INFERENTIAL STATISTICS

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Inferential statistics is a branch of statistics that involves using sample data to make inferences or draw conclusions about a larger population. The main goal of inferential statistics is to use statistical methods to make predictions, estimate parameters, or test hypotheses about a population based on a subset of data from that population.

Inferential statistics typically involves the use of probability theory to determine the likelihood of different outcomes or events, and statistical tests to evaluate the strength of the evidence in support of a hypothesis or claim. Some common techniques used in inferential statistics include hypothesis testing, confidence intervals, and regression analysis.

USES IN VARIOUS SOME FIELDS

Social Sciences: Inferential statistics is used in social sciences to study human behavior, attitudes, and preferences. It is used to test hypotheses and make predictions about social phenomena, such as the impact of education on income or the relationship between social class and health.
Business and Economics: Inferential statistics is used in business and economics to analyze data and make predictions about market trends and consumer behavior. It is used to test the effectiveness of marketing strategies and to determine the success of business decisions.
Medicine and Health: Inferential statistics is used in medicine and health to analyze data from clinical trials and observational studies. It is used to determine the effectiveness of medical treatments, evaluate the risk factors for diseases, and make predictions about patient outcomes.
Engineering: Inferential statistics is used in engineering to analyze data and make predictions about the performance of systems and processes. It is used to test the reliability of products, optimize manufacturing processes, and evaluate the impact of environmental factors on infrastructure.
Environmental Science: Inferential statistics is used in environmental science to analyze data and make predictions about the impact of human activities on the environment. It is used to evaluate the effectiveness of environmental policies and to predict the future state of the environment.
Education: Inferential statistics is used in education to analyze student performance data and evaluate the effectiveness of teaching methods. It is also used to identify factors that influence student achievement and to make predictions about future academic outcomes.
Sports: Inferential statistics is used in sports to analyze player and team performance data and to make predictions about future performance. It is also used to evaluate the effectiveness of different coaching strategies and to identify factors that influence athletic success.
Government and Public Policy: Inferential statistics is used in government and public policy to evaluate the effectiveness of programs and policies. It is used to analyze data on social, economic, and environmental factors, and to make predictions about the impact of policy decisions.
Market Research: Inferential statistics is used in market research to analyze data on consumer behavior and preferences. It is used to make predictions about market trends, evaluate the effectiveness of advertising campaigns, and identify factors that influence consumer buying decisions.
Psychology: Inferential statistics is used in psychology to study the human mind and behavior. It is used to test hypotheses about the causes of psychological disorders, evaluate the effectiveness of psychotherapy treatments, and make predictions about behavior in different contexts.
Farming: Inferential statistics can be used in farming to make decisions based on data analysis and to test hypotheses related to agricultural practices.
THE PARAMETRIC TEST
Parametric tests are statistical tests that are based on assumptions about the underlying distribution of the data. These assumptions typically include the normality (i.e., bell-shaped) of the distribution and the equality of variances between groups.
Parametric tests are useful when the data meet the assumptions, as they tend to have higher statistical power (i.e., ability to detect true differences or relationships) compared to non-parametric tests. Some common examples of parametric tests include t-tests, ANOVA (analysis of variance), and linear regression.
Here's a brief explanation of a few commonly used parametric tests:
1. Student's t-test: This test is used to compare the means of two groups when the sample sizes are small (typically less than 30) and the population standard deviations are unknown. There are two types of t-tests: one-sample t-test (to compare a sample mean to a known population mean) and independent-samples t-test (to compare the means of two independent samples).
2. Analysis of Variance (ANOVA): This test is used to compare the means of three or more groups. There are several types of ANOVA tests, including one-way ANOVA (when there is only one independent variable) and factorial ANOVA (when there are multiple independent variables).
3. Linear Regression: This test is used to examine the relationship between two continuous variables. It involves fitting a line to the data and assessing the significance of the slope of the line. Multiple linear regression can be used when there are multiple independent variables.
SAMPLE PROBLEMS
Problem 1:
A local coffee shop wants to determine if there is a significant difference in the amount of coffee that customers purchase on weekdays versus weekends. They randomly select 50 customers and record the amount of coffee they purchase on a weekday and the amount of coffee they purchase on a weekend. The mean amount of coffee purchased on weekdays is 12 ounces with a standard deviation of 2 ounces, and the mean amount of coffee purchased on weekends is 14 ounces with a standard deviation of 3 ounces. Is there a significant difference in the amount of coffee purchased on weekdays versus weekends at this coffee shop?
Solution:
Step 1: Hypotheses
We need to set up the null and alternative hypotheses. The null hypothesis (H0) is that there is no significant difference in the amount of coffee purchased on weekdays versus weekends. The alternative hypothesis (Ha) is that there is a significant difference in the amount of coffee purchased on weekdays versus weekends.
H0: μweekday = μweekend Ha: μweekday ≠ μweekend
Step 2: Level of Significance
We need to determine the level of significance, which is the probability of rejecting the null hypothesis when it is actually true. Let's choose a level of significance of 0.05, which is a commonly used level in statistical testing.
@ α = 0.05
Step 3: Test Statistic
We will use a two-sample t-test to determine if there is a significant difference in the amount of coffee purchased on weekdays versus weekends. The test statistic is calculated as:
t-test formula:
where:
x̄ = 12-14= -2 ; the sample mean
$S_{1}^{2}=2^{2}=4$ ; the sample variance
$S_{2}^{2}=3^{2}=9$ ; the sample variance
n = 50; the sample size
Using the values given in the problem, we get:
$t=\frac{12-14} {\sqrt{(\frac{2^{2}}{50}+\frac{3^{2}}{50}})}={\color{Red} -2.23}$
therefore: t = -2.23
Step 4: p-value
We need to calculate the p-value, which is the probability of obtaining a test statistic as extreme or more extreme than the one we calculated, assuming the null hypothesis is true. We will use a two-tailed test, since the alternative hypothesis is that the means are not equal.
Using a t-distribution table or calculator with degrees of freedom (df) = n1 + n2 - 2 = 98, we find that the p-value for a t-statistic of -2.23 is 0.027. This means that if the null hypothesis is true (i.e., there is no significant difference in the amount of coffee purchased on weekdays versus weekends), there is a 2.7% chance of obtaining a test statistic as extreme or more extreme than the one we calculated.
Step 5: Conclusion
Since the p-value (0.027) is less than the level of significance (0.05), we reject the null hypothesis and conclude that there is a significant difference in the amount of coffee purchased on weekdays versus weekends at this coffee shop. We can interpret the results to mean that, on average, customers purchase more coffee on weekends than on weekdays at this coffee shop.
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Problem 2:
A company produces light bulbs and claims that the average lifespan of their bulbs is 1200 hours with a standard deviation of 150 hours. A sample of 25 bulbs is randomly selected and tested, and the mean lifespan is found to be 1250 hours. Conduct a hypothesis test to determine if there is evidence to suggest that the company's claim is incorrect.
Solution:
This problem involves testing a hypothesis about a population mean using a sample mean and standard deviation. The null hypothesis in this case is that the population mean lifespan is equal to the claimed value of 1200 hours, and the alternative hypothesis is that it is greater than 1200 hours.
To test this hypothesis, we can use a t-test for a single sample. We will calculate the t-value using the formula:
$t=\frac{(\overline{x}-\mu) }{\frac{s}{\sqrt{n}}}$
where:
x̄ is the sample mean,
μ is the hypothesized population mean,
s is the sample standard deviation, and
n is the sample size.
Plugging in the values from the problem, we get:
$t=\frac{(\overline{x}-\mu) }{\frac{s}{\sqrt{n}}}=\frac{1250-1200}{\frac{150}{\sqrt{25}}}=\ {\color{Red} 2.5}$
Using a t-table with 24 degrees of freedom (n - 1), we can find the p-value associated with a t-value of 2.5. Assuming a significance level of 0.05, the p-value would need to be less than 0.05 for us to reject the null hypothesis.
Looking at the t-table, we can see that the closest value to 2.5 with 24 degrees of freedom is 2.492. The corresponding p-value is 0.016, which is less than 0.05. Therefore, we can reject the null hypothesis and conclude that there is evidence to suggest that the average lifespan of the company's light bulbs is greater than the claimed value of 1200 hours.
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Problem 3:
A bakery claims that the average weight of their croissants is 4 ounces with a standard deviation of 0.2 ounces. A random sample of 50 croissants is taken and the average weight is found to be 3.8 ounces. Conduct a hypothesis test to determine if there is evidence to suggest that the bakery's claim is incorrect at a significance level of 0.01.
Solution:
This problem involves testing a hypothesis about a population mean using a sample mean and standard deviation. The null hypothesis in this case is that the population mean weight of croissants is equal to the claimed value of 4 ounces, and the alternative hypothesis is that it is less than 4 ounces.
To test this hypothesis, we can use a z-test for a single sample. We will calculate the z-value using the formula:
$z=\frac{\overline{x}-\mu } {\frac{\sigma }{\sqrt{n}}}$
where:
x̄ is the sample mean,
μ is the hypothesized population mean,
σ is the population standard deviation (since we know it), and
n is the sample size.
Plugging in the values from the problem, we get:
$z=\frac{\overline{x}-\mu } {\frac{\sigma }{\sqrt{n}}}=\frac{3.8-4} {\frac{0.2}{\sqrt{50}}}=\ {\color{Red} -2.236}$
Using a z-table, we can find the p-value associated with a z-value of -2.236. Assuming a significance level of 0.01, the p-value would need to be less than 0.01 for us to reject the null hypothesis.
Looking at the z-table, we can see that the closest value to -2.236 is -2.24. The corresponding p-value is 0.0129, which is less than 0.01. Therefore, we can reject the null hypothesis and conclude that there is evidence to suggest that the average weight of the bakery's croissants is less than the claimed value of 4 ounces.
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Problem 4:
A manufacturer of light bulbs claims that the mean life of their bulbs is 800 hours. To test this claim, a sample of 50 bulbs is selected and their mean life is found to be 775 hours with a standard deviation of 50 hours.
a) Is there evidence to suggest that the mean life of the bulbs is different from 800 hours?
b) What is the p-value for the test?
c) What is the 95% confidence interval for the mean life of the bulbs?

Solution:
a) Hypothesis Testing:
We will use a two-tailed t-test to determine if there is evidence to suggest that the mean life of the bulbs is different from 800 hours. The null hypothesis is that the mean life of the bulbs is equal to 800 hours, while the alternative hypothesis is that the mean life of the bulbs is different from 800 hours.
Null hypothesis: H0: μ = 800 Alternative hypothesis: H1: μ ≠ 800
We will use a significance level of α = 0.05.
The formula for calculating the t-value is:
$t=\frac{\overline{x}-\mu } {\frac{s}{\sqrt{n}}}$
Where:
x̄ = sample mean
μ = population mean
s = sample standard deviation
n = sample size
Substituting the values in the formula, we get:
$t=\frac{\overline{x}-\mu } {\frac{s}{\sqrt{n}}}=\frac{775-800} {\frac{50}{\sqrt{50}}}=\ {\color{Red} -3.54}$
Therefore, t = -3.54
The degrees of freedom (df) for the t-test is (n-1), which is 49 in this case. Using a t-distribution table or a calculator, we find that the p-value is less than 0.001.
Since the p-value is less than the significance level of 0.05, we reject the null hypothesis. There is sufficient evidence to suggest that the mean life of the bulbs is different from 800 hours.
b) Calculation of p-value:
The p-value is the probability of obtaining a test statistic as extreme or more extreme than the one observed, assuming that the null hypothesis is true. Since this is a two-tailed test, the p-value is the area under the t-distribution curve to the left of -3.54 and to the right of 3.54.
Using a t-distribution table or a calculator, we find that the area to the left of -3.54 is approximately 0.0003 and the area to the right of 3.54 is also approximately 0.0003. Therefore, the p-value is the sum of these two areas, which is 0.0006.
c) Calculation of 95% Confidence Interval:
We can calculate the 95% confidence interval for the mean life of the bulbs using the formula:
CI = x̄ ± tα/2 (s / √n)
Where:
x̄ = sample mean
tα/2 = the t-value from the t-distribution table with a degree of freedom of (n-1) and a significance level of α/2
s = sample standard deviation
n = sample size
Substituting the values in the formula, we get:
CI = 775 ± 2.01 (50 / √50)
CI = (757.46, 792.54)
Therefore, we can say with 95% confidence that the mean life of the bulbs is between 757.46 and 792.54 hours.
Conclusion:
Based on the results of the t-test, we can conclude that there is sufficient evidence to suggest that the mean life of the bulbs is different from 800 hours. The p-value for the test is 0.0006, which is less than the significance

TRY IT YOURSELF
1. A manufacturer claims that their product has a mean weight of 500 grams with a standard deviation of 20 grams. A sample of 25 products is taken and the mean weight is found to be 490 grams. Test the hypothesis that the mean weight of the products is less than 500 grams at a significance level of 0.05.
2. A survey of 500 people found that 280 of them support a particular political candidate. Test the hypothesis that the proportion of people who support the candidate is different from 0.5 at a significance level of 0.01.
3. A researcher claims that the mean IQ score for a population is at least 110 with a standard deviation of 10. A sample of 36 people is taken and the mean IQ score is found to be 105. Test the hypothesis that the mean IQ score is less than 110 at a significance level of 0.1.
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