Ifrah-Khalid-BUSI 1013E-UNIT Exercise 3

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May 13, 2024

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Unit Exercise 3 Q1: a) The researcher took a random sample of 64 students from a population of 5000 students who use an e-learning platform. The hours spent on the platform per week by the sample of students were gathered. This provides sample data on usage hours that can estimate the overall average hours for the population. Some key details:  There are 5000 student’s total.  64 students were randomly sampled.  The data is on weekly e-learning platform usage. A statistical range that is likely to include the true population mean is called a 95% confidence interval. If one were to take 100 different samples and calculate a confidence interval from each, approximately 95 of those intervals would be expected to include the true population mean. In other words, there is a 0.95 probability (95% chance) that the researcher's computed confidence interval captures the true average platform usage for all 5000 students. The researcher states they are 95% confident the true population mean hours falls between 36.2 and 40.8 hours per week. This signifies that based on analyzing the sample statistics, they calculated an interval estimate they can be 95% confident contains the actual average weekly usage hours for the full student group. Q2: a) The null hypothesis ( H 0 ) in this case is that the mean score of customer services at the shopping mall is equal to or greater than 85 ( H 0 : μ≥ 85 ¿ The alternative hypothesis hypothesis ( H 1 ) is that the mean score is less than 85 ( H 1 : μ≥ 85 ¿ b) Since the sample size n=25 and the population standard deviation is not known, the student’s t distribution is used when the sample size is small and / Or the population standard deviation is unknown, making it a more robust choice for these types of situations. c) In simpler terms, the property manager of the shopping mall claims that their customer service has a mean score of at least 85. To test this claim, they randomly selected 25 customers and asked them to evaluate customer service. After analyzing the results, it was found that there's strong evidence against the claim (p−value<0.002). This means that there's less than a 0.2% chance that an average customer service score of at least 85 occurred by chance, based on this sample.

In other words, it's highly unlikely that the true customer service score is as high as 85, and there's strong evidence to suggest that it might be lower than what the property manager claims. Q3: a) Construct a 95% confidence interval for the mean unit cost of the population: The formula to calculate the confidence interval for the population mean is: For a 95% confidence interval, the critical value for a two-tailed test is approximately 1.96. Given: Sample Mean x = 25 Population standard deviation ( σ )$10.50 sample size (n) = 120 confidence level = 95% critical value ( Z c ) = 1.96 Using excel function = NORM.S.INV (0.025) Substitute the values into the formula: CI = x±Z c × σ √ n ¿ $ 25 ± 1.96 × $ 10.50 √ 120 ¿ $ 25 ±$ 1.87868 ¿ $ 23.121 , $ 26.879 b) For a 90% confidence interval, the critical value is approximately 1.645 Using excel function = NORM.S.INV (0.05) ME = Z c × σ √ n ¿ 1.645 × $ 10.50 √ 120 = $1.57675 = $1.5768 The margin of error at the 90% confidence level is approximately $1.5768 . c) Using the same formula as in part a: Confidence interval

CI = $ 25 ± 1.96 × $ 10.50 √ 120 d) Based on the result from a and c what can say about the effect of larger sample size on the length of the confidence interval s: As the sample size increases from (120 to 200), the standard error decreases. This reduction is standard error leads to a narrower confidence interval. In other words, a larger sample size results in a more precise estimate of the population mean. The confidence interval becomes narrowest because with a larger sample size, there is more information about the population’s chartists', which reduces the uncertainty associated with estimate. Q4: The statement made by the statistician contains a misinterpretation of the Central Limit Theorem (CLT) and the behavior of sample means in large samples. The CLT asserts that as the sample size increases, the sampling distribution of the sample mean tends to follow a normal distribution, regardless of the shape of the population distribution, provided that the sample size is sufficiently large. This implies that, for large samples, the distribution of sample means becomes approximately normal and centers around the population mean. However, the probability that the sample mean is higher or lower than the population mean is not necessarily equal due to random sampling variability. Variability exists around the central value in the sampling distribution of the sample mean. Additionally, if there is bias in the sampling process or if the population distribution is skewed, bias in one direction (either higher or lower) relative to the population mean may persist, even in large samples, impacting the probabilities of the sample mean being higher or lower than the population mean. Let's assume a population with a mean of μ and a standard deviation of σ. With large samples (e.g. n = 30), the sampling distribution of the sample mean, according to the CLT, will be approximately normal with a mean of μ . However, due to sampling variability, some sample means may still deviate from μ . This clarification highlights the nuances in the behavior of sample means in large samples and the potential impact of bias or skewed population distributions. Q5: A) Test the hypothesis that the population mean length is greater than 145 min:  Null hypothesis ( H 0 : μ≥ 145 )  Alternative hypothesis ( H 1 : μ≥ 145 ) Given:  Sample mean (x̄) = 150 min  Population standard deviation (σ) = 10 min  Sample size (n) = 25

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