Suppose that the amount of time teenagers spend on the Internet is normally distributed, with a standard deviation of 1.5 hours. A sample of 100 teenagers is selected at random, and the sample mean is...

Question 1(a)
Let X be the random variable representing the amount of time (in hours) teenagers spend on the
Internet.
Hence X~ N (μ, 1.5
2
) where μ is the true population mean of the amount of time (in hours)
teenagers spend on the Internet.
Let X be the sample mean of the amount of time (in hours) teenagers spend on the Internet.
and n be the sample size.
(i) Now X = 6.5 and n= 100
Since the population is normally distributed and the value of the population variance is known,
we can use the z-test to find the confidence interval for true population mean μ.
Let α be the level of significance. In the general, the (1-α)% confidence interval for μ is
μ ϵ ( X – zα/2 * σ/n1/2 , X + z α/2 * σ/n1/2 ) where σ is the population standard deviation.
Hence the 99% confidence interval would be
μ ϵ ( X – z0.005 * σ/n1/2 , X + z0.005 * σ/n1/2 )
Here X = 6.5, σ = 1.5 , n=100 , z0.005 ( from the z distribution table ) = 2.58
Hence μ ϵ (6.5– 2.58 * 1.5/1001/2 , 6.5+ 2.58 * 1.5/1001/2 ) => μ ϵ (6.11, 6.89)
The interpretation of the above confidence interval is that out of such 100 random intervals
constructed, 99 will contain the true population mean, μ.
Hence this interval (6.1,6.9) will contain μ with probability 99%.
(ii) Here n=300 and α=0.05
Hence the 95% confidence interval would be
μ ϵ ( X – z0.025 * σ/n1/2 , X + z0.025 * σ/n1/2 )
Hence μ ϵ (6.5– 1.95 * 1.5/3001/2 , 6.5+ 1.95 * 1.5/3001/2 ) => μ ϵ (6.33, 6.67)
The interpretation of the above confidence interval is that out of such 100 random intervals
constructed, 95 will contain the true population mean, μ.
Hence this interval (6.3,6.7) will contain μ with probability 95%.
(iii) There are two things to consider here, first the sample size and second the level of
confidence. Let’s consider the each one by one:
 When the size of the sample increases, keeping the level of confidence the
same, the width of the confidence interval decreases which means we get the
more precise confidence interval.
Consider an example below:
μ ϵ (6.5– 2.58 * 1.5/3001/2 , 6.5+ 2.58 * 1.5/3001/2 ) => μ ϵ (6.28, 6.72)

Here we have changed just the sample size, rest is the same as in part(i). We
can see that this confidence interval due to lower width has more precision.
 When the level of confidence decreases, keeping the sample size the same, the
width of the confidence interval decreases which means we get the more
precise confidence interval.
Consider an example below:
Here we just change the level of confidence to 95% from 99% in part (i)
μ ϵ (6.5– 1.95 * 1.5/1001/2, 6.5+ 1.95 * 1.5/1001/2) => μ ϵ (6.21, 6.79)

We can see that this confidence interval due to lower width has more
precision. Although the probability that this interval contains the true mean, μ
would decrease.
Hence increasing the sample size and decreasing the level of confidence together would decrease
the width of the interval, along with decreasing the probability of the interval containing the true
parameter.
Question 1(b)
Let Y be the random variable representing the time spent (in hours) by Australian adults playing
sports per day.
Hence Y ~ N( 4, 1.25
2
) where μ= 4 and σ= 1.25
(i) P( Y> 5) = P[z> (5-4)/1.25] where z is a standard normal variate
 P(z> 0.8) = 0.2119 (from the z table)
(ii) Let the average number of hours spent paying sport by four Adults Australian is Y
`
Since Y~ N(4, 1.25
2
) => Y ~N(4, 1.25
2
4)
(If a random variable is normally distributed, the sample mean of it would also be
normally distributed with same mean and variance divided by the sample size. This
means the sample mean has lower variability)
P(Y >5) = P[z> (5-4)/(1.25/2)]
 = P(z>1.6) = 0.0548(from the z table)
(iii) Suppose Y1 be the time spent by player 1 on the sports, Y2 be the time spent by player
2 on the sports, Y3 be the time spent by player 3 on the sports and Y4 be the time
spent by player 1 on the sports.
We assume all these to be independent of one another.
P(all four play sports for more than 5 hours per day)= P(Y1 >5, Y2 >5, Y3 >5, Y4 >5)
Since they are independent of one another, we can write
P(Y1 >5, Y2 >5, Y3 >5, Y4 >5) = P(Y1 >5)*P(Y2 >5)*P(Y3 >5)*P(Y4 >5)
Since each Yi is normally distributed, P(Y1 >5)*P(Y2 >5)*P(Y3 >5)*P(Y4 >5) = [P(Y1 >5)]
4
Now P(Y1 >5) = P[z> (5-4)/1.25] where z is a standard normal variate
 P(z> 0.8) = 0.2119 (from the z table)
Hence P(all four play sports for more than 5 hours per day) = 0.2119
5
= 0.460326
Question 2 (a)
Let p be the proportion of students passing the course after the adoption of practice of having
lecture recordings.
Let p be the estimate of p and n be the sample size.
We have p = 11/25= 0.44 and se( p )=[p*(1-p)/n]
0.5
= [0.44*(1-0.44)/25]
0.5
= 0.099277
Our hypothesize population proportion is 80%
Setting up the hypotheses:
Ho(null hypothesis): p >= 0.8
Ha(alternative hypothesis): p < 0.8
This is a left tailed test.
Under the null hypothesis, our test statistic would be (p-p) / se( p ) which follows standard
normal distribution.

Hence the value of the test statistic = (0.44-0.8)/ 0.099277= -3.6262
Since α= 0.01, hence the critical value (from the z-table) is -2.33.
Since the value of the test statistic is less than the critical value, we reject the null hypothesis.
Hence we can conclude that at 1% level of significance, the proportion of students passing the
course after the adoption of practice of having lecture recordings has significantly reduced from
0.8.
Question 2 (b)
Let μ be the true value of the mean of additional hours of sleep due to the new drug
Hence our hypotheses would be
Ho(null hypothesis): μ >=2
Ha(alternative hypothesis): μ < 2
This is a left tailed test.
We have sample...