Math / Data & Statistics

QuestionIn a random sample of 16 residents of the state of Washington, the mean waste recycled per person per day was 2.8 pounds with a standard deviation of 0.24 pounds. Determine the $80 \%$ confidence interval for the mean waste recycled per person per day for the population of Washington. Assume the population is approximately normal.
Step 2 of 2: Construct the $80 \%$ confidence interval. Round your answer to one decimal place.
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Lower endpoint: $\square$ Upper endpoint: $\square$

Studdy Solution

STEP 1

1. The sample size $n = 16$ .
2. The sample mean $\bar{x} = 2.8$ pounds.
3. The sample standard deviation $s = 0.24$ pounds.
4. The population is approximately normal.
5. We are constructing an $80\%$ confidence interval.

STEP 2

1. Determine the critical value for the $80\%$ confidence level.
2. Construct the confidence interval using the critical value.

STEP 3

Since the sample size is small ( $n < 30$ ), we use the t-distribution. The degrees of freedom $df = n - 1 = 15$ .

STEP 4

For an $80\%$ confidence interval, the level of significance $\alpha = 1 - 0.80 = 0.20$ . The critical value $t_{\alpha/2}$ corresponds to $\alpha/2 = 0.10$ in each tail. Using a t-table or calculator, find $t_{0.10, 15}$ .

STEP 5

Calculate the standard error of the mean (SEM):
$SEM = \frac{s}{\sqrt{n}} = \frac{0.24}{\sqrt{16}} = 0.06$

STEP 6

Calculate the margin of error (ME):
$ME = t_{\alpha/2} \times SEM$
Assuming $t_{0.10, 15} \approx 1.341$ (value from t-table),
$ME = 1.341 \times 0.06 \approx 0.08046$

STEP 7

Construct the confidence interval:
$\text{Lower endpoint} = \bar{x} - ME = 2.8 - 0.08046 \approx 2.7$
$\text{Upper endpoint} = \bar{x} + ME = 2.8 + 0.08046 \approx 2.9$
Round to one decimal place:
$\text{Lower endpoint} = 2.7, \quad \text{Upper endpoint} = 2.9$
The $80\%$ confidence interval for the mean waste recycled per person per day is:
$(2.7, 2.9)$

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Studdy Solution

STEP 1

1. The sample size $n = 16$ .
2. The sample mean $\bar{x} = 2.8$ pounds.
3. The sample standard deviation $s = 0.24$ pounds.
4. The population is approximately normal.
5. We are constructing an $80\%$ confidence interval.

STEP 2

1. Determine the critical value for the $80\%$ confidence level.
2. Construct the confidence interval using the critical value.

STEP 3

Since the sample size is small ( $n < 30$ ), we use the t-distribution. The degrees of freedom $df = n - 1 = 15$ .

STEP 4

For an $80\%$ confidence interval, the level of significance $\alpha = 1 - 0.80 = 0.20$ . The critical value $t_{\alpha/2}$ corresponds to $\alpha/2 = 0.10$ in each tail. Using a t-table or calculator, find $t_{0.10, 15}$ .

STEP 5

Calculate the standard error of the mean (SEM):
$SEM = \frac{s}{\sqrt{n}} = \frac{0.24}{\sqrt{16}} = 0.06$

STEP 6

Calculate the margin of error (ME):
$ME = t_{\alpha/2} \times SEM$
Assuming $t_{0.10, 15} \approx 1.341$ (value from t-table),
$ME = 1.341 \times 0.06 \approx 0.08046$

STEP 7

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Studdy Solution

STEP 1

1. The sample size n=16 n = 16 n=16.2. The sample mean xˉ=2.8 \bar{x} = 2.8 xˉ=2.8 pounds.3. The sample standard deviation s=0.24 s = 0.24 s=0.24 pounds.4. The population is approximately normal.5. We are constructing an 80% 80\% 80% confidence interval.

STEP 2

1. Determine the critical value for the 80% 80\% 80% confidence level.2. Construct the confidence interval using the critical value.

STEP 3

Since the sample size is small (n<30 n < 30 n<30), we use the t-distribution. The degrees of freedom df=n−1=15 df = n - 1 = 15 df=n−1=15.

STEP 4

STEP 5

Calculate the standard error of the mean (SEM):SEM=sn=0.2416=0.06SEM = \frac{s}{\sqrt{n}} = \frac{0.24}{\sqrt{16}} = 0.06SEM=n​s​=16​0.24​=0.06

STEP 6

Calculate the margin of error (ME):ME=tα/2×SEMME = t_{\alpha/2} \times SEMME=tα/2​×SEMAssuming t0.10,15≈1.341 t_{0.10, 15} \approx 1.341 t0.10,15​≈1.341 (value from t-table),ME=1.341×0.06≈0.08046ME = 1.341 \times 0.06 \approx 0.08046ME=1.341×0.06≈0.08046

STEP 7

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1. The sample size $n = 16$ .
2. The sample mean $\bar{x} = 2.8$ pounds.
3. The sample standard deviation $s = 0.24$ pounds.
4. The population is approximately normal.
5. We are constructing an $80\%$ confidence interval.

1. Determine the critical value for the $80\%$ confidence level.
2. Construct the confidence interval using the critical value.

Since the sample size is small ( $n < 30$ ), we use the t-distribution. The degrees of freedom $df = n - 1 = 15$ .

Calculate the standard error of the mean (SEM):
$SEM = \frac{s}{\sqrt{n}} = \frac{0.24}{\sqrt{16}} = 0.06$

Calculate the margin of error (ME):
$ME = t_{\alpha/2} \times SEM$
Assuming $t_{0.10, 15} \approx 1.341$ (value from t-table),
$ME = 1.341 \times 0.06 \approx 0.08046$