Question

A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each one recorded. The results are given below. Assume the percentages of students' absences are approximately normally distributed. Use Excel to estimate the mean percentage of absences per tutorial over the past 5 years with 90% confidence. Round your answers to two decimal places and use increasing order. Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

Answer 1

Answer:

The 90% confidence interval using Student's t-distribution is (9.22, 11.61).

Step-by-step explanation:

Since we know the sample is not big enough to use a z-distribution, we use student's t-distribution instead.

The formula to calculate the confidence interval is given by:

$\bar{x}\pm t_{n-1} \times s/\sqrt{n}$

Where:

$\bar{x}$ is the sample's mean,

$t_{n-1}$ is t-score with n-1 degrees of freedom,

$s$ is the standard error,

$n$ is the sample's size.

This part of the equation is called margin of error:

$s/\sqrt{n}$

We know that:

$n=28$

$\bar{x}=10.41$

degrees of freedom $= 27$

$1-\alpha = 0.90 \Rightarrow \alpha = 0.10$

$t_{n-1} = 1.703$

$s = 3.71$

Replacing in the formula with the corresponding values we obtain the confidence interval:

$\bar{x}\pm t_{n-1} \times s/\sqrt{n} = 10.41 \pm 1.70 \times 3.71/\sqrt{28} = (9.22, 11.61)$

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Answer 2

Answer:

The calculation for the sample mean and sample standard deviation by use of Excel is given:

Number of Absent in ascending order

1.7   4.4  4.5  4.8  5.7  8.2  8.4  8.8  9.7  9.7  10  10.3  10.3  10.6  10.8  10.9  11.5  11.7  11.9  12.2  12.3  12.4  13.2  13.9  15.2  15.9  16.2  16.4

Average 10.41428571

Variance 13.7768254

Step-by-step explanation: