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3 years ago

15

can someone help me?, I had the notes in my notebook but the notebook got full, so I threw it away not thinking I was gonna need

these notes again ;(

1 answer:

Natalija [7]3 years ago

5 0

Step-by-step explanation:

hope this helps you

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18 = (240/100) x what;
what = 18 x (100/240);
what = 1800/240;
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An automobile company wants to determine the average amount of time it takes a machine to assemble a car. A sample of 40 times y

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A 98% confidence interval for the mean assembly time is [21.34, 26.49] .

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We are given that a sample of 40 times yielded an average time of 23.92 minutes, with a sample standard deviation of 6.72 minutes.

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 average time = 23.92 minutes

s = sample standard deviation = 6.72 minutes

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$\mu$ = population mean assembly time

<em> Here for constructing a 98% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

<u>So, a 98% confidence interval for the population mean, </u> $\mu$ <u> is; </u>

P(-2.426 < $t_3_9$ < 2.426) = 0.98 {As the critical value of z at 1% level

of significance are -2.426 & 2.426}

P(-2.426 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.426) = 0.98

P( $-2.426 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

<u>98% confidence interval for</u> $\mu$ = [ $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $23.92-2.426 \times {\frac{6.72}{\sqrt{40} } }$ , $23.92+2.426 \times {\frac{6.72}{\sqrt{40} } }$ ]

= [21.34, 26.49]

Therefore, a 98% confidence interval for the mean assembly time is [21.34, 26.49] .

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