the following data give the one - way commuting times (in minutes) from home to work for a random sample of 30 workers.
23 17 34 26 18 33 46 42 13 37
45 14 22 20 28 32 18 39 40 48
25 36 23 39 42 46 29 17 24 31
round your answers to 2 decimal places.
a. calculate the value of the point estimate of the mean one - way commuting time from home to work for all workers.
907 i
b. construct a 99% confidence interval for the mean one - way commuting time from home to work for all workers.
hint
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Accepted Answer

# Explanation:
## Step1: Calculate the sample mean
First, sum up all the data values:
$$
\begin{align*}
&23 + 17+34 + 26+18+33+46+42+13+37+45+14+22+20+28+32+18+39+40+48+25+36+23+39+42+46+29+17+24+31\
=& 907
\end{align*}
$$
The sample size $n = 30$. The sample mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}=\frac{907}{30}\approx30.23$

## Step2: Calculate the sample standard deviation
The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$

First, calculate $(x_{i}-\bar{x})^{2}$ for each $x_{i}$:
$$
\begin{align*}
&(23 - 30.23)^{2}+(17-30.23)^{2}+(34 - 30.23)^{2}+(26-30.23)^{2}+(18-30.23)^{2}+(33-30.23)^{2}+(46-30.23)^{2}+(42-30.23)^{2}+(13-30.23)^{2}+(37-30.23)^{2}+(45-30.23)^{2}+(14-30.23)^{2}+(22-30.23)^{2}+(20-30.23)^{2}+(28-30.23)^{2}+(32-30.23)^{2}+(18-30.23)^{2}+(39-30.23)^{2}+(40-30.23)^{2}+(48-30.23)^{2}+(25-30.23)^{2}+(36-30.23)^{2}+(23-30.23)^{2}+(39-30.23)^{2}+(42-30.23)^{2}+(46-30.23)^{2}+(29-30.23)^{2}+(17-30.23)^{2}+(24-30.23)^{2}+(31-30.23)^{2}\
\end{align*}
$$
$$
\begin{align*}
\sum_{i = 1}^{30}(x_{i}-\bar{x})^{2}&=(- 7.23)^{2}+(-13.23)^{2}+(3.77)^{2}+(-4.23)^{2}+(-12.23)^{2}+(2.77)^{2}+(15.77)^{2}+(11.77)^{2}+(-17.23)^{2}+(6.77)^{2}+(14.77)^{2}+(-16.23)^{2}+(-8.23)^{2}+(-10.23)^{2}+(-2.23)^{2}+(1.77)^{2}+(-12.23)^{2}+(8.77)^{2}+(9.77)^{2}+(17.77)^{2}+(-5.23)^{2}+(5.77)^{2}+(-7.23)^{2}+(8.77)^{2}+(11.77)^{2}+(15.77)^{2}+(-1.23)^{2}+(-13.23)^{2}+(-6.23)^{2}+(0.77)^{2}\
&=52.2729+175.0329 + 14.1129+17.8929+149.5729+7.6729+248.7929+138.5329+296.8729+45.8329+218.1529+263.4129+67.7329+104.6529+4.9729+3.1329+149.5729+76.9129+95.4529+315.7729+27.3529+33.2929+52.2729+76.9129+138.5329+248.7929+1.5129+175.0329+38.8129+0.5929\
&=3193.7
\end{align*}
$$
$s=\sqrt{\frac{3193.7}{30 - 1}}=\sqrt{\frac{3193.7}{29}}\approx\sqrt{110.13}\approx10.50$

## Step3: Find the critical value
For a $99\%$ confidence interval and $n-1 = 29$ degrees of freedom, the critical value $t_{\alpha/2}$ from the $t -$distribution table. $\alpha=1 - 0.99=0.01$, $\alpha/2=0.005$. $t_{0.005,29}=2.756$

## Step4: Calculate the margin of error
The margin of error $E=t_{\alpha/2}\frac{s}{\sqrt{n}}=2.756\times\frac{10.50}{\sqrt{30}}$
$$
\begin{align*}
E&=2.756\times\frac{10.50}{5.477}\
&=2.756\times1.917\
&\approx5.29
\end{align*}
$$

## Step5: Calculate the confidence interval
The confidence interval is $\bar{x}-E<\mu<\bar{x} + E$
$30.23-5.29<\mu<30.23 + 5.29$
$24.94<\mu<35.52$

# Answer:
a. The point - estimate (sample mean) is $\boldsymbol{30.23}$
b. The $99\%$ confidence interval is $(\boldsymbol{24.94},\boldsymbol{35.52})$

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the sample mean

Step2: Calculate the sample standard deviation

Step3: Find the critical value

Step4: Calculate the margin of error

Step5: Calculate the confidence interval

Answer: