I need help quick!! you can have brainliest this is algerbra btw
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Answer:
80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].
Step-by-step explanation:
We are given that a college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of hours they watch per week. The results are provided below;
<u>Hours of TV per week (X)</u>: 6, 14, 13, 6, 16, 10, 19, 4, 5, 5, 18, 8, 7, 14, 8, 8, 9, 12, 6, 5.
Firstly, the Pivotal quantity for 80% confidence interval for the true average is given by;
P.Q. = ~
where, = sample mean number of hours of TV watched per week =
= 9.65
s = sample standard deviation = = 4.61
n = sample of people = 20
= true average number of hours of TV watched per week
<em>Here for constructing 80% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.</em>
<u>So, 80% confidence interval for the true average, </u><u> is ;</u>
P(-1.33 < < 1.33) = 0.80 {As the critical value of t at 19 degrees of
freedom are -1.33 & 1.33 with P = 10%}
P(-1.33 < < 1.33) = 0.80
P( <
<
) = 0.80
P( <
<
) = 0.80
<u>80% confidence interval for</u> = [
,
]
= [ ,
]
= [8.28 hours, 11.02 hours]
Therefore, 80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].
Tony walks 116.7 m diagonally across the lot.
Solution:
Length of the rectangle = 80 m
Width of the rectangle = 85 m
<u>To find the diagonal of the rectangle:</u>
Diagonal of the rectangle =
Diagonal of the rectangle = 116.7 m
Hence Tony walks 116.7 m diagonally across the lot.