Mathematics, 23.09.2020 17:01 fatherbamboo

The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. 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 25.7.1)The manufacturer wants to be certain that the mean battery life exceeds 25 h. What conclusions can be drawn from these data(useα=0.05)2)Please construct a 95% lower confidence interval on mean battery life. Why would the manufacturer be interested in a one-sided confidence interval?

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Answers

(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

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.

(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

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

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

= [25.06 hr]

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

Answer from: Quest

Hello!

the answer is:

the correct option:

b. $x=2.23$

why?

to solve the problem, we need to remember how to operate to simplify exponential expressions using the logarithms.

we are given the equation:

$7^{x}=77$

then, using placing logarith with base 7 on both sides of the equation, we have:

$7^{x}=77log_{7}(7^{x})=log_{7}(77)x=log_{7}(77)x=2.2322$

so, rounding the obtained result to two decimal places (nearest hundreth), we have:

$x=2.23$

hence, we have that the correct option:

b. $x=2.23$

have a nice day!

Answer from: Quest

The answer should be 5.

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