Question

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.
883
b. construct a 99% confidence interval for the mean one - way commuting time from home to work for all workers.
to
hint
assistance used
assistance used

Explanation:

Step1: Calculate the point estimate (mean)

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
First, find \(\sum_{i=1}^{30}x_{i}\):
\[ \begin{align*} &23 + 17+34 + 26+18+33+46+42+13+37+45+14+22+20+28+32+18+39+40+48+25+36+23+39+42+46+29+17+24+31\\ =&(23\times2)+(17\times2)+(34)+(26)+(18\times2)+(33)+(46\times2)+(42\times2)+(13)+(37)+(45)+(14)+(22)+(20)+(28)+(32)+(39\times2)+(40)+(48)+(25)+(36)+(29)+(24)+(31)\\ =&46+34 + 26+36+33+92+84+13+37+45+14+22+20+28+32+78+40+48+25+36+29+24+31\\ =&(46+34)+(26+36)+(33+92)+(84+13)+(37+45)+(14+22)+(20+28)+(32+78)+(40+48)+(25+36)+(29+24)+31\\ =&80+62+125+97+82+36+48+110+88+61+53+31\\ =&(80 + 62)+(125+97)+(82+36)+(48+110)+(88+61)+(53+31)\\ =&142+222+118+158+149+84\\ =&(142+222)+(118+158)+(149+84)\\ =&364+276+233\\ =&873 \end{align*} \]
Since \(n = 30\), \(\bar{x}=\frac{873}{30}=29.1\)

Step2: Calculate the standard deviation \(s\)

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\[ \begin{align*} \sum_{i=1}^{n}(x_{i}-\bar{x})^{2}&=\sum_{i = 1}^{30}(x_{i}-29.1)^{2}\\ \end{align*} \]
\[ \begin{align*} &(23 - 29.1)^{2}+(17-29.1)^{2}+(34 - 29.1)^{2}+(26-29.1)^{2}+(18-29.1)^{2}+(33-29.1)^{2}+(46-29.1)^{2}+(42-29.1)^{2}+(13-29.1)^{2}+(37-29.1)^{2}+(45-29.1)^{2}+(14-29.1)^{2}+(22-29.1)^{2}+(20-29.1)^{2}+(28-29.1)^{2}+(32-29.1)^{2}+(18-29.1)^{2}+(39-29.1)^{2}+(40-29.1)^{2}+(48-29.1)^{2}+(25-29.1)^{2}+(36-29.1)^{2}+(23-29.1)^{2}+(39-29.1)^{2}+(42-29.1)^{2}+(46-29.1)^{2}+(29-29.1)^{2}+(17-29.1)^{2}+(24-29.1)^{2}+(31-29.1)^{2}\\ =&(-6.1)^{2}+(-12.1)^{2}+(4.9)^{2}+(-3.1)^{2}+(-11.1)^{2}+(3.9)^{2}+(16.9)^{2}+(12.9)^{2}+(-16.1)^{2}+(7.9)^{2}+(15.9)^{2}+(-15.1)^{2}+(-7.1)^{2}+(-9.1)^{2}+(-1.1)^{2}+(2.9)^{2}+(-11.1)^{2}+(9.9)^{2}+(10.9)^{2}+(18.9)^{2}+(-4.1)^{2}+(6.9)^{2}+(-6.1)^{2}+(9.9)^{2}+(12.9)^{2}+(16.9)^{2}+(-0.1)^{2}+(-12.1)^{2}+(-5.1)^{2}+(1.9)^{2}\\ =&37.21+146.41 + 24.01+9.61+123.21+15.21+285.61+166.41+259.21+62.41+252.81+228.01+50.41+82.81+1.21+8.41+123.21+98.01+118.81+357.21+16.81+47.61+37.21+98.01+166.41+285.61+0.01+146.41+26.01+3.61\\ =&(37.21\times2)+(146.41\times2)+(24.01)+(9.61)+(123.21\times2)+(15.21)+(285.61\times2)+(166.41\times2)+(259.21)+(62.41)+(252.81)+(228.01)+(50.41)+(82.81)+(1.21)+(8.41)+(98.01\times2)+(118.81)+(357.21)+(16.81)+(47.61)+(3.61)\\ =&74.42+292.82+24.01+9.61+246.42+15.21+571.22+332.82+259.21+62.41+252.81+228.01+50.41+82.81+1.21+8.41+196.02+118.81+357.21+16.81+47.61+3.61\\ =&(74.42+292.82)+(24.01+9.61)+(246.42+15.21)+(571.22+332.82)+(259.21+62.41)+(252.81+228.01)+(50.41+82.81)+(1.21+8.41)+(196.02+118.81)+(357.21+16.81)+(47.61+3.61)\\ =&367.24+33.62+261.63+904.04+321.62+480.82+133.22+9.62+314.83+374.02+51.22\\ =&(367.24+33.62)+(261.63+904.04)+(321.62+480.82)+(133.22+9.62)+(314.83+374.02)+51.22\\ =&400.86+1165.67+802.44+142.84+688.85+51.22\\ =&(400.86+1165.67)+(802.44+142.84)+(688.85+51.22)\\ =&1566.53+945.28+740.07\\ =&3251.88 \end{align*} \]
\(s=\sqrt{\frac{3251.88}{30 - 1}}=\sqrt{\frac{3251.88}{29}}\approx\sqrt{112.1338}\approx10.59\)

Step3: Find the \(t\) - value

For a \(99\%\) confidence interval and \(n-1 = 29\) degrees of freedom, using a \(t\) - distribution table or calculator, \(t_{\alpha/2}=2.756\)

Step4: Calculate the margin of error \(E\)

The formula for the margin of error \(E = t_{\alpha/2}\frac{s}{\sqrt{n}}\)
\(E=2.756\times\frac{10.59}{\sqrt{30}}\)
\(\sqrt{30}\approx5.477\), \(\frac{10.59}{5.477}\approx1.933\)
\(E=2.756\times1.933\approx5.33\)

Step5: Calculate the confidence interval

The confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\)
\(29.1-5.33<\mu<29.1 + 5.33\)
\(23.77<\mu<34.43\)

Answer:

a. The point estimate is \(29.10\)
b. The \(99\%\) confidence interval is \((23.77,34.43)\)