Questions: A data set includes weights (in grams) of 36 Reese's Peanut Butter Cup Miniatures. The accompanying Statdisk display shows results from using all 36 weights to test the claim that the sample is from a population with a mean equal to 8.953 g. Test the given claim by using the display provided from Statdisk. Use a 0.01 significance level. Identify the null and alternative hypotheses.

A data set includes weights (in grams) of 36 Reese's Peanut Butter Cup Miniatures. The accompanying Statdisk display shows results from using all 36 weights to test the claim that the sample is from a population with a mean equal to 8.953 g. Test the given claim by using the display provided from Statdisk. Use a 0.01 significance level.

Identify the null and alternative hypotheses.
Transcript text: A data set includes weights (in grams) of 36 Reese's Peanut Butter Cup Miniatures. The accompanying Statdisk display shows results from using all 36 weights to test the claim that the sample is from a population with a mean equal to 8.953 g . Test the given claim by using the display provided from Statdisk. Use a 0.01 significance level. Identify the null and alternative hypotheses. $\mathrm{H}_{0}$ : $\square$ $\square$ $\square$ $\mathrm{H}_{1}$ : $\square$ $\square$ $\square$
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Solution

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Solution Steps

Step 1: Define Hypotheses

We are testing the claim that the sample is from a population with a mean equal to μ0=8.953g \mu_0 = 8.953 \, g . The null and alternative hypotheses are defined as follows:

H0:μ=8.953 \mathrm{H}_{0}: \mu = 8.953 H1:μ8.953 \mathrm{H}_{1}: \mu \neq 8.953

Step 2: Calculate Standard Error

The standard error SE SE is calculated using the formula:

SE=σn=0.13236=0.022 SE = \frac{\sigma}{\sqrt{n}} = \frac{0.132}{\sqrt{36}} = 0.022

Step 3: Calculate Test Statistic

The test statistic Ztest Z_{\text{test}} is calculated using the formula:

Ztest=xˉμ0SE=8.9418.9530.022=0.5455 Z_{\text{test}} = \frac{\bar{x} - \mu_0}{SE} = \frac{8.941 - 8.953}{0.022} = -0.5455

Step 4: Calculate P-value

For a two-tailed test, the p-value P P is calculated as:

P=2×(1T(z))=0.5854 P = 2 \times (1 - T(|z|)) = 0.5854

Step 5: Decision Rule

We compare the p-value to the significance level α=0.01 \alpha = 0.01 :

  • If P<α P < \alpha , we reject the null hypothesis.
  • If Pα P \geq \alpha , we fail to reject the null hypothesis.

In this case:

0.58540.01 0.5854 \geq 0.01

Step 6: Conclusion

Since the p-value is greater than the significance level, we fail to reject the null hypothesis.

Final Answer

The conclusion is that there is not enough evidence to reject the claim that the sample is from a population with a mean equal to 8.953g 8.953 \, g .

Fail to reject H0\boxed{\text{Fail to reject } H_0}

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