Question

es The table below is based on data from a poll. Use a 0.05 significance level to test the claim that gender and left-handedn
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Answer #1

Solution:

Given:

1590166793857_image.png

Expected values can be found :

Expected Values Left-handed Not Left-handed Total Women 235 x 293 402 171.281 167 x 293 402 121.719 293 Men 235x 109 = 63.719

We have to calculate the term,

\frac{(E-O)^{2}}{E}

Therefore,

Squared Distances Left-handed Not Left-handed Women (178-171.281 171.281 0.264 (115-121.719) 121.719 -0.371 Men (57-63.719) =

The following null and alternative hypotheses need to be tested:

H0​: The two variables are independent

Ha​: The two variables are dependent

This corresponds to a Chi-Square test of independence.

Based on the information provided, the significance level is α=0.05 , the number of degrees of freedom is df=(2−1)×(2−1)=1, so then the rejection region for this test is R={χ2:χ2>3.841}.

The Chi-Squared statistic is computed as follows:

1590166984575_image.png

Since it is observed that χ2=2.34 ≤ χc2​=3.841, it is then concluded that the null hypothesis is not rejected.

Conclusion:

No, there is not sufficient evidence to warrant rejection of the claim of independence between gender and left-handedness (optionD)

Done

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