).
table_8_7 <- c(28, 30, 26, 26, 23, 25, 30, 25, 26, 32, 27, 23, 23, 25, 29, 28,
25, 20, 30, 30, 30, 25, 22, 22, 28, 26, 29, 31, 24, 33, 23, 28,
20, 27, 21, 23)Solution
The R output follows. The standard deviation is assumed to be \(3.5\), the value of \(M\) computed from the data is \(26.194\), the computed test statistic is \(z = 0.33\), and the p-value is \(0.369\). The null hypothesis is not rejected.
BSDA::z.test(table_8_7, sigma.x = 3.5, mu = 26, alternative = "greater")##
## One-sample z-Test
##
## data: table_8_7
## z = 0.33333, p-value = 0.3694
## alternative hypothesis: true mean is greater than 26
## 95 percent confidence interval:
## 25.23495 NA
## sample estimates:
## mean of x
## 26.19444Table 8.8 gives the food consumption of \(n = 50\) rats given a appetite suppressant. Use R to test \(H_0: \mu = 12 \quad H_1: \mu < 12\) with \(\alpha = 0.05\). The standard deviation is assumed to be \(3\).
table_8_8 <- c(9, 10, 16, 10, 5, 16, 9, 11, 7, 8, 9, 9, 8, 14, 19, 9, 6, 8, 11, 9,
14, 12, 15, 12, 10, 10, 14, 13, 5, 9, 5, 11, 11, 15, 15, 11, 10,
14, 9, 10, 18, 13, 14, 12, 9, 12, 13, 11, 12, 14)Solution
The R output follows. The standard deviation is assumed to be 3, the value of \(M\) computed from the data is \(11.12\), the computed test statistic is \(z = -2.07\), and the p-value is \(0.019\). The null hypothesis is rejected.
BSDA::z.test(table_8_8, sigma.x = 3.0, mu = 12, alternative = "less")##
## One-sample z-Test
##
## data: table_8_8
## z = -2.0742, p-value = 0.01903
## alternative hypothesis: true mean is less than 12
## 95 percent confidence interval:
## NA 11.81785
## sample estimates:
## mean of x
## 11.12The next table gives the test scores from a sample of 50 students in a Statistics for Psychology course that uses statistical software on a weekly basis. Use R to test \(H_0: \mu = 70 \quad H_1: \mu \neq 70\) with \(\alpha = 0.05\). The standard deviation is assumed to be 5.
t47 <- c(75, 66, 76, 72, 78, 68, 70, 78, 71, 74, 75, 82, 67, 69, 74, 69, 65, 74,
75, 77, 73, 73, 70, 83, 68, 67, 74, 73, 75, 69, 74, 65, 69, 70, 72, 64,
74, 74, 65, 72, 67, 74, 76, 72, 74, 74, 73, 75, 61, 73)Solution
You should reject \(H_0\) if \(t\) is not between -2.0 and 2.0:
c(qt(p = 0.05/2, df = length(t47) - 1), qt(p = 1 - 0.05/2, df = length(t47) - 1))## [1] -2.009575 2.009575The R output follows. The standard deviation is assumed to be 5, the value of \(M\) computed from the data is 71.96, the computed test statistic is \(z = 2.77\), and the p-value is 0.006. The null hypothesis is rejected.
BSDA::z.test(t47, sigma.x = 5.0, mu = 70, alternative = "two.sided")##
## One-sample z-Test
##
## data: t47
## z = 2.7719, p-value = 0.005574
## alternative hypothesis: true mean is not equal to 70
## 95 percent confidence interval:
## 70.5741 73.3459
## sample estimates:
## mean of x
## 71.96The nex table gives the weights of 50 rats taking a growth hormone. Use R to test \(H_0: \mu = 930 \quad H_1: \mu > 930\) with \(\alpha = 0.05\). The standard deviation is assumed to be 25
t48 <- c(947, 972, 896, 981, 973, 976, 948, 902, 966, 963, 958, 934, 934, 935,
968, 960, 948, 933, 874, 917, 930, 928, 925, 932, 916, 893, 980, 953,
926, 935, 922, 938, 901, 902, 962, 909, 936, 926, 920, 953, 970, 916,
933, 913, 937, 934, 958, 967, 936, 916)Solution
The R output follows. The standard deviation is assumed to be 25, the value of M computed from the data is 937.04, the computed test statistic is \(z = 1.99\), and the p-value is 0.023. The null hypothesis is rejected.
BSDA::z.test(t48, sigma.x = 25.0, mu = 930, alternative = "greater")##
## One-sample z-Test
##
## data: t48
## z = 1.9912, p-value = 0.02323
## alternative hypothesis: true mean is greater than 930
## 95 percent confidence interval:
## 931.2246 NA
## sample estimates:
## mean of x
## 937.04