Question

The nonprofit group Public Agenda conducted telephone

interviews with three randomly selected groups of parents of high school children. There

were 202 black parents, 202 Hispanic parents, and 201 white parents. One question asked,

“Are the high schools in your state doing an excellent, good, fair, or poor job, or don’t you

know enough to say?” Here are the survey results:

Do these data provide convincing evidence that the distributions of opinion about high

schools differ for the three populations of parents?

Short answer

There's persuading prove that conveyances of opinion approximately tall schools contrast for the three populaces of guardians.

Step-by-step solution

Given information

We have been given the opinions of black, Hispanic, and white parents' opinions about high schools.

Explanation

a =significanceposition(supposition) The invalidenterprisecountriesthat there is nodiscrepancywithin thedispersionof the categorical variable for each of the populations/ treatments. Theoptionalpropositioncountriesthat there is adifference.

Ho Thedissipationofsuppositionis thesamefor eachrace.
Ha The dissipation of supposition is not the same for each race. The anticipated frequentness E are the item of the column and push add up to, separated by the table add up to.

$$\begin{gathered} E\_11=(r\_1\times c\_1)/n=(68\times 202)/605\approx 22.70 \\ E\_12=(r\_1\times c\_2)/n=(68\times 202)/605\approx 22.70 \\ E\_13=(r\_1\times c\_3)/n=(68\times 201)/605\approx 22.59 \\ E\_21=(r\_2\times c\_1)/n=(205\times 202)/605\approx 68.45 \\ E\_22=(r\_2\times c\_2)/n=(205\times 202)/605\approx 68.45 \\ E\_23=(r\_2\times c\_3)/n=(205\times 201)/605\approx 68.11 \\ E\_31=(r\_3\times c\_1)/n=(196\times 202)/605\approx 65.44 \\ E\_32=(r\_3\times c\_2)/n=(196\times 202)/605\approx 65.44 \\ E\_33=(r\_3\times c\_3)/n=(196\times 201)/605\approx 65.12 \\ E\_41=(r\_4\times c\_1)/n=(72\times 202)/605\approx 24.04 \\ E\_42=(r\_4\times c\_2)/n=(72\times 202)/605\approx 24.04 \\ E\_43=(r\_4\times c\_3)/n=(72\times 201)/605\approx 23.92 \\ E\_51=(r\_5\times c\_1)/n=(64\times 202)/605\approx 21.37 \\ E\_52=(r\_5\times c\_2)/n=(64\times 202)/605\approx 21.37 \\ E\_53=(r\_5\times c\_3)/n=(64\times 201)/605\approx 21.26) \end{gathered}$$

Conditions The conditions for performing a ki-square test of unity/ independence are Arbitrary, Autonomous, Huge counts Random Fulfilled, since the bunches of guardians were aimlessly named. Independent Fulfilled, since the individualities are lower than of all individualities( since there are further than people). Large checks Fulfilled, since all anticipated censuses are at fewest. Since all conditions are fulfilled, it's fitting to carry out a test of unity/ independence.

The value of the test- statistic is also the sum of the ki-square subtotals

$$\begin{gathered} \chi ^2\&=\sum \left(\right(O-E)^2)/E \\ =\left(\right(12-22.70)^2)/22.70+\left(\right(34-22.70)^2)/22.70+\left(\right(22-22.59)^2)/22.59 \\ +\left(\right(69-68.45)^2)/68.45+\left(\right(55-68.45)^2)/68.45+\left(\right(81-68.11)^2)/68.11 \\ +\left(\right(75-65.44)^2)/65.44+\left(\right(61-65.44)^2)/65.44+\left(\right(60-65.12)^2)/65.12+\left(\right(24-24.04)^2)/24.04+\left(\right(24-24.04)^2)/24.04+\left(\right(24-23.92)^2)/23.92 \\ +\left(\right(22-21.37)^2)/21.37+\left(\right(28-21.37)^2)/21.37+\left(\right(14-21.26)^2)/21.26 \\ \approx 22.4261 \end{gathered}$$

The- value is the liability of getting the regard of the test dimension, or a regard more extraordinary. The- value is the number( or interval) within the column title of the ki-square vehicle table within the reference section containing the value $X^2$in the row$df=(r-1)(c-1)=(5-1)(3-1)=8$

$0.0025<P<0.005$

Since the value of $P$is less than $0.05$

The thesis is rejected