A multinational firm wants to estimate the average number of hours in a month that theiremployees spend on social media while on the job. A random sample of 83 employees showedthat they spent an average of 21.5 hours per month on social media, with a standard deviationof 2.5. Construct and interpret a 95% confidence interval for the population mean hours spenton social media per month.
2 answers:
Answer:
95% of the population spent between 16.8 hours to 26.2 hours on social media per month
Step-by-step explanation:
subtracting 1 from sample size: 83-1= 82 Note Mean=21.5 and standard deviation=2.5 Subtracting confidence interval from 1 and dividing by 2: (1-0.95)/2=0.025 At α=0.025 and dF=82, the t-distribution table gives 1.984 Divide standard deviation by square root of sample size: 21.5/√83= 2.36 Multiplying answer in step 4 by answer in step 5: 2.36×1.984= 4.68 For lower end: 21.5-4.68= 16.82 for upper end: 21.5+4.68=26.18
Answer:
20.96-22.0378
Step-by-step explanation:
For a 95% confidence interval the z value is 1.96
The confidence interval will be given by
Mean±z×б/√n
21.5-1.96×2.5/√83 =20.96
21.5+1.96×2.5/√83 =22.0378
With 95% confidence the average hours spent by employees on social media per hour is between 20.96 to 22.0378 based on this sample data
You might be interested in
Answer:
Volume is 21
Explanation
Base length is 4.5
Widgth is 3.5
and height is 4 (I can't tell if its 2 because picture is blurry)
V=lwh/3=4.5·3.5·4= 21
Answer:
77 maybe
sorry if I get it wrong
Answer:
6+12y
Step-by-step explanation:
first, distribute the 2 to the parenthesis:
-multiply the 2 to the 3 and 6y
We get 6+12y
hope this helps :)
Answer:
yes they are the same
Step-by-step explanation:
-2x + 8 = -4x + 1 - 1 -2x 7 = -6x — -6x 7/-6 or -7/6