Distance from the x-axis is 2

We are given that a recent survey of 51 students reported that the average amount of time they spent listening to music was 11.5 hours per week, with a sample standard deviation of 9.2 hours.

Firstly, the pivotal quantity for 90% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average amount of time spent listening to music = 11.5

s = sample standard deviation = 9.2 hours

n = sample of students = 51

$\mu$ = population mean per week spent listening to the radio

Here for constructing 90% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.676 < $t_5_0$ < 1.676) = 0.90 {As the critical value of t at 50 degree of

freedom are -1.676 & 1.676 with P = 5%}

P(-1.676 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.676) = 0.90

P( $-1.676 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.676 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.676 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.676 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.676 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.676 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $11.5-1.676 \times {\frac{9.2}{\sqrt{51} } }$ , $11.5+1.676 \times {\frac{9.2}{\sqrt{51} } }$ ]

= [9.34 hours , 13.66 hours]

Therefore, 90% confidence interval for the mean time per week spent listening to the radio is [9.34 hours , 13.66 hours].

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4 years ago