Math / Data & Statistics

Question $v 1\rangle=2$ 3 4 5 6 7 8 9 10 11 12 Espafic
Let's go to the movies: A random sample of 44 Hollywood movies made in the last 10 years had a mean length of 131.1 minutes, with a standard deviation of 12.7 minutes.
Part: $0 / 2$ $\square$
Part 1 of 2 (a) Construct a $99 \%$ confidence interval for the true mean length of all Hollywood movies in the last 10 years. Round the answers to at least one decimal place.
A $99 \%$ confidence interval for the true mean length of all Hollywood movies made in the last 10 years is $\square$ $<\mu<$ $\square$ .

Studdy Solution

STEP 1

1. The sample size is $n = 44$ .
2. The sample mean is $\bar{x} = 131.1$ minutes.
3. The sample standard deviation is $s = 12.7$ minutes.
4. We are constructing a $99\%$ confidence interval for the population mean.
5. The sample size is large enough to use the t-distribution for the confidence interval.

STEP 2

1. Determine the critical value for a $99\%$ confidence interval.
2. Calculate the standard error of the mean.
3. Construct the confidence interval using the sample mean, critical value, and standard error.

STEP 3

Determine the degrees of freedom, which is $n - 1 = 44 - 1 = 43$ .

STEP 4

Find the critical value $t^$ for a $99\%$ confidence interval with 43 degrees of freedom. Using a t-distribution table or calculator, $t^ \approx 2.697$ .

STEP 5

Calculate the standard error of the mean (SEM):
$\text{SEM} = \frac{s}{\sqrt{n}} = \frac{12.7}{\sqrt{44}} \approx 1.913$

STEP 6

Calculate the margin of error (ME):
$\text{ME} = t^* \times \text{SEM} = 2.697 \times 1.913 \approx 5.157$

STEP 7

Construct the confidence interval:
$\bar{x} - \text{ME} < \mu < \bar{x} + \text{ME}$
$131.1 - 5.157 < \mu < 131.1 + 5.157$
$125.9 < \mu < 136.3$
The $99\%$ confidence interval for the true mean length of all Hollywood movies made in the last 10 years is:
$\boxed{125.9} < \mu < \boxed{136.3}$

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Studdy Solution

STEP 1

STEP 2

1. Determine the critical value for a $99\%$ confidence interval.
2. Calculate the standard error of the mean.
3. Construct the confidence interval using the sample mean, critical value, and standard error.

STEP 3

Determine the degrees of freedom, which is $n - 1 = 44 - 1 = 43$ .

STEP 4

Find the critical value $t^$ for a $99\%$ confidence interval with 43 degrees of freedom. Using a t-distribution table or calculator, $t^ \approx 2.697$ .

STEP 5

Calculate the standard error of the mean (SEM):
$\text{SEM} = \frac{s}{\sqrt{n}} = \frac{12.7}{\sqrt{44}} \approx 1.913$

STEP 6

Calculate the margin of error (ME):
$\text{ME} = t^* \times \text{SEM} = 2.697 \times 1.913 \approx 5.157$

STEP 7

Construct the confidence interval:
$\bar{x} - \text{ME} < \mu < \bar{x} + \text{ME}$
$131.1 - 5.157 < \mu < 131.1 + 5.157$
$125.9 < \mu < 136.3$
The $99\%$ confidence interval for the true mean length of all Hollywood movies made in the last 10 years is:
$\boxed{125.9} < \mu < \boxed{136.3}$

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Studdy Solution

STEP 1

STEP 2

1. Determine the critical value for a 99% 99\% 99% confidence interval.2. Calculate the standard error of the mean.3. Construct the confidence interval using the sample mean, critical value, and standard error.

STEP 3

Determine the degrees of freedom, which is n−1=44−1=43 n - 1 = 44 - 1 = 43 n−1=44−1=43.

STEP 4

Find the critical value t∗ t^* t∗ for a 99% 99\% 99% confidence interval with 43 degrees of freedom. Using a t-distribution table or calculator, t∗≈2.697 t^* \approx 2.697 t∗≈2.697.

STEP 5

Calculate the standard error of the mean (SEM):SEM=sn=12.744≈1.913 \text{SEM} = \frac{s}{\sqrt{n}} = \frac{12.7}{\sqrt{44}} \approx 1.913 SEM=n​s​=44​12.7​≈1.913

STEP 6

Calculate the margin of error (ME):ME=t∗×SEM=2.697×1.913≈5.157 \text{ME} = t^* \times \text{SEM} = 2.697 \times 1.913 \approx 5.157 ME=t∗×SEM=2.697×1.913≈5.157

STEP 7

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1. Determine the critical value for a $99\%$ confidence interval.
2. Calculate the standard error of the mean.
3. Construct the confidence interval using the sample mean, critical value, and standard error.

Determine the degrees of freedom, which is $n - 1 = 44 - 1 = 43$ .

Find the critical value $t^$ for a $99\%$ confidence interval with 43 degrees of freedom. Using a t-distribution table or calculator, $t^ \approx 2.697$ .

Calculate the standard error of the mean (SEM):
$\text{SEM} = \frac{s}{\sqrt{n}} = \frac{12.7}{\sqrt{44}} \approx 1.913$

Calculate the margin of error (ME):
$\text{ME} = t^* \times \text{SEM} = 2.697 \times 1.913 \approx 5.157$