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A random sample of 59 second graders in a certain school district are given a standardized mathematics skills test. The sample mean score is x=56. Assume the standard deviation of test scores is o=15. The nationwide average score on this test is 50. The school superintendent wants to know whether the second graders in her school district have greater math skills than the nationwide average. Use the a=0.05 level of significance and the P-value method .

Hv0 : [?]
Hv1 : [?]

This is a left/right or two-tailed test

7 answers

To determine the hypotheses for this scenario, we'll first clarify what we are testing. The superintendent wants to know if the second graders in her school district have greater math skills than the nationwide average. This implies that we are testing whether the mean score in the school district is greater than the nationwide average score.

Step 1: Define the Hypotheses

  • Null Hypothesis (\(H_0\)): The mean score of second graders in the school district is equal to the nationwide average, which is 50. \[ H_0: \mu = 50 \]

  • Alternative Hypothesis (\(H_1\)): The mean score of second graders in the school district is greater than the nationwide average. \[ H_1: \mu > 50 \]

Step 2: Type of Test

This is a right-tailed test because we are testing if the mean score of the second graders is greater than the nationwide average.

Summary of the Test

  • \(H_0: \mu = 50\)
  • \(H_1: \mu > 50\)
  • Test Type: Right-tailed test

Additional Information

Since you want to use the P-value method and have the necessary sample data, you would calculate the test statistic using the sample mean, population mean, standard deviation, and sample size, then compare the P-value to your significance level (\(\alpha = 0.05\)) to make a decision. If the P-value is less than \(\alpha\), you would reject the null hypothesis in favor of the alternative hypothesis.

(b) Compute the value of the test statistic. Round the answer to two decimal places.

To compute the test statistic for the given scenario, we will use the formula for the z-test statistic, since the sample size is large (n = 59) and the population standard deviation (\( \sigma \)) is known.

Formula for the Z-Test Statistic

The formula for the z-test statistic is:

\[ z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \]

Where:

  • \( \bar{x} \) = sample mean
  • \( \mu_0 \) = population mean under the null hypothesis
  • \( \sigma \) = population standard deviation
  • \( n \) = sample size

Given Values

  • Sample mean (\( \bar{x} \)) = 56
  • Population mean (\( \mu_0 \)) = 50
  • Population standard deviation (\( \sigma \)) = 15
  • Sample size (\( n \)) = 59

Step 1: Calculate the Standard Error (SE)

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{59}} \]

Calculating \( \sqrt{59} \):

\[ \sqrt{59} \approx 7.68 \]

Thus,

\[ SE \approx \frac{15}{7.68} \approx 1.95 \]

Step 2: Calculate the Z-Test Statistic

Now substituting the values into the z-formula:

\[ z = \frac{56 - 50}{1.95} = \frac{6}{1.95} \approx 3.08 \]

Final Answer

The value of the test statistic (z) is approximately 3.08, rounded to two decimal places.

(c) Compute the p-value of the test statistic. Round the answer to four decimal places.

To compute the p-value for the test statistic \( z = 3.08 \) in a right-tailed test, we will use the standard normal distribution.

Step 1: Find the P-value

The p-value is the probability of observing a test statistic as extreme as \( z = 3.08 \) or more extreme in the direction of the alternative hypothesis. For a right-tailed test, we need to find \( P(Z > 3.08) \).

Using a standard normal distribution table or calculator, we first find \( P(Z < 3.08) \) and then use it to find the right-tail probability:

  1. Look up the z-value of \( 3.08 \) in the standard normal distribution table or use a calculator:

    • Using a z-table or calculator, we find: \[ P(Z < 3.08) \approx 0.9990 \]

  2. Now, calculate the right-tail probability (p-value): \[ P(Z > 3.08) = 1 - P(Z < 3.08) = 1 - 0.9990 = 0.0010 \]

Final Answer

Thus, the p-value for the test statistic \( z = 3.08 \) is approximately 0.0010, rounded to four decimal places.

Should we reject Hv0 use the a=0.05 level

To determine whether to reject the null hypothesis (\(H_0\)), we need to compare the p-value with the significance level \(\alpha = 0.05\).

Given:

  • P-value: 0.0010
  • Significance level (\(\alpha\)): 0.05

Decision Rule:

  • If the p-value \(\leq \alpha\), reject \(H_0\).
  • If the p-value \(> \alpha\), fail to reject \(H_0\).

Comparison:

  • Since \(0.0010 \leq 0.05\), we reject \(H_0\).

Conclusion:

There is sufficient evidence to conclude that the mean math score of second graders in the school district is greater than the nationwide average of 50 at the 0.05 significance level.