Question

In a study of children with a particular disorder, parents were asked to rate their child on a variety of items related to how well their child performs different tasks. One item was "Has difficulty organizing work," rated on a five-point scale of 0 to 4 with 0 corresponding to "not at all" and 4 corresponding to "very much." The mean rating for 275 boys with the disorder was reported as 2.21 with a standard deviation of 1.04. (Round your answers to four decimal places.)

Compute the 90% confidence interval.  

Compute the 95% confidence interval.

Compute the 99% confidence interval.

Answer 1

We are given,

The mean rating $=\bar x=2.21$

standard deviation $=s=1.04$

Number of boys observed $=n=275$

Now as the population standard deviation is unknown, but the sample size is large ( >30 ) we will use z-test statistic.

$\frac{\bar x-\mu}{\frac{s}{\sqrt{n}}}\sim N(0,1)$

The 90% confidence interval of population mean i.e., $\mu$ is

$\bar x+Z_{0.05}\times \frac{s}{\sqrt{n}}<\mu<\bar x+Z_{0.95}\times \frac{s}{\sqrt{n}}$

$2.21-1.645\times \frac{1.04}{\sqrt{275}}<\mu<2.21+1.645\times \frac{1.04}{\sqrt{275}}$

$\Rightarrow 2.1068<\mu<2.3132$

This is the 90% confidence interval of $\mu$

The 95% confidence interval of $\mu$ is

$\bar x+Z_{0.025}\times \frac{s}{\sqrt{n}}<\mu<\bar x+Z_{0.975}\times \frac{s}{\sqrt{n}}$

$2.21-1.960\times \frac{1.04}{\sqrt{275}}<\mu<2.21+1.960\times \frac{1.04}{\sqrt{275}}$

$\Rightarrow 2.0871<\mu<2.3329$

This is the 95% confidence interval of $\mu$

The 99% confidence interval of $\mu$ is

$\bar x+Z_{0.005}\times \frac{s}{\sqrt{n}}<\mu<\bar x+Z_{0.995}\times \frac{s}{\sqrt{n}}$

$2.21-2.576\times \frac{1.04}{\sqrt{275}}<\mu<2.21+2.576\times \frac{1.04}{\sqrt{275}}$

$\Rightarrow 2.0484<\mu<2.3716$

This is the 99% confidence interval of $\mu$