Question

you can calculate the p-value for a chi-square test using technology. after calculating the standardized test statistic, use the cumulative distribution function (cdf) to calculate the area under the curve. use the p-value method to test the claim.
a school administrator claims that the standard deviation for eighth-grade students on a test is greater than 35 points. a random sample of 27 eighth-grade students has a standard deviation of 36.9 points. at α = 0.01, is there enough evidence to support the administrator’s claim?
identify the null and alternative hypotheses.
a. h₀: σ ≥ 35 hₐ: σ < 35
b. h₀: σ ≤ 35 hₐ: σ > 35
c. h₀: σ < 35 hₐ: σ ≥ 35
d. h₀: σ > 35 hₐ: σ ≤ 35
identify the standardized test statistic.
χ² = (round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for chi - square test statistic for standard deviation

The formula for the chi - square test statistic \(\chi^{2}\) when testing a claim about the standard deviation \(\sigma\) is \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}\), where \(n\) is the sample size, \(s\) is the sample standard deviation, and \(\sigma\) is the hypothesized population standard deviation.

Step2: Identify the values

We are given that \(n = 27\), \(s=36.9\), and \(\sigma = 35\).

Step3: Substitute the values into the formula

First, calculate \((n - 1)\): \(n-1=27 - 1=26\).
Then, calculate \(s^{2}\): \(s^{2}=(36.9)^{2}=36.9\times36.9 = 1361.61\).
And \(\sigma^{2}=(35)^{2}=1225\).

Now, substitute these values into the formula:
\(\chi^{2}=\frac{26\times1361.61}{1225}\)
First, calculate the numerator: \(26\times1361.61 = 26\times1361.61=35401.86\)
Then, divide by the denominator: \(\chi^{2}=\frac{35401.86}{1225}\approx28.223\)

Answer:

28.223