&nbsp;Brand of Smartphone&nbsp;&nbsp;Male&nbsp;A43B34C48A. What is the value for the degrees of freedom?B. If the expected number of males with smartphone brand B is 32.5, what is the chi-square component?C. If the test statistic (chi-square) = 6.40, then the p-value =D. If the significance level was 0.05, should the researcher reject or fail to reject the null?

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&amp;nbsp;Brand of Smartphone&amp;nbsp;&amp;nbsp;Male&amp;nbsp;A43B34C48A. What is the value for the degrees of freedom?B. If the expected number of males with smartphone brand B is 32.5, what is the chi-square component?C. If the test statistic (chi-square) = 6.40, then the p-value =D. If the significance level was 0.05, should the researcher reject or fail to reject the null?

husudiwareh · Accepted Answer

Chi-Square Test: In statistical hypothesis test, a chi-square test, or specifically Pearson&#039;s chi-square test is performed when the test statistic is distributed as chi-square under the null hypothesis. This test is used to determine whether there is a significant difference between the observed and the expected frequencies in one or more categories of a contingency table.In the standard applications of this test, the observations are classified into mutually exclusive classes. If the null hypothesis is true i.e. there are no significant differences between the classes in the population, the test statistic which is computed from the observations follows a chi-square frequency distribution. The purpose of the test is to determine how likely the observed frequencies would be assuming the null hypothesis is true.Solution AWe have, for a chi-square test, the degrees of freedom=number of classes - 1Therefore, here the value of degrees of freedom=3−1=2(Answer)Solution BThe chi-square component of the ith class is given by,(Oi−Ei)2Eiwhere Oi is the observed frequency of the ith classand Ei is the expected frequency of the ith classHere, for the smartphone brand B,Oi=34,Ei=32.5Therefore, the chi-square component for the smartphone brand B is(Oi−Ei)2Ei=(34−32.5)232.5=0.0692(rounded to 4 decimal places)(Answer)Solution CFrom the chi-square value, we calculate p-value for this test using MS-Excel,p-value=CHISQ.DIST(chi-square value,n-1,TRUE) Here, the chi-square value is 6.40.Therefore,p-value=CHISQ.DIST(chi-square value, n-1, TRUE)=CHISQ.DIST(6.40,2,TRUE)=0.9592(rounded to 4 decimal places)(Answer)Solution DThe null hypothesis is rejected ifp&amp;lt;αand accept otherwise,where α is the level of significance.Here,α=0.05Clearly,p&amp;gt;αTherefore, the researcher should accept the null hypothesis.(Answer)

\(\begin {array}{|c|c|} \hline \text { Brand of Smartphone }&\text{ Male }\\ \hline A&43\\ \hline B&

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