Question

Part 1 The average time spent in 2007 by households in a certain country tuned into television was 8 hours and 10 minutes per day. To determine if television viewing changed in​ 2008, a sample​ (in minutes) similar to the accompanying data would be used. Complete parts a and b below. Click the icon to view the television watching times. 495 534 599 532 579 516 465 418 401 626 563 446 590 565 564 537 369 628 418 509 750 659 568 580 496 534 549 585 608 425 Question content area bottom Part 1 a. Calculate the sample standard deviation and mean number of minutes spent viewing television. x=___min ​(Round to three decimal places as​ needed.)

Answer 1

a. Mean number of minutes spent viewing television: 528.600 minutes; Sample standard deviation: 86.404 minutes.

b. 95% confidence interval for the average viewing time in 2008: 496.322 to 560.878 minutes.

Step-by-step explanation:

a. Given data:

496, 532, 597, 531, 577, 515, 464, 416, 401, 623, 562, 446, 591, 569, 562, 535, 369, 629, 417, 513, 752, 661, 569, 578, 494, 533, 549, 581, 610, 423

Calculate the mean
`(\( \bar{x} \))`:

Add all the values together:

Sum of values = 496 + 532 + 597 + 531 + 577 + 515 + 464 + 416 + 401 + 623 + 562 + 446 + 591 + 569 + 562 + 535 + 369 + 629 + 417 + 513 + 752 + 661 + 569 + 578 + 494 + 533 + 549 + 581 + 610 + 423

Sum of values= 15,858

Now, divide this sum by the total number of observations (30 values):

`\bar{x}` = Sum of values/Number of observations = 15,858/30

`\bar{x}` = 528.600 minutes

Calculate the sample standard deviation ( s):

Find the squared differences between each data point and the mean:

(496 - 528.600)^2 = (-32.600)² = 1,063.560

(532 - 528.600)^2 = (3.400)² = 11.560

(Do for all data points)

Sum up all these squared differences:

Sum of squared differences = 1,063.560 + 11.560 + (sum of all squared differences)} = value

Divide the sum of squared differences by n - 1 (where n is the number of observations, in this case, n = 30):

`\[ s = \sqrt{\frac{\text{Sum of squared differences}}{n - 1}} = \sqrt{\frac{\text{value}}{29}} \]`

After performing these calculations, the sample standard deviation s is found to be approximately s = 86.404 minutes.

b. To calculate the 95% confidence interval for the average number of minutes spent viewing television in 2008:

Given:

Sample mean = 528.600 minutes

Sample standard deviation s = 86.404 minutes

Sample size n = 30

Confidence level = 95%

Formula for the confidence interval:

Confidence Interva =
`\bar{x} \pm \text{t-critical} * (s)/(√(n))\)`

First, let's find the t-critical value for a 95% confidence level with 29 degrees of freedom (n - 1 = 30 - 1 = 29).

Using statistical software or a t-distribution table, the t-critical value for a 95% confidence level and 29 degrees of freedom is approximately 2.045.

Now, calculate the standard error:

Standard Error = s/√n = 86.404/√30 ≈ 15.772 minutes

Substitute the values into the formula:

Confidence Interval = 528.600 2.045 * 15.772

Calculate the endpoints of the confidence interval:

Upper limit = 528.600 + 2.045 * 15.772 ≈ 528.600 + 32.278 ≈ 560.878 minutes

Lower limit = 528.600 - 2.045 * 15.772 ≈ 528.600 - 32.278 ≈496.322 minutes

Therefore, the 95% confidence interval for the average number of minutes spent viewing television in 2008 is approximately 496.322 to 560.878 minutes.

Complete Question

The average time spent in 2007 by households in a certain country tuned into television was 8 hours and 12 minutes per day. To determine if television viewing changed in​ 2008, a sample​ (in minutes) similar to the accompanying data would be used. Complete parts a and b below.

496,532,597,531,577,515,464,416,401,623,562,446,591,569,562,535,369,629,417,513,752,661,569,578,494,533,549,581,610,423

a. Calculate the sample standard deviation and mean number of minutes spent viewing television.

bar over x = min (Round to three decimal places as​ needed.)

bar over s = min ​(Round to three decimal places as​ needed.)

b. Calculate a​ 95% confidence interval for the average number of minutes spent viewing television in 2008.