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Use the sign test at a 0.05 level of significance to test the null hypothesis [tex]\mu = 0.55[/tex] against the alternative hypothesis [tex]\mu \neq 0.55[/tex].

Given a random sample of 15 measurements of the octane rating of a certain kind of gasoline: 99, 102.3, 99.8, 100.5, 99.7, 96.2, 99.1, 102.5, 103.3, 97.4, 100.4, 98.9, 98.3, 98, 101.6.

Answer :

To use the sign test to test the hypothesis, we're performing a non-parametric test suitable when data doesn't necessarily meet the assumptions required for parametric tests like the t-test. This can be relevant here, as we are analyzing whether the median octane rating differs from a specified value, which is suitable for sign tests.

Step-by-step Process:

  1. Objective and Hypotheses:

    • Null Hypothesis (Hâ‚€): The median of the gasoline octane ratings is 0.55.
    • Alternative Hypothesis (H₁): The median of the gasoline octane ratings is not 0.55.

  2. Data Collection:

    • Sample: 99, 102.3, 99.8, 100.5, 99.7, 96.2, 99.1, 102.5, 103.3, 97.4, 100.4, 98.9, 98.3, 98, 101.6.

  3. Calculate Differences:

    • Calculate which sample values are above or below 0.55.
    • Each value in the sample exceeds 0.55. Thus, we will count the 'positive' differences because all values are larger than 0.55, and zero values do not exist in this context.

  4. Sign Test Calculation:

    • Determine the number of positive and negative signs:
      • Positive Signs = 15 (since all observed values are greater than 0.55)
      • Negative Signs = 0
    • Total sample size, [tex]N = 15[/tex].

  5. Significance Level:

    • Use a significance level of 0.05.

  6. Decision Rule:

    • For a two-tailed test, if the number of positive or negative signs is extreme, relative to the binomial distribution for [tex]N[/tex] with success probability 0.5, reject the null hypothesis.

  7. Critical Values and Conclusion:

    • With [tex]N = 15[/tex], refer to a sign test table for a two-tailed test:
      • For [tex]\alpha/2 = 0.025[/tex], we check critical values on both extremes.
    • Since we have 15 positive signs, this is an extreme result.
    • The result is usually significant, and we reject [tex]Hâ‚€[/tex] at the 0.05 significance level.

Conclusion:

Given our data, we conclude there is significant evidence to reject the null hypothesis that the median octane rating is 0.55. The median is likely to be different from 0.55.

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Rewritten by : Jeany

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