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2 years ago

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A supervisor records the repair cost for 22 randomly selected VCRs. A sample mean of $75.50 and standard deviation of $18.07 are

subsequently computed. Determine the 99% confidence interval for the mean repair cost for the VCRs. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

1 answer:

VashaNatasha [74]2 years ago

8 0

Answer:

The t value for 99% CI for 21 df is 2.831.

The critical value that should be used in constructing the confidence interval is (64.593, 86.407).

Step-by-step explanation:

Now the sample size is less than 30 and also population standard deviation is not known.

Then we will use t distribution to find CI

t value for 99% CI for 21 df is TINV(0.01,21)=2.831

The margin of error is $E=t\times\frac{s}{\sqrt{n}}\\\\=2.831\times\frac{18.07}{\sqrt{22}}\\\\=10.907$

Hence CI is $CI=\overline{x} \pm E\\\\ =75.50 \pm 10.907\\\\=(64.593,86.407 )$

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3 years ago

what is the probability of seeing a sample mean for 21 observations less or equal to the sample mean that we observed

kramer

Answer:

$P(x \le 21) = 0.69146$

Step-by-step explanation:

The missing parameters are:

$n = 64$ --- population

$\mu = 20$ --- population mean

$\sigma = 16$ -- population standard deviation

Required

$P(x \le 21)$

First, calculate the sample standard deviation

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{16}{\sqrt {64}}$

$\sigma_x = \frac{16}{8}$

$\sigma_x = 2$

Next, calculate the sample mean $\bar_x$

$\bar x = \mu$

So:

$\bar x = 20$

So, we have:

$\sigma_x = 2$

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Calculate the z score

$x = \frac{x - \mu}{\sigma}$

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So, we have:

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