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A group of researchers estimates the mean length of time (in minutes) the average U.S. adult spends watching television using digital video recorders (DVRs) and other forms of time-shifted television each day. To do so, the group takes a random sample of 30 U.S. adults and obtains the times (in minutes) below. From past studies, the research council assumes that ĂŹĆ’ is 6.5 minutes. Use the information to construct 90% and 99% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.

          A group of researchers estimates the mean length of time (in minutes) the average U.S. adult spends watching television using digital video recorders (DVRs) and other forms of time-shifted television each day. To do so, the group takes a random sample of 30 U.S. adults and obtains the times (in minutes) below. From past studies, the research council assumes that ĂŹĆ’ is 6.5 minutes. Use the information to construct 90% and 99% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.
        
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Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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A group of researchers estimates the mean length of time (in minutes) the average U.S. adult spends watching television using digital video recorders (DVRs) and other forms of time-shifted television each day. To do so, the group takes a random sample of 30 U.S. adults and obtains the times (in minutes) below. From past studies, the research council assumes that ĂŹĆ’ is 6.5 minutes. Use the information to construct 90% and 99% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.
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Use the information to construct the $90 \%$ and $99 \%$ confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. DVRs A research council wants to estimate the mean length of time (in minutes) the average U.S. adult spends watching television using digital video recorders (DVRs) each day. To determine this estimate, the research council takes a random sample of 20 U.S. adults and obtains the times (in minutes) below. $$\begin{array}{llllllllll}24 & 27 & 26 & 29 & 33 & 21 & 18 & 24 & 23 & 34 \\17 & 15 & 19 & 23 & 25 & 29 & 36 & 19 & 18 & 22\end{array}$$ From past studies, the research council assumes that $\sigma$ is 4.3 minutes and that the population of times is normally distributed.

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Confidence Intervals

Confidence Intervals for the Mean ( $\sigma$ Known)

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A research council wants to estimate the mean length of time (in minutes) the average U.S. adult spends watching television using digital video recorders (DVRs) each day. To determine this estimate, the research council takes a random sample of 20 U.S. adults and obtains the times (in minutes) below: 24, 26, 29, 33, 23, 34, 15, 19, 23, 25, 29, 36, 19, 18, 22. Find the 90% and 99% confidence intervals for the population mean, knowing that from past studies the population standard deviation is 4.3 minutes and that the population of times is normally distributed. Interpret the confidence intervals you found in part a.

Dominador T.

suppose-that-the-average-time-spent-per-day-with-digital-media-several-years-ago-was-3-hours-and-42-minutes-for-last-year-a-random-sample-of-20-adults-in-a-certain-region-spent-the-numbers-o-72606

Suppose that the average time spent per day with digital media several years ago was 3 hours and 42 minutes. For last year, a random sample of 20 adults in a certain region spent the numbers of hours per day with digital media given in the accompanying table. Preliminary data analyses indicate that the t-interval procedure can reasonably be applied. Find and interpret a 95% confidence interval for last year's mean time spent per day with digital media by adults of the region. (Note: x̄ equals 4.50 hr and s equals 2.18 hr.)

Robin C.


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Transcript

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00:01 To solve this problem, we have been given mean mu equal to 52 ,950 and standard deviation sigma equal to 4 ,370.
00:21 So in part a, we need to find probability where x is greater than 55 ,000, that is equal to p.
00:41 Divided by sigma is greater than 55 ,000 minus 52 ,950 divided by 4 ,370.
01:03 So it is equal to p z is greater than 0 .4691.
01:19 The value is 0 .3192.
01:29 From normal probability table.
01:51 So the answer is p x is greater than 50 ,000, 55 ,000 is equal to 0 .3192.
02:11 So this is the answer of part a.
02:16 In part b we need to find p where x is greater than 55 ,000 and less than 57 ,000, and less than 57 ,000...
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