Mathematics, 15.04.2020 05:18, officialgraciela67

A toy maker claims his best product has an average lifespan of exactly 18 years. A skeptical product evaluator asks for evidence (data) that might be used to evaluate this claim. The product evaluator was provided data collected from a random sample of 35 people who used the product. Using the data, an average product lifespan of 13 years and a standard deviation of 2 years was calculated. Select the 99%, confidence interval for the true mean lifespan of this product.

Answers: 3

Show answers

Answers

answered: ciera51

99%, confidence interval for the true mean lifespan of this product is [12.08 years , 13.92 years].

Step-by-step explanation:

We are given that a toy maker claims his best product has an average lifespan of exactly 18 years.

The product evaluator was provided data collected from a random sample of 35 people who used the product. Using the data, an average product lifespan of 13 years and a standard deviation of 2 years was calculated.

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 product lifespan = 13 years

n = sample of people = 35

s = sample standard deviation = 2 years

$\mu$ = true mean lifespan

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

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

P(-2.728 < t_3_4 < 2.728) = 0.99 {As the critical value of t at 34 degree

of freedom are -2.728 & 2.728 with P = 0.5%}

P(-2.728 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.728) = 0.99

P( $-2.728 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.728 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.728 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.728 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-2.728 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.728 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $13-2.728 \times {\frac{2}{\sqrt{35} } }$ , $13+2.728 \times {\frac{2}{\sqrt{35} } }$ ]

= [12.08 , 13.92]

Therefore, 99% confidence interval for the true mean lifespan of this product is [12.08 years , 13.92 years].

answered: Guest

A. p = 2(l + w)p/2 - l = ww = p/2 - l b. while isolating the variable w in the equation, i first divided both sides by 2 and then i subtracted l from both sides. i hope this .

answered: Guest

Abit confused about what you mean, but i think you might be referring to slope intercept form: y = mx + bin that case b, or the y intercept is already given.(0,-1)find the slope by using the formulay2 - y1/x2 - x1the final slope will be 4/2, equivalent to 2.now your equation should look like thisy = 4/2x -1sorry if this didn’t you out!