The 96% confidence interval estimate for the mean daily number of minutes that BYU students spend on their phones in fall 2019 is between 306.65 minutes and 317.35 minutes.

Step-by-step explanation:

Confidence interval normal

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

$\alpha = \frac{1 - 0.96}{2} = 0.02$

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

That is z with a pvalue of $1 - 0.02 = 0.98$ , so Z = 2.054.

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.

$M = 2.054\frac{54}{\sqrt{430}} = 5.35$

The lower end of the interval is the sample mean subtracted by M. So it is 312 - 5.35 = 306.65 minutes

The upper end of the interval is the sample mean added to M. So it is 312 + 5.35 = 317.35 minutes

The 96% confidence interval estimate for the mean daily number of minutes that BYU students spend on their phones in fall 2019 is between 306.65 minutes and 317.35 minutes.

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2 years ago

1. Solve the quadratic equation using quadratic formula :x2 – 7x + 12 = 0​

1. Solve the quadratic equation using quadratic formula :x2 – 7x + 12 = 0