Representative sample? For a class project, a group of statistics students is required to take an SRS of students from their large high school to take part in a survey. The students’ sample consists of $54$freshmen, $66$sophomores, $56$juniors, and $30$seniors. The school roster shows that $29\%$of the students enrolled at the school are freshmen, $27\%$are sophomores, $25\%$ are juniors, and $19\%$ are seniors.

(a) Construct a well-labeled bar graph that shows the distribution of grade levels (in percents) for the sample data. Do these data give you any reason to suspect that the statistics students’ sample is unusual? Explain. (b) Use an appropriate test to determine whether the sample data differ significantly from the actual distribution of students by grade level at the school.

Percentage of freshman $=29\%$.

Percentage of sophomores $=27\%$.

Percentage of juniors $=25\%$.

Percentage of seniors $=19\%$.

Number of freshmen in a sample $=54$.

Number of sophomores in a sample $=66$.

Number of juniors in a sample $=56$.

Number of seniors in a sample $=30$.

Total number of students in a sample $=54+66+56+30$

$=206$.

Finding percentages:

Freshman: $\frac{54}{206}\times 100=26.21\%$.

Sophomores: $\frac{66}{206}\times 100=32.05\%$.

Juniors: $\frac{56}{206}\times 100=27.18\%$.

Seniors: $\frac{30}{206}\times 100=14.56\%$.

Graph:

For sample data

Interpretation:

The preceding graph shows that the majority of students are in the sophomores and then juniors groups, however the claimed statistic indicates that the majority of students are in the freshman group.

The school roster shows that $29\%$ of the students enrolled at the school are freshmen, $27\%$ are sophomores, $25\%$ are juniors, and $19\%$ are seniors.

The expected count will be

The observed (O) and expected (E) counts are

The null and alternate hypotheses are

$H_0:p_1=0.29,$

$p_2=0.27,$

$p_3=0.25,$

$p_4=0.19$

$H_1$: At least one of the proportions is not equal.

Test statistic

$\frac{(54-59.74{)}^2}{59.74}+\frac{(66-55.62{)}^2}{55.62}+\frac{(56-51.5{)}^2}{51.5}+\frac{(30-39.14{)}^2}{39.14}~{\chi }_3$

The test value is $5.016$.

P value

$P\left({\chi }_3>5.016\right)=0.170629$

Conclusion:

Because the p value is bigger than the significance level, there is insufficient evidence to reject the null hypothesis at the $5\%$level of significance, leading to the conclusion that the actual and sample distributions are identical, and the sample data represents the real data.