Ionizing radiation is being given increasing attention as a method for preserving horticultural products. A study reports that 153 of 180 ir

Question

Ionizing radiation is being given increasing attention as a method for preserving horticultural products. A study reports that 153 of 180 irradiated garlic bulbs were marketable (no external sprouting, rotting, or softening) 240 days after treatment, whereas only 117 of 180 untreated bulbs were marketable after this length of time.

Does this data suggest that ionizing radiation is beneficial as far as marketability is concerned? Use a=.05.

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Nevaeh 2 weeks 2021-11-22T05:27:01+00:00 1 Answer 0 views 0

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    2021-11-22T05:28:53+00:00

    Answer:

    Yes, there is  enough evidence to support the claim that ionizing radiation is beneficial in terms of marketability.

    Step-by-step explanation:

    This is a hypothesis test for the difference between proportions.

    The claim is that ionizing radiation is beneficial in terms of marketability.

    Then, the null and alternative hypothesis are:

    H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2> 0

    Being π1: the true proportion of treated bulbs that can be comercialized after 240 days, and π2: the true proportion of untreated bulbs that can be comercialized after 240 days.

    The significance level is 0.05.

    The sample 1, of size n1=180 has a proportion of p1=0.85.

    p_1=X_1/n_1=153/180=0.85

    The sample 2, of size n2=180 has a proportion of p2=0.65.

    p_2=X_2/n_2=117/180=0.65

    The difference between proportions is (p1-p2)=0.2.

    p_d=p_1-p_2=0.85-0.65=0.2

    The pooled proportion, needed to calculate the standard error, is:

    p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{153+117}{180+180}=\dfrac{270}{360}=0.75

    The estimated standard error of the difference between means is computed using the formula:

    s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.75*0.25}{180}+\dfrac{0.75*0.25}{180}}\\\\\\s_{M_d}=\sqrt{0.001+0.001}=\sqrt{0.002}=0.0456

    Then, we can calculate the z-statistic as:

    z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.2-0}{0.0456}=\dfrac{0.2}{0.0456}=4.382

    This test is a right-tailed test, so the P-value for this test is calculated as (using a z-table):

    P-value=P(t>4.382)=0.000008

    As the P-value (0.000008) is smaller than the significance level (0.05), the effect is significant.

    The null hypothesis is rejected.

    There is  enough evidence to support the claim that ionizing radiation is beneficial in terms of marketability.

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