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victus00 [196]
3 years ago
8

The central limit theorem states that sampling distributions are always the same shape as the population distribution from whenc

e the data came. True or False
Mathematics
1 answer:
sergejj [24]3 years ago
3 0

Explanation:

The sample mean is not always equal to the population mean but if we take more and more number of samples from the population then the mean of the sample would become equal to the population mean.

The Central Limit Theorem states that we can have a normal distribution of sample means even if the original population doesn't follow normal distribution, But we have to take a lot of samples.

Suppose a population doesn't follow normal distribution and is very skewed then we can still have sampling distribution that is completely normal if we take a lot of samples.

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Answer:

1. The 99% confidence interval for the difference in average is -6.47377 < μ₁ - μ₂ < -11.34623

2. The possible issues in the calculations includes;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

Step-by-step explanation:

1. The number of adults with insomnia in the sample = 45

The number of adults that participated in the therapy, n₁ = 22

The number of candidates that served as control group, n₂ = 23

The average score for the for the 22 participants of the program, \overline x_1 = 6.59

The standard deviation for the 22 participants of the program, s₁ = 4.10

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The confidence interval for unknown standard deviation, σ, is given by the following expression;

\left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

α = 1 - 0.99 = 0.01

α/2 = 0.005

The degrees of freedom, df = 22 - 1 = 21

t_{\alpha /2} = t_{0.005, \, 21} = 1.721

Therefore, we have;

\left (6.59- 15.5  \right )\pm1.721 \cdot \sqrt{\dfrac{4.10^{2}}{22}+\dfrac{5.34^{2}}{23}}

The 99% confidence interval for the difference in average is therefore given as follows;

-6.47377 < μ₁ - μ₂ < -11.34623

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2. The possible issues in the calculations are;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

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