Question

(a) A sample of 12 of bags of Calbie Chips were weighed (to the nearest gram), and listed here as follows. [9 marks]

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.

(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 performances 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. [8 marks]

Answer 1

a) A sample of 12 of bags of Calbie Chips were weighed

They are,

219, 226, 217, 224, 223, 216, 221, 228, 215, 229, 225, 229

Mean of the sample = (219+226+...+229)/12 = 222.67

Variance of the sample =[ (219-222.67)2 + (226-222.67)2 +....+(229-222.67)2​​​​​​​ ]/12= 278.67/12 = 23.22

Standard deviation = 4.82

95% confidence interval of the mean is given by,

   , z* = 1.96

  

= ( 219.9 , 225.4) (ans)

b)  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.

Difference = mu (1) - mu (2) = 78.8 - 86 = -7.20
Estimate for difference: -7.2

  T-Test of difference; t = -1.85 , (obtained in minitab)

p-value = 0.081

90% confidence interval for difference of mean is,

=( -13.7 , -0.70) (ans)

Here we accept the null hypothesis. so, based on the evidence, we can conclude that two sections are equally smarter.

c) Mortality rate = ( number of deaths due to COVID-19) / (number of confirmed cases of COVID 19)

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.

Mortality rate(US) = 33288/671493 = 0.0496 = p1

Mortality rate (UK) = 14606 / 109754 = 0.1331 = p2

  Difference = p (1) - p (2)
Estimate for difference: -0.0835

  Test for difference = 0 (vs not = 0): Z = -78.86 P-Value = 0.000

92% Confidence Interval for difference betweern two mortality rate.

  , Here z* = 1.75

  

= ( -0.0854 , -0.0817) (ans)

We conclude that the difference in the mortality rates between the US and the UK are not equal.

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