Question

a statistics teacher claims that, on average, 20% of students in his class get an a, 25% get a b, 35% get a c, 10% get a d and the rest get an f. the grades of a random sample of his students over the years is recorded. test the claim at 5% significance. round to 4 decimals when appropriate. test statistic: degrees of freedom: p - value: decision rule: select an answer did something significant happen? select an answer there select an answer enough evidence to conclude select an answer

Explanation:

Step1: Calculate the total observed frequency

$$ 49 + 39+88 + 10+59=245 $$

Step2: Calculate the expected frequencies

• For \(A\): \(E_A=0.2\times245 = 49\)
• For \(B\): \(E_B=0.25\times245=61.25\)
• For \(C\): \(E_C = 0.35\times245=85.75\)
• For \(D\): \(E_D=0.1\times245 = 24.5\)
• For \(F\): \(E_F=(1-(0.2 + 0.25+0.35 + 0.1))\times245=0.1\times245=24.5\)

Step3: Calculate the test - statistic (\(\chi^2\))

$$ \chi^2=\sum\frac{(O - E)^2}{E} $$

$$
LATEXBLOCK0
$$

Step4: Calculate the degrees of freedom

The formula for degrees of freedom in a goodness - of - fit test is \(df = k - 1\), where \(k\) is the number of categories. Here \(k = 5\), so \(df=5 - 1=4\)

Step5: Calculate the p - value

Using a \(\chi^2\) distribution table or a calculator (\(\chi^2\) calculator), for \(\chi^2 = 65.3047\) and \(df = 4\), the p - value is approximately \(P(\chi^2>65.3047)\approx0\)

Step6: Decision Rule

Since the significance level \(\alpha = 0.05\) and \(p - value\approx0<0.05\), we reject the null hypothesis.

Answer:

• Test Statistic: \(65.3047\)
• Degrees of Freedom: \(4\)
• p - value: \(0.0000\)
• Decision Rule: Reject the null hypothesis
• Did something significant happen: Yes
• There is enough evidence to conclude that the teacher's claim about the grade distribution is not correct.