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NNADVOKAT [17]
3 years ago
6

Twenty-five students were randomly selected and asked "How many hours of TV did you watch this weekend?" The sample mean for the

data is 5 hours.
The dotplot below shows the distribution of the sample mean hours of TV watched for 500 random samples of size 25 taken with replacement from the
original sample.
45
6
S.S
Simulated sample mean
Distribution of Simulated Mean
• samples mean SD
500 4.969 0.288
Use the results of the simulation to approximate the margin of error for the estimate of the mean number of hours of television watched. (answer in the
format 0.123)

Mathematics
1 answer:
Tpy6a [65]3 years ago
3 0

Using a confidence level of 95%, the margin of error for the estimate of the mean number of television hours watched will be 1.129

<u>The</u><u> </u><u>approximate</u><u> </u><u>margin</u><u> </u><u>of</u><u> </u><u>error</u><u> </u><u>can</u><u> </u><u>be</u><u> </u><u>calculated</u><u> </u><u>using</u><u> </u><u>the</u><u> </u><u>relation</u><u> </u><u>:</u>

  • Margin of Error = Z_{\frac{α}{2}} \frac{σ}{\sqrt{n}}

  • <em>Sample</em><em> </em><em>size</em><em>,</em><em> </em><em>n</em><em> </em><em>=</em><em> </em><em>25</em><em> </em>
  • <em>Standard</em><em> </em><em>deviation</em><em>,</em><em> </em><em>=</em><em> </em><em>2.88</em>
  • <em>Confidence</em><em> </em><em>level</em><em> </em><em>=</em><em> </em><em>95</em><em>%</em>

  • Z_{\frac{α}{2}} = 1.96

Margin of Error = 1.96 \frac{2.88}{\sqrt{25}}

Margin of Error = 1.96 \frac{2.88}{5}

Margin of Error = 1.96 \times 0.576

Margin of Error = 1.129

Therefore, the margin of error for the estimate of the mean number of <em>television hours watched</em> is 1.129.

Learn more : brainly.com/question/25630111

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