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for randomly selected adults, IQ scores are normally distributed with a standard deviation of 15. the scores of 14 randomly selected college students are listed below. use a 0.10 significance level to test the claim that the standard deviation of IQ scores of college students is less than 15. round the smaple standard deviation to three decimal places.

115, 128, 107, 109, 116, 124, 135
127, 115, 104, 118, 126, 129, 133

  • stat - ,

    Null and alternative hypotheses:
    Ho: The population standard deviation is = 15
    Ha: The population standard deviation is < 15

    Data given:
    sample standard deviation = ? (Note: standard deviation is the square root of the variance; therefore, variance is standard deviation squared).
    level of significance = 0.10
    n = 14

    Calculate the standard deviation from the data given.

    The test statistic used in making a decision is the chi-square.
    Here's the formula:

    chi-square = (sample size - 1)(sample variance)/(value specified in the null hypothesis)

    Note: Convert both standard deviations to variances when working with the formula.

    Degrees of freedom is equal to n - 1, which is 13.

    Checking a chi-square table using alpha = 0.10 with 13 degrees of freedom, find the critical value.

    If the test statistic exceeds the critical value from the chi-square table, then the null is rejected and you can accept the alternative hypothesis. If the test statistic does not exceed the critical value from the table, then the null cannot be rejected and you cannot conclude a difference.

    I hope this will help get you started.

  • stat - ,

    Test statistic: chi 2 = 5.571. Critical value: chi 2 = 7.042.

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