Question

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. Calculate the 80% confidence interval of the true average number of hours of TV watched per week.
P.S: excel formola needed only. For lower and Upper Bound
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Answer 1

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;

Hours of TV per week (X): 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. =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample mean number of hours of TV watched per week = $\frac{\sum X}{n}$ = 9.65

            s = sample standard deviation = $\sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }$  = 4.61

            n = sample of people = 20

           $\mu$ = true average number of hours of TV watched per week

Here for constructing 80% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the true average, $\mu$ is ;

P(-1.33 < $t_1_9$ < 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 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.33) = 0.80

P( $-1.33 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ]

                                         = [ $9.65-1.33 \times {\frac{4.61}{\sqrt{20} } }$ , $9.65+1.33 \times {\frac{4.61}{\sqrt{20} } }$ ]

                                         = [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].