What are the proofs

The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of 10 batteries

Nat2105 [25]

Answer:

(1) We are conducting the one-sample t-test and will be testing for the hypothesis of mean battery greater than 25 h. So, we will reject the null hypothesis for mean battery life equal to 25 hr against the alternative hypothesis for mean battery life greater than 25 hr.

(2) A 95% lower confidence interval on mean battery life is 25.06 hr.

Step-by-step explanation:

We are given that a random sample of 10 batteries is subjected to an accelerated life test by running them continuously at an elevated temperature until failure, and the following lifetimes (in hours) are obtained: 25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

Firstly, the pivotal quantity for finding the 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 mean BAC = $\frac{\sum X}{n}$ = 26 hr

s = sample standard deviation = $\sqrt{\frac{\sum(X - \bar X)^{2} }{n-1} }$ = 1.625 hr

n = sample of batteries = 10

$\mu$ = population mean battery life

<em> Here for constructing a 95% lower confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

(1) It is stated that the manufacturer wants to be certain that the mean battery life exceeds 25 h.

Since we are conducting the one-sample t-test and will be testing for the hypothesis of mean battery greater than 25 h. So, we will reject the null hypothesis for mean battery life equal to 25 hr against the alternative hypothesis for mean battery life greater than 25 hr.

(2) So, a 95% lower confidence interval for the population mean, $\mu$ is;

P(-1.833 < $t_9$ ) = 0.95 {As the lower critical value of t at 9

degrees of freedom is -1.833 with P = 5%}

P(-1.833 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $-1.833 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ ) = 0.95

P( $\bar X-1.833 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ ) = 0.95

<u>95% lower confidence interval for</u> $\mu$ = [ $\bar X-1.833 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $26-1.833 \times {\frac{1.625}{\sqrt{10} } }$ ]

= [25.06 hr]

Therefore, a 95% lower confidence interval on mean battery life is 25.06 hr.

5 0

3 years ago

The coordinates of the points bellow represent the vertices of a rectangle: P: (2,2). Q:(6,2) R:(6,5) S:(2,5) What is the perime

skad [1K]

Answer:

Therefore the value of perimeter of the rectangle is 14 in.

Step-by-step explanation:

The opposite sides of a rectangle are congruent.

The perimeter of a rectangle is 2(length + width)

The area of a rectangle is (length×width).

Given the vertices of a rectangle are P(2,2), Q(6,2), R(6,5) and S(2,5).

To find the the perimeter, first we need the sides of the rectangle.

The distance between two points(x₁,y₁) and (x₂,y₂) is

$=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The distance between P and Q is

$=\sqrt{(6-2)^2+(2-2)^2}$ in

$=\sqrt {4^2+0^2}$ in

=4 in

The distance between Q and R is

$=\sqrt{(6-6)^2+(5-2)^2$ in

$=\sqrt{0^2+3^2}$ in

=3 in

The distance between R and S is

$=\sqrt{(2-6)^2+(5-5)^2}$ in

$=\sqrt{ (-4)^2}$ in

=4 in

The distance between S and P is

$=\sqrt{(2-2)^2+(2-5)^2}$ in

$=\sqrt{(-3)^2}$ in

=3 in

Therefore the length of the rectangle is = 4 in.

an the width of the rectangle is = 3 in

Therefore the perimeter of the rectangle is =2(4+3) in

=14 in

7 0

3 years ago

How do I order these numbers from least to greatest?

natita [175]

Answer:

least—> greater ( 1/2 , √0.28 , 0.58 )

Step-by-step explanation:

o 1/2 = 0.5

o √0.28 = 0.52

o 0.58 = 0.58

I hope I helped you^_^

8 0

3 years ago

Can someone please answer this question please help

daser333 [38]

Here we are looking for the percentage of beads out of the total that Paco received that were orange.

The total number of beads Paco received is equal to orange + the different color. 45 + 5 = 50 total beads

Then, we'll need to create a fraction for the number of orange beads Paco received over the total number of beads. 45 / 50

Lastly, we can use our calculator to find a decimal number for the proportion of orange beads. 0.90

Multiply 0.90 by 100 to get the percentage. 90% orange beads

Hope this helps!! :)

3 0

3 years ago

Read 2 more answers

Nancy buys 15 m of ribbon. Her sister buys 6 m less ribbon but pays twice as much per metre. Together they pay R66. What is the

Katarina [22]

Answer: 44/9 ----> 22/9

I hope this helps you

4 0

3 years ago