The 90% confidence interval for the mean score of all takers of this test is between 59.92 and 64.08. The lower end is 59.92, and the upper end is 64.08.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.645*\frac{4}{\sqrt{10}} = 2.08$

The lower end of the interval is the sample mean subtracted by M. So it is 62 - 2.08 = 59.92

The upper end of the interval is the sample mean added to M. So it is 62 + 2.08 = 64.08.

The 90% confidence interval for the mean score of all takers of this test is between 59.92 and 64.08. The lower end is 59.92, and the upper end is 64.08.

5 0

2 years ago

I really need help please!

pashok25 [27]

As we know that angle of semi circle is 180°.

Here,

FLG = 40°

GLH = ?

HLJ = 60°

FLG + GLH + HLG = 180°

40° + GLH + 60° = 180°

GLH + 100° = 180°