Question

1. we have iq test scores of 31 seventh - grade girls in a midwest school district. we have calculated that sample mean is 105.84 and the standard deviation is 14.27. give a 99% confidence interval for the average score in the population. (1 point)

97, 114
96, 115
98, 113
99, 112

Explanation:

Step1: Identify the formula

For a confidence interval when the population standard - deviation is unknown (we use the sample standard - deviation $s$ instead), the formula is $\bar{x}\pm t_{\alpha/2}\frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $s$ is the sample standard - deviation, $n$ is the sample size, and $t_{\alpha/2}$ is the critical value. Here, $\bar{x} = 105.84$, $s=14.27$, $n = 31$. The degrees of freedom $df=n - 1=31 - 1 = 30$. For a 99% confidence interval, $\alpha=1 - 0.99 = 0.01$, and $\alpha/2=0.005$.

Step2: Find the critical value

Looking up the $t$ - value in the $t$ - distribution table with $df = 30$ and $\alpha/2=0.005$, we get $t_{0.005,30}=2.750$.

Step3: Calculate the margin of error

The margin of error $E=t_{\alpha/2}\frac{s}{\sqrt{n}}=2.750\times\frac{14.27}{\sqrt{31}}\approx2.750\times\frac{14.27}{5.568}\approx2.750\times2.563\approx7.05$.

Step4: Calculate the confidence interval

The lower limit is $\bar{x}-E=105.84 - 7.05=98.79\approx98$. The upper limit is $\bar{x}+E=105.84 + 7.05=112.89\approx113$.

Answer:

98, 113