Assume That The Marks Of Students In Statistical Methods

Assume That The Marks Of Students In Statistical Methods Are Normally

Assume that the marks of students in Statistical Methods are normally distributed. A random sample of 25 students’ marks yields a sample mean of 70 and a sample standard deviation of 10.

(a) Estimate the population mean of the marks with 95% confidence.

(b) If the population mean of the marks is 67 and the probability that the sample mean of 25 students’ marks is greater than 70 is 3%, find the population standard deviation of the marks.

(c) A child psychologist was interested in the difference in age (in years) between a boy and a girl when they first learn to ride a two-wheeled bicycle. The psychologist calculated a 99% confidence interval for the difference in age to be (-0.58, 0.71). Can he conclude that there is no difference in age, on the average, at which boys and girls learn to ride a two-wheeled bicycle?

Paper For Above instruction

The analysis of statistical data often involves estimating population parameters based on sample data and testing hypotheses to infer more about the underlying distributions. This paper addresses three interconnected statistical problems based on the assumption that certain variables follow a normal distribution. The first problem involves estimating the population mean of students' marks with a known sample mean and standard deviation, while the second involves deducing the population standard deviation given a specific mean, probability, and sample data. The third problem examines confidence intervals concerning the age at which children learn to ride bicycles, aiming to infer whether no difference exists between genders regarding this developmental milestone.

Part (a): Estimating the Population Mean with 95% Confidence

Given a sample size (n) of 25 students, with a sample mean (x̄) of 70 and a sample standard deviation (s) of 10, the objective is to estimate the population mean (μ) with 95% confidence. Because the sample size is small (n

The standard error of the mean (SEM) is calculated as:

SEM = s / √n = 10 / √25 = 10 / 5 = 2

The critical t-value (t*) for a 95% confidence level with 24 degrees of freedom (n-1) can be obtained from t-tables or statistical software. It is approximately 2.064.

Hence, the confidence interval (CI) for the population mean is:

CI = x̄ ± t* × SEM = 70 ± 2.064 × 2 = 70 ± 4.128

Thus, the 95% confidence interval for the population mean is approximately (65.872, 74.128). Therefore, we are 95% confident that the true average marks of students in Statistical Methods lie within this interval.

Part (b): Determining Population Standard Deviation

The problem states that if the population mean (μ) is 67, then the probability P( sample mean > 70 ) is 3%. We are asked to find the population standard deviation (σ), given the sample size and the probability constraint.

First, note that the sampling distribution of the sample mean (x̄) is normal with mean μ = 67 and standard deviation σ / √n. We know that:

P( x̄ > 70 ) = 3%

Expressed in terms of the standard normal variable Z:

P( Z > (70 - μ) / (σ / √n) ) = 0.03

Calculate the z-score corresponding to the upper tail probability of 3%. Using standard normal tables or software, Z ≈ 1.88.

Set up the equation:

(70 - 67) / (σ / √25) = 1.88

Simplify:

3 / (σ / 5) = 1.88

Solving for σ:

σ = (3 × 5) / 1.88 ≈ 15 / 1.88 ≈ 7.98

Therefore, the population standard deviation is approximately 7.98.

Part (c): Confidence Interval for the Difference in Ages

The psychologist computed a 99% confidence interval for the difference in age (D) between boys and girls learning to ride a two-wheeled bicycle as (-0.58, 0.71). The key question is whether this interval suggests that there is no significant difference.

A confidence interval that includes zero indicates that the difference might be zero, implying no statistically significant difference at the given confidence level. Since the interval spans from negative to positive (-0.58 to 0.71), zero is within this interval.

Therefore, based on the data and at the 99% confidence level, there is insufficient evidence to conclude that there is a significant difference in the mean age at which boys and girls learn to ride the bicycle. The interval's inclusion of zero implies the difference could be null.

This conclusion aligns with standard interpretations of confidence intervals, where the presence of zero within the interval indicates the possibility of no effect (or difference), thus suggesting that the actual difference could be zero.

Conclusion

This analysis underscores how confidence intervals and hypothesis testing facilitate understanding of population parameters from sample data. In part (a), the estimated mean with a 95% confidence interval provides a range where the true mean likely resides. Part (b) demonstrates how probability and sample data can be used to estimate variance parameters, with the computed population standard deviation being approximately 7.98. Finally, part (c) highlights the interpretive power of confidence intervals in testing hypotheses about differences between groups, where the inclusion of zero indicates no significant difference at the specified confidence level. These techniques are fundamental to statistical inference, enabling researchers and practitioners to make informed decisions based on sample data.

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