Math  /  Data & Statistics

QuestionThe Health Department states the mean pregnancy term is 8.60 months. You conduct a study of 35 women in the Reedley area and find the mean pregnancy term is 8.68 months and a sample standard deviation of 0.2 months. Using a .01 level of significance, can you conclude that the mean pregnancy term in Reedley is different than 8.6 months?
Be sure to include all of the sections listed below. Use the picture provided to choose how your curve should look and be labeled.
State H0\mathrm{H}_{0} and H1\mathrm{H}_{1}. Use the picture and tell me the letter of the curve that should be used. State your critical value(s). Find the test statistic.

Studdy Solution

STEP 1

1. The sample size is n=35 n = 35 .
2. The sample mean is xˉ=8.68 \bar{x} = 8.68 months.
3. The sample standard deviation is s=0.2 s = 0.2 months.
4. The population mean under the null hypothesis is μ0=8.60 \mu_0 = 8.60 months.
5. The significance level is α=0.01 \alpha = 0.01 .
6. We are conducting a two-tailed test.

STEP 2

1. State the null and alternative hypotheses.
2. Determine the appropriate curve and critical values.
3. Calculate the test statistic.
4. Make a decision based on the test statistic and critical values.

STEP 3

State the null and alternative hypotheses:
\begin{align*} \mathrm{H}_0 &: \mu = 8.60 \\ \mathrm{H}_1 &: \mu \neq 8.60 \end{align*}

STEP 4

Since this is a two-tailed test with a significance level of α=0.01 \alpha = 0.01 , we will use a standard normal distribution curve (Z-distribution). The critical values for a two-tailed test at α=0.01 \alpha = 0.01 are typically ±2.576 \pm 2.576 .
The curve that should be used is labeled with the critical values at ±2.576 \pm 2.576 .

STEP 5

Calculate the test statistic using the formula:
z=xˉμ0snz = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}
Substitute the given values:
z=8.688.600.235z = \frac{8.68 - 8.60}{\frac{0.2}{\sqrt{35}}}
Calculate the standard error:
Standard Error=0.2350.0338\text{Standard Error} = \frac{0.2}{\sqrt{35}} \approx 0.0338
Calculate the test statistic:
z0.080.03382.366z \approx \frac{0.08}{0.0338} \approx 2.366

STEP 6

Compare the test statistic to the critical values:
The test statistic z2.366 z \approx 2.366 does not exceed the critical values ±2.576 \pm 2.576 .
Since 2.366 2.366 is within the range 2.576-2.576 to 2.5762.576, we fail to reject the null hypothesis.
Conclusion: At the α=0.01 \alpha = 0.01 significance level, we do not have enough evidence to conclude that the mean pregnancy term in Reedley is different from 8.60 months.

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