A consumer products company wants to increase the absorption capacity of a sponge. Based on past data, the average sponge could absorb 3.5 o

Question

A consumer products company wants to increase the absorption capacity of a sponge. Based on past data, the average sponge could absorb 3.5 ounces. After the redesign, the absorption amounts of a sample of sponges were (in ounces): 4.1, 3.7, 3.3, 3.5, 3.8, 3.9, 3.6, 3.8, 4.0, and 3.9. At the 0.01 level of significance, what is the decision rule to test if the new design increased the absorption of the sponge?

a. Reject the null hypothesis if computed t is less than 2.580.
b. Reject the null hypothesis if computed t is greater than 2.821.
c. Reject the null hypothesis if computed z is 1.96 or larger.
d. Reject the null hypothesis if computed t is less than 2.764.

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Charlie 4 weeks 2021-11-08T11:37:04+00:00 1 Answer 0 views 0

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    2021-11-08T11:38:07+00:00

    Answer:

    b. Reject the null hypothesis if computed t is greater than 2.821.

    Step-by-step explanation:

    Hello!

    The consumer products company wants to increase the capacity of absorption of a sponge, so they redesigned the sponges and then took a sample of 10 new sponges to test if they absorb more than 3.5 ounces.

    Historically the average sponge absorbs 3.5 ounces.

    The hypotheses for this test are:

    H₀: μ ≤ 3.5

    H₁: μ > 3.5

    α: 0.01

    This hypothesis test is one-tailed to the right, this means that you have only one critical value and will reject the null hypothesis to large values of the statistic.

    There is no information about the variable distribution and the sample is too small to use CTL to approximate the sample mean to normal. So I’ve conducted a normality test using the sample data, using the same significance level as the exercise, with p-value 0.8476 the decision is to not reject the null hypothesis, so the variable X: Absorption capacity of the new sponge design has a normal distribution.

    The statistic to use considering the distribution of the variable, the sample size and the fact that the population variance is unknown is a Students t.

    t= \frac{(X[bar]-Mu)}{\frac{S}{\sqrt{n} } } ~~t_{n-1}

    The critical value is:

    t_{n-1: 1 - \alpha } = t_{9: 0.99}= 2.821

    Decision rule:

    If t_{H_0} ≥ 2.821, then you reject the null hypothesis.

    If t_{H_0} < 2.821, then you do not reject the null hypothesis.

    t_{H_0}= \frac{3.76-3.5}{\frac{0.24}{\sqrt{10} } }= 3.425

    The value of the statistic is greater than the critival value, so the decision is to reject the null hypothesis.

    I hope you have a SUPER day!

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