What is the thirty-second term of the arithmetic sequence -12, -7, -2, 3,…? plz with sulution 6 grade pl i will mark brainliest and like plzzz

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rachellynn02

16.02.2021

Mathematics

99% confidence interval for the true mean resale value of a 5-year-old car of this model is [$12,173.24 , $13,306.76].

Step-by-step explanation:

We are given that you manage to obtain data on 17 recently resold 5-year-old foreign sedans of the same model.

These 17 cars were resold at an average price of $ 12 comma 740 with a standard deviation of $ 800.

Firstly, the pivotal quantity for 99% confidence interval for the true mean is given by;

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

where, $\bar X$ = sample average price = $12,740

s = sample standard deviation = $800

n = sample of cars = 17

$\mu$ = true population mean

Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.921 < t_1_6 < 2.921) = 0.99 {As the critical value of t at 16 degree of

freedom are -2.921 & 2.921 with P = 0.5%}

P(-2.921 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.921) = 0.99

P( $-2.921 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.921 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.921 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.921 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-2.921 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.921 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $12,740-2.921 \times {\frac{800}{\sqrt{17} } }$ , $12,740+2.921 \times {\frac{800}{\sqrt{17} } }$ ]

= [$12,173.24 , $13,306.76]

Therefore, 99% confidence interval for the true mean resale value of a 5-year-old car of this model is [$12,173.24 , $13,306.76].

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