Question

question 1/ (1 point)
a researcher is curious if age makes a difference in whether or not students make use of the gym at a university. he takes a random sample of 30 days and counts the number of upperclassmen (group 1) and underclassmen (group 2) that use the gym each day. the data are below. the population standard deviation for underclassmen is known to be 22.57 and the population standard deviation for upperclassmen is known to be 13.57.
upper classmen average = 202.4, population sd = 13.57, n = 30
under classmen average = 191.3, population sd = 22.57, n = 30
is there evidence to suggest that a difference exists in gym usage based on age? construct a confidence interval for the data above to decide. use α=0.10.
confidence interval (round to 4 decimal places):
_<μ1 - μ2<_

blank # 1

blank # 2

Explanation:

Step1: Identify the z - value

For a 90% confidence interval ($\alpha = 0.10$), the two - tailed z - value $z_{\alpha/2}=z_{0.05}$. From the standard normal distribution table, $z_{0.05} = 1.645$.

Step2: Calculate the standard error

The formula for the standard error of the difference between two means when the population standard deviations $\sigma_1$ and $\sigma_2$ are known is $SE=\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}$. Here, $\sigma_1 = 13.57$, $n_1 = 30$, $\sigma_2=22.57$, $n_2 = 30$.
\[

$$\begin{align*} SE&=\sqrt{\frac{13.57^{2}}{30}+\frac{22.57^{2}}{30}}\\ &=\sqrt{\frac{184.1449}{30}+\frac{509.4049}{30}}\\ &=\sqrt{\frac{184.1449 + 509.4049}{30}}\\ &=\sqrt{\frac{693.5498}{30}}\\ &=\sqrt{23.118327}\\ &\approx4.8081 \end{align*}$$

\]

Step3: Calculate the margin of error

The margin of error $E = z_{\alpha/2}\times SE$. Substituting $z_{\alpha/2}=1.645$ and $SE = 4.8081$, we get $E=1.645\times4.8081\approx7.9093$.

Step4: Calculate the confidence interval

The difference in sample means is $\bar{x}_1-\bar{x}_2=202.4 - 191.3=11.1$. The confidence interval for $\mu_1-\mu_2$ is $(\bar{x}_1-\bar{x}_2)-E<\mu_1 - \mu_2<(\bar{x}_1-\bar{x}_2)+E$.
\[

$$\begin{align*} 11.1-7.9093&<\mu_1-\mu_2<11.1 + 7.9093\\ 3.1907&<\mu_1-\mu_2<19.0093 \end{align*}$$

\]

Answer:

$3.1907<\mu_1-\mu_2<19.0093$