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The claim is a mean is 212 and you want to prove it is more. Test the hypothesis within a 10% level of significance.Your sample of 59 had a mean of 214.12 and standard deviation of 63.6. Calculate the Critical Decision Rule.

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Final answer:

The critical z-score for a right-tailed test at the 10% level of significance is 1.28. After calculating the test statistic (0.256), it is determined to be less than the critical z-score, meaning that the null hypothesis cannot be rejected, and there isn't enough evidence to prove the mean is greater than 212.

Step-by-step explanation:

To test the hypothesis that the mean is greater than 212, we will perform a one-tailed test. The null hypothesis (℃₀) states that the population mean (μ) is equal to 212, while the alternative hypothesis (℃₁) posits that the mean is greater than 212.

We begin by identifying the level of significance (α), which is 10%, or 0.10. Because this is a right-tailed test, we will find the critical z-score that corresponds to the 90th percentile of the standard normal distribution.

Using statistical tables or software, the critical z-score corresponding to an α of 0.10 in a right-tailed test is approximately 1.28. This means that if our test statistic calculates to a z-score greater than 1.28, we will reject the null hypothesis.

Next, we calculate the test statistic using the formula:

z = (X - μ) / (σ / sqrt(n))

where X is the sample mean, μ is the population mean under the null hypothesis, σ is the sample standard deviation, and n is the sample size.

Substituting the given values:

z = (214.12 - 212) / (63.6 / sqrt(59))

z = 2.12 / (63.6 / 7.685)

z = 2.12 / 8.278

z = 0.256

Since 0.256 is less than the critical z-score of 1.28, we do not reject the null hypothesis. Therefore, we do not have enough evidence to claim that the mean is more than 212 at the 10% level of significance.

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