Many elementary school students in a school district currently have ear infections. A random sample of children in two different schools fou

Question

Many elementary school students in a school district currently have ear infections. A random sample of children in two different schools found that 16 of 40 at one school and 13 of 30 at the other had this infection. Conduct a test to answer if there is sufficient evidence to conclude that a difference exists between the proportion of students who have ear infections at one school and the other. Find the test statistic. [Suggestion: try to use a TI 83 or a similar calculator.]

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Gerda 7 months 2021-07-28T03:01:59+00:00 1 Answers 10 views 0

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    2021-07-28T03:03:58+00:00

    Answer:

    The test statistic is z = -0.28

    Step-by-step explanation:

    First, before finding the test statistic, we need to understand the central limit theorem and difference between normal variables.

    Central Limit Theorem

    The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

    For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

    For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

    Subtraction between normal variables:

    When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

    A random sample of children in two different schools found that 16 of 40 at one school

    This means that:

    p_1 = \frac{16}{40} = 0.4, s_1 = \sqrt{\frac{0.4*0.6}{40}} = 0.0775

    13 of 30 at the other had this infection.

    This means that:

    p_2 = \frac{13}{30} = 0.4333, s_2 = \sqrt{\frac{0.4333*0.5667}{30}} = 0.0905

    Conduct a test to answer if there is sufficient evidence to conclude that a difference exists between the proportion of students who have ear infections at one school and the other.

    At the null hypothesis, we test if there is no difference, that is:

    H_0: p_1 - p_2 = 0

    And at the alternate hypothesis, we test if there is difference, that is:

    H_a: p_1 - p_2 \neq 0

    The test statistic is:

    z = \frac{X - \mu}{s}

    In which X is the sample mean, \mu is the value tested at the null hypothesis and s is the standard error

    0 is tested at the null hypothesis:

    This means that \mu = 0

    From the two samples:

    p = p_1 - p_2 = 0.4 - 0.4333 = -0.0333

    s = \sqrt{s_1^2+s_2^2} = \sqrt{0.0775^2+0.0905^2} = 0.1191

    Value of the test statistic:

    z = \frac{X - \mu}{s}

    z = \frac{-0.0333 - 0}{0.1191}

    z = -0.28

    The test statistic is z = -0.28

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