The situation is as follows: Smart phones have changed the world and how we spend our time. A group of researchers at BYU want to estimate the mean daily number of minutes that BYU students spend on their phones. In fall 2019, they took a random sample of 430 BYU students and found that on average, students spend 312 minutes on their phones with a standard deviation of 54 minutes per day. The plot of the sample data showed no extreme skewness or outliers. Calculate a 96% confidence interval estimate for the mean daily number of minutes that BYU students spend on their phones in fall 2019. Do not round off in the intermediate steps, only the final answer should be rounded off to two decimal places.

Question

04-03-2021
Mathematics

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The situation is as follows: Smart phones have changed the world and how we spend our time. A group of researchers at BYU want to estimate the mean daily number of minutes that BYU students spend on their phones. In fall 2019, they took a random sample of 430 BYU students and found that on average, students spend 312 minutes on their phones with a standard deviation of 54 minutes per day. The plot of the sample data showed no extreme skewness or outliers. Calculate a 96% confidence interval estimate for the mean daily number of minutes that BYU students spend on their phones in fall 2019. Do not round off in the intermediate steps, only the final answer should be rounded off to two decimal places.

Sagot :

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Joaobezerraidn Joaobezerraidn · Answer 1 · 2021-03-05T19:29:25-05:00

Answer:

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 [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.96}{2} = 0.02[/tex]

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

That is z with a pvalue of [tex]1 - 0.02 = 0.98[/tex], so Z = 2.054.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 2.054\frac{54}{\sqrt{430}} = 5.35[/tex]

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.

Sagot :

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