Question

you can calculate the p-value for a chi-square test using technology. after calculating the cumulative distribution function (cdf) to calculate the area under the curve. use t claim.
a hospital spokesperson claims that the standard deviation of the waiting times exper emergency department is no more than 0.5 minutes. a random sample of 20 waiting ti 0.6 minutes. at α=0.05, is there enough evidence to reject the spokespersons claim?
identify the null and alternative hypotheses.
a. h₀: σ≤0.5 hₐ: σ>0.5
b. h₀: σ≥0.5 hₐ: σ<0.5
c. h₀: σ>0.5 hₐ: σ≤0.5
d. h₀: σ<0.5 hₐ: σ≥0.5
identify the standardized test statistic.
χ² = (round to two decimal places as needed.)
identify the p-value.
p = (round to four decimal places as needed.)

Explanation:

Step1: Identify Hypotheses

The claim is $\sigma \leq 0.5$. Null hypothesis ($H_0$) contains the claim or equality, alternative ($H_a$) is opposite. So $H_0: \sigma \leq 0.5$, $H_a: \sigma > 0.5$ (Option A).

Step2: Calculate Test Statistic

Formula for chi - square test statistic for standard deviation: $\chi^{2}=\frac{(n - 1)s^{2}}{\sigma_{0}^{2}}$. Here, $n = 20$, $s = 0.6$, $\sigma_{0}=0.5$.
Substitute values: $(n - 1)=19$, $s^{2}=0.36$, $\sigma_{0}^{2}=0.25$.
$\chi^{2}=\frac{19\times0.36}{0.25}=\frac{6.84}{0.25}=27.36$.

Step3: Calculate P - value

Degrees of freedom $df=n - 1 = 19$. We need $P(\chi^{2}>27.36)$ with $df = 19$. Using chi - square distribution table or calculator, $P\approx0.0803$.

Answer:

• Null and Alternative Hypotheses: A. $H_0:\sigma\leq0.5$, $H_a:\sigma > 0.5$
• Standardized Test Statistic: $\chi^{2}=\boxed{27.36}$
• P - value: $P=\boxed{0.0803}$