Last school year student body local university consisted of 35% freshmen 24% sophomores 26% juniors and 15% seniors. A sample of 300 Students taken from this year’s student body showed the following number of students in each classification. Freshmen 90 Sophomores 60 Juniors 90 Seniors 60 What is the expected frequency of seniors?A- 65B- 45 C - 35 D- 55

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Ouhamiptkg · Accepted Answer

A Chi-Square for goodness of fit test is a test used to assess whether the observed data can be claimed to reasonably fit the expected data. Sometimes, a Chi-Square test for goodness of fit is referred as a test for multinomial experiments, because there is a fixed number of N categories, and each of the outcomes of the experiment falls in exactly one of those categories. Then, based on sample information, the test uses a Chi-Square statistic to assess if the expected proportions for all categories reasonably fit the sample data. The main properties of a one sample Chi-Square test for goodness of fit are:1. The distribution of the test statistic is the Chi-Square distribution, with n-1 degrees of freedom, where n is the number of categories2. The Chi-Square distribution is one of the most important distributions in statistics, together with the normal distribution and the F-distribution3. The Chi-Square test of goodness of fit is right-tailedThe formula for a Chi-Square statistic isχ2=∑{ij=1}n(Oij−Eij)2EijOne of the most common uses for this test is to assess whether two categorical variables are significantly related or not.Usually, the Chi-Square test for independence is referred to as a 2-way cross-tabulation test. If we have a one-way crosstabulation, we should use a Chi-square test for goodness of fit.For ease of calculations, construct the following table:&amp;nbsp;COLLEGE GRADE LEVEL&amp;nbsp;&amp;nbsp;PERCENTAGE&amp;nbsp;&amp;nbsp;NUMBER OF STUDENTS&amp;nbsp;&amp;nbsp;FRESHMEN&amp;nbsp;35%105&amp;nbsp;SOPHOMORES&amp;nbsp;24%72&amp;nbsp;&amp;nbsp;JUNIORS&amp;nbsp;26%78&amp;nbsp;&amp;nbsp;SENIORS&amp;nbsp;15%45&amp;nbsp;&amp;nbsp;&amp;nbsp;TOTAL&amp;nbsp;300&amp;nbsp;Categories&amp;nbsp;&amp;nbsp;Observed&amp;nbsp;&amp;nbsp;Expected&amp;nbsp;&amp;nbsp;(fo-fe)^2/fe&amp;nbsp;&amp;nbsp;Category 1&amp;nbsp;90300⋆0.35=105(90−105)2/105=2.143&amp;nbsp;Category 2&amp;nbsp;60300⋆0.24=72(60−72)2/72=2&amp;nbsp;Category 3&amp;nbsp;90300⋆0.25=78(90−78)2/78=1.846&amp;nbsp;Category 4&amp;nbsp;60300⋆0.15=45(60−45)2/45=5&amp;nbsp;Sum =&amp;nbsp;30030010.989Thus, the expected number of senior is 45. Option (b) is correct.

Last school year student body local university consisted

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