Question

For a school, the principal expected that a student takes 45 minutes to travel to school on average. Suppose the time taken to travel to school by a student follows a normal distribution. 100 students are randomly selected from the school for interview and they took on average 44 minutes to travel to school with a standard deviation of 15 minutes. 6 of the selected students had lateness record in the last month.

1. Construct a 95% confidence interval for the true mean time taken to travel to school by a student. Correct the final answers to 2decimal places.
2. How many additional sample is needed to ensure the estimate of μ is accurate within 0.8 minutes with a confidence of 94% and σ is known to be 10 minutes.
3. By using critical value approach to perform a hypothesis testing for the principal’s claim that the true mean time taken to travel to school by student was 45minutes at level of significant α= 0.05. (if σ is unknown) Correct the final answers to 2 decimal places.
4. Construct a 92.5% confidence interval for the true proportion (p) of students who had lateness record in the last month. Correct the final answers to 4decimal places.
5. How large a sample is required to ensure the estimate of p is accurate within 2% with a confidence level of 97%?
6. By using critical value approach to perform a hypothesis testing for the principal’s claim that under (10 of all students had lateness record in the last month at level of significant α = 0.01. Correct the test statistics to 2 decimal places.

Answer 1

Sample Mean = 44

Std Dev = 15

n = 100

Std Error = 15/100^1/2 = 1.5

a)

At alpha = 0.05,

Z Critical = 1.96

95% CI = Mean +/- ZCritical * Std Error = 44 +/- 1.96 * 1.5 = {41.06,46.94}

b)

Margin of Error = 0.8

Alpha = 0.06

Z_Critical = 1.881

Std Dev = 10

Margin of Error = Z_Critical ­* SE = Z_Critical * Std Dev/n^1/2

n = (1.881*10/0.8)² = 552.84 ~ 553

So, Additional Sample = 553-100 = 453

c)

Alpha = 0.05,

N = 100

T Critical = +/- 1.984

Null and Alternate Hypothesis

H₀: µ = 45

H_a: µ <> 45

Test Statistic

t = (44-45)/1.5 = - 0.67

Since the t value lies in the interval {-1.984,1.984} we fail to reject the null hypothesis ie true mean time taken to travel to school by student was 45minutes.

d)

Alpha = 0.075

Z Critical ~ 1.75

Proportion (p) = 6/100 = 0.06

Std Dev = {p*(1-p)/n}^1/2 = 0.024

Hence 92.5% CI = 0.06 +/- 1.75*0.024 = {0.018,0.102}

PS: We are only allowed to answer 4 parts per question