Question

A simple random sample of 60 items resulted in a sample mean of 65. The population standard deviation is 14. a. Compute the 95% confidence interval for the population mean (to 1 decimal).

Answer 1

Answer:

The 95% confidence interval for the population mean is between 61.5 and 68.5.

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:

$\alpha = \frac{1-0.95}{2} = 0.025$

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 $1-0.025 = 0.975$, so $z = 1.96$

Now, find 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.

$M = 1.96*\frac{14}{\sqrt{60}} = 3.5$

The lower end of the interval is the sample mean subtracted by M. So it is 65 - 3.5 = 61.5

The upper end of the interval is the sample mean added to M. So it is 65 + 3.5 = 68.5

The 95% confidence interval for the population mean is between 61.5 and 68.5.

Answer 2

Answer:

95% confidence interval for the population mean is  [61.5 , 68.5].

Step-by-step explanation:

We are given that a random sample of 60 items resulted in a sample mean of 65. The population standard deviation is 14.

So, the pivotal quantity for 95% confidence interval for the average age is given by;

            P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean = 65

           $\sigma$ = population standard deviation = 14

           n = sample of items = 60

           $\mu$ = population mean

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 196) = 0.95

P(-1.96 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.95

P( $\bar X - 1.96 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X + 1.96 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X - 1.96 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X + 1.96 \times {\frac{\sigma}{\sqrt{n} }$ ]

                                                 = [ $65 - 1.96 \times {\frac{14}{\sqrt{60} }$ , $65 + 1.96 \times {\frac{14}{\sqrt{60} }$ ]

                                                 = [61.5 , 68.5]

Therefore, 95% confidence interval for the population mean is [61.5 , 68.5].