What is 3/9 eqvailent to?

In a test of weight loss programs, 40 randomly selected adults used the Atkins weight loss program. After 12 months, their mean

natulia [17]

Answer:

90% confidence interval for the mean weight loss for all such subjects is [0.82 lbs, 3.38 lbs].

Step-by-step explanation:

We are given that in a test of weight loss programs, 40 randomly selected adults used the Atkins weight loss program.

After 12 months, their mean weight loss was found to be 2.1 lbs., with a standard deviation of 4.8 lbs.

Firstly, the pivotal quantity for 90% confidence interval for the true mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight loss of 40 adults = 2.1 lbs

s = sample standard deviation = 4.8 lbs

n = sample of adults = 40

$\mu$ = population mean weight loss

Here for constructing 90% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.685 < $t_3_9$ < 1.685) = 0.90 {As the critical value of t at 39 degree

of freedom are -1.685 & 1.685 with P = 5%}

P(-1.685 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.685) = 0.90

P( $-1.685 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.685 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.685 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X +1.685 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.685 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.685 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $2.1-1.685 \times {\frac{4.8}{\sqrt{40} } }$ , $2.1+1.685 \times {\frac{4.8}{\sqrt{40} } }$ ]

= [0.82 , 3.38]

Therefore, 90% confidence interval for the true mean weight loss for all such subjects is [0.82 lbs, 3.38 lbs].

Interpretation of this confidence interval is that we are 90% confident that the mean weight loss for all such subjects will lie between 0.82 lbs and 3.38 lbs.

So, if our true mean weight will be between 0.82 lbs and 2.1 lbs then we can say that the Atkins program appear to be effective. But it is somewhat appear to be practical kind of aspect.

5 0

3 years ago

The distribution of GPAs is normally distributed with a mean of 2.7 pounds and a standard deviation of 0.3. Approximately what p

serious [3.7K]

The best answer to this statistic question would be 15.7% or, if rounded off, 16%. This is because those students who have GPAs are approximately at least 1 standard deviation away from the mean and more - which means that 84% of the GPAs are less then 3 pounds. I subtracted 84 from 99.7 (which is the whole distribution's percentage) and came up with 15.7.

3 0

3 years ago

laney pays $0.99 for each song she downloads from the internet. in total she paid $24.75 for songs from the internet. how many s

dangina [55]

Laney would have downloaded 25 songs

6 0

3 years ago

Half a number increased by 14

gregori [183]

Answer:

x/2 +14

Step-by-step explanation: