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denpristay [2]
3 years ago
14

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.

Mathematics
2 answers:
elena55 [62]3 years ago
7 0

Answer:

95% of the population spent between 16.8 hours to 26.2 hours on social media per month

Step-by-step explanation:

  1. subtracting 1 from sample size: 83-1= 82
  2. Note Mean=21.5 and standard deviation=2.5
  3. Subtracting confidence interval from 1 and dividing by 2: (1-0.95)/2=0.025
  4. At α=0.025 and dF=82,  the t-distribution table gives  1.984
  5. Divide standard deviation by square root of sample size: 21.5/√83= 2.36
  6. Multiplying answer in step 4 by answer in step 5: 2.36×1.984= 4.68
  7. For lower end: 21.5-4.68= 16.82
  8. for upper end: 21.5+4.68=26.18

sasho [114]3 years ago
4 0

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

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Answer:

A) mPQ = 71º

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Step-by-step explanation:

We know that mST is 19º and QR is also 19º because they're opposite angles. We also know that angle PUR is 90º.

If we subtract 19º from 90º we get 71º for mPQ.

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