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solve 1，2 and7 1 In a certain country, the average number of children per family...
Jun 2, 2024
solve 1，2
Solution by Steps
step 1
Define the null hypothesis (H0H_0) and the alternative hypothesis (HaH_a)
step 2
The null hypothesis (H0H_0) states that the average number of children per family has not decreased. Thus, H0:μ=2.4H_0: \mu = 2.4
step 3
The alternative hypothesis (HaH_a) states that the average number of children per family has decreased. Thus, H_a: \mu < 2.4
Answer
H0:μ=2.4H_0: \mu = 2.4 and H_a: \mu < 2.4
Key Concept
Hypothesis Testing
Explanation
The null hypothesis represents the status quo or no change, while the alternative hypothesis represents the researcher's claim that there has been a decrease.
Question 2
step 1
Define the null hypothesis (H0H_0) and the alternative hypothesis (HaH_a)
step 2
The null hypothesis (H0H_0) states that the new program has no effect on the GPA. Thus, H0:μ=2.66H_0: \mu = 2.66
step 3
The alternative hypothesis (HaH_a) states that the new program has increased the GPA. Thus, H_a: \mu > 2.66
Answer
H0:μ=2.66H_0: \mu = 2.66 and H_a: \mu > 2.66
Key Concept
Hypothesis Testing
Explanation
The null hypothesis represents the status quo or no change, while the alternative hypothesis represents the researcher's claim that the new program has improved the GPA.
solve 3 4 5 6 7 8
Solution by Steps
step 1
Define the null hypothesis (H0H_0) and the alternative hypothesis (H1H_1)
step 2
The null hypothesis (H0H_0) states that the mean weight of fish in the Merry River is equal to 60 g. Thus, H0:μ=60H_0: \mu = 60
step 3
The alternative hypothesis (H1H_1) states that the mean weight of fish in the Merry River is greater than 60 g. Thus, H_1: \mu > 60
step 4
Calculate the test statistic using the formula: z=xˉ−μσ/nz = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}. Here, xˉ=65.8\bar{x} = 65.8, μ=60\mu = 60, σ=4.5\sigma = 4.5, and n=10n = 10
step 5
Substitute the values into the formula: z=65.8−604.5/10=5.81.423≈4.08z = \frac{65.8 - 60}{4.5 / \sqrt{10}} = \frac{5.8}{1.423} \approx 4.08
step 6
Find the pp-value corresponding to the zz-score of 4.08 using standard normal distribution tables or a calculator
step 7
The pp-value for z=4.08z = 4.08 is very small, typically less than 0.0001
Answer
The pp-value is less than 0.0001.
Key Concept
Hypothesis Testing
Explanation
The pp-value indicates the probability of observing a sample mean as extreme as 65.8 g if the true mean weight of fish is 60 g.
Question 4
step 1
Define the null hypothesis (H0H_0) and the alternative hypothesis (H1H_1)
step 2
The null hypothesis (H0H_0) states that the mean concentration of the pollutant is equal to 34 ppm. Thus, H0:μ=34H_0: \mu = 34
step 3
The alternative hypothesis (H1H_1) states that the mean concentration of the pollutant is less than 34 ppm. Thus, H_1: \mu < 34
step 4
Calculate the test statistic using the formula: z=xˉ−μσ/nz = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}. Here, xˉ=32.5\bar{x} = 32.5, μ=34\mu = 34, σ=8\sigma = 8, and n=50n = 50
step 5
Substitute the values into the formula: z=32.5−348/50=−1.51.131≈−1.33z = \frac{32.5 - 34}{8 / \sqrt{50}} = \frac{-1.5}{1.131} \approx -1.33
step 6
Find the pp-value corresponding to the zz-score of -1.33 using standard normal distribution tables or a calculator
step 7
The pp-value for z=−1.33z = -1.33 is approximately 0.0918
Answer
The pp-value is approximately 0.0918.
Key Concept
Hypothesis Testing
Explanation
The pp-value indicates the probability of observing a sample mean as extreme as 32.5 ppm if the true mean concentration is 34 ppm.
Question 5
step 1
Interpret the pp-value in terms of the strength of evidence against the null hypothesis
step 2
For pp-value = 0.033: This provides moderate evidence against the null hypothesis
step 3
For pp-value = 0.245: This provides weak evidence against the null hypothesis
step 4
For pp-value = 0.003: This provides very strong evidence against the null hypothesis
step 5
For pp-value = 0.0049: This provides strong evidence against the null hypothesis
step 6
For pp-value = 0.0008: This provides extremely strong evidence against the null hypothesis
Answer
Interpretations provided for each pp-value.
Key Concept
pp-value Interpretation
Explanation
The pp-value helps determine the strength of evidence against the null hypothesis.
Question 6
step 1
Define the null hypothesis (H0H_0) and the alternative hypothesis (H1H_1)
step 2
The null hypothesis (H0H_0) states that the mean is equal to 50. Thus, H0:μ=50H_0: \mu = 50
step 3
The alternative hypothesis (H1H_1) states that the mean is less than 50. Thus, H_1: \mu < 50
step 4
Given the pp-value of 0.0355, compare it to a significance level (commonly α=0.05\alpha = 0.05)
step 5
Since pp-value (0.0355) < α\alpha (0.05), we reject the null hypothesis
Answer
We reject the null hypothesis.
Key Concept
pp-value Comparison
Explanation
If the pp-value is less than the significance level, we reject the null hypothesis.
Question 7
step 1
Define the null hypothesis (H0H_0) and the alternative hypothesis (H1H_1)
step 2
The null hypothesis (H0H_0) states that the mean is equal to 10. Thus, H0:μ=10H_0: \mu = 10
step 3
The alternative hypothesis (H1H_1) states that the mean is greater than 10. Thus, H_1: \mu > 10
step 4
Given the pp-value of 0.099, compare it to a significance level (commonly α=0.05\alpha = 0.05)
step 5
Since pp-value (0.099) > α\alpha (0.05), we fail to reject the null hypothesis
Answer
We fail to reject the null hypothesis.
Key Concept
pp-value Comparison
Explanation
If the pp-value is greater than the significance level, we fail to reject the null hypothesis.
Question 8
step 1
Define the null hypothesis (H0H_0) and the alternative hypothesis (H1H_1)
step 2
The null hypothesis (H0H_0) states that the mean is equal to 40. Thus, H0:μ=40H_0: \mu = 40
step 3
The alternative hypothesis (H1H_1) states that the mean is less than 40. Thus, H_1: \mu < 40
step 4
Given the pp-value of 0.013, compare it to a significance level (commonly α=0.05\alpha = 0.05)
step 5
Since pp-value (0.013) < α\alpha (0.05), we reject the null hypothesis
Answer
We reject the null hypothesis.
Key Concept
pp-value Comparison
Explanation
If the pp-value is less than the significance level, we reject the null hypothesis.
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