Question

A sample of 12 of bags of Calbie Chips were weighed (to the nearest gram), and listed here as follows.

219, 226, 217, 224, 223, 216, 221, 228, 215, 229, 225, 229 Find a 95% confidence interval for the mean mass of bags of Calbie Chips.

Find a 95% confidence interval for the mean mass of bags of Calbie Chips

2. (b) Professor GeniusAtCalculus has two lecture sections (A and B) of the same 4th year Advanced Calculus (AMA 4301) course in Semester 2. She wants to investigate whether section A students maybe ”smarter” than section B students by comparing their perfor- mances in the midterm test. A random sample of 12 students were taken from section A, with mean midterm test score of 78.8 and standard deviation 8.5; and a random sample of 9 students were taken from section B, with mean midterm test score of 86 and standard deviation 9.3. Assume the population standard deviations of midterm test scores for both sections are the same. Construct the 90% confidence interval for the difference in midterm test scores of the two sections. Based on the sample midterm test scores from the two sections, can Professor GeniusAtCalculus conclude that there is any evidence that one section of students are ”smarter” than the other section? Justify your conclusions.

   [8 marks]

3. (c) The COVID-19 (coronavirus) mortality rate of a country is defined as the ratio of the number of deaths due to COVID-19 divided by the number of (confirmed) cases of COVID-19 in that country. Suppose we want to investigate if there is any difference between the COVID-19 mortality rate in the US and the UK. On April 18, 2020, out of a sample of 671,493 cases of COVID-19 in the US, there was 33,288 deaths; and out of a sample of 109,754 cases of COVID-19 in the UK, there was 14,606 deaths. What is the 92% confidence interval in the true difference in the mortality rates between the two countries? What can you conclude about the difference in the mortality rates between the US and the UK? Justify your conclusions.

fidence interval for the mean mass of bags of Calbie Chips. [9 marks] (b) Professor GeniusAtCalculus has two lecture sections (A and B) of the same 4th year Advanced Calculus (AMA 4301) course in Semester 2. She wants to investigate whether section A students maybe ”smarter” than section B students by comparing their perfor- mances in the midterm test. A random sample of 12 students were taken from section A, with mean midterm test score of 78.8 and standard deviation 8.5; and a random sample of 9 students were taken from section B, with mean midterm test score of 86 and standard deviation 9.3. Assume the population standard deviations of midterm test scores for both sections are the same. Construct the 90% confidence interval for the difference in midterm test scores of the two sections. Based on the sample midterm test scores from the two sections, can Professor GeniusAtCalculus conclude that there is any evidence that one section of students are ”smarter” than the other section? Justify your conclusions. [8 marks] (c) The COVID-19 (coronavirus) mortality rate of a country is defined as the ratio of the number of deaths due to COVID-19 divided by the number of (confirmed) cases of COVID-19 in that country. Suppose we want to investigate if there is any difference between the COVID-19 mortality rate in the US and the UK. On April 18, 2020, out of a sample of 671,493 cases of COVID-19 in the US, there was 33,288 deaths; and out of a sample of 109,754 cases of COVID-19 in the UK, there was 14,606 deaths. What is the 92% confidence interval in the true difference in the mortality rates between the two countries? What can you conclude about the difference in the mortality rates between the US and the UK? Justify your conclusions.

Answer 1

a.)

The provided sample mean is and the sample standard deviation is s = 5.03. The size of the sample is n = 12 and the required confidence level is 95%.

The number of degrees of freedom are df = 12 - 1 = 11, and the significance level is α=0.05.

Based on the provided information, the critical t-value for α=0.05 and df = 11 degrees of freedom is .

The 95% confidence for the population mean μ is computed using the following expression

Therefore, based on the information provided, the 95 % confidence for the population mean μ is

=(222.67−3.196,222.67+3.196)

= (219.474, 225.866)