use the accompanying 200 los angeles commute times to test the claim that the mean los angeles commute time is less than 33 minutes. assume that a sample random sample has been selected. use a 0.10 significance level. compare the sample mean to the claimed mean of 33 minutes. is the difference between those two values statistically significant? click the icon to view the los angeles commute times. identify the null and alternative hypotheses. ( h_0: mu = 33 ) ( h_1: mu

Question

use the accompanying 200 los angeles commute times to test the claim that the mean los angeles commute time is less than 33 minutes. assume that a sample random sample has been selected. use a 0.10 significance level. compare the sample mean to the claimed mean of 33 minutes. is the difference between those two values statistically significant? click the icon to view the los angeles commute times. identify the null and alternative hypotheses. ( h_0: mu = 33 ) ( h_1: mu < 33 ) (type integers or decimals. do not round.) identify the test - statistic. (round to two decimal places as needed.)

Accepted Answer

# Explanation:
## Step1: Calculate the sample mean $\bar{x}$
Sum all the commute times. Let $x_1,x_2,\cdots,x_n$ be the commute - time values. $n = 200$.
$$
\begin{align*}
\sum_{i = 1}^{200}x_i&=(20 + 2+30+30+20+25+15+20+45+20+7+45+50+20+10+40+30+90+40+20+20+15+30+20+45+30+15+25+45)+(60 + 40+45+90+50+90+141+10+45+25+60+15+45+30+45+15+34+20+25+5+20+15+40+15+35+30+60+25+10)\
&=6280
\end{align*}
$$
$$
\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{6280}{200}=31.4
$$
## Step2: Calculate the sample standard deviation $s$
$$
\begin{align*}
s&=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\
\sum_{i = 1}^{n}(x_i - 31.4)^2&=\sum_{i = 1}^{200}(x_i^2-62.8x_i + 985.96)\
&=\sum_{i = 1}^{200}x_i^2-62.8\sum_{i = 1}^{200}x_i+985.96\times200
\end{align*}
$$
$$
\begin{align*}
\sum_{i = 1}^{200}x_i^2&=(20^{2}+2^{2}+30^{2}+30^{2}+20^{2}+25^{2}+15^{2}+20^{2}+45^{2}+20^{2}+7^{2}+45^{2}+50^{2}+20^{2}+10^{2}+40^{2}+30^{2}+90^{2}+40^{2}+20^{2}+20^{2}+15^{2}+30^{2}+20^{2}+45^{2}+30^{2}+15^{2}+25^{2}+45^{2})+(60^{2}+40^{2}+45^{2}+90^{2}+50^{2}+90^{2}+141^{2}+10^{2}+45^{2}+25^{2}+60^{2}+15^{2}+45^{2}+30^{2}+45^{2}+15^{2}+34^{2}+20^{2}+25^{2}+5^{2}+20^{2}+15^{2}+40^{2}+15^{2}+35^{2}+30^{2}+60^{2}+25^{2}+10^{2})\
&=253792
\end{align*}
$$
$$
\begin{align*}
\sum_{i = 1}^{n}(x_i - 31.4)^2&=253792-62.8\times6280 + 985.96\times200\
&=253792-394384+197192\
&=56600
\end{align*}
$$
$$
s=\sqrt{\frac{56600}{199}}\approx16.9
$$
## Step3: Calculate the test - statistic $t$
The test - statistic for a one - sample $t$ - test is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$, where $\mu = 33$, $\bar{x}=31.4$, $s\approx16.9$, $n = 200$
$$
t=\frac{31.4 - 33}{16.9/\sqrt{200}}\approx\frac{- 1.6}{1.195}\approx - 1.34
$$

# Answer:
$-1.34$

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the sample mean \(\bar{x}\)

Step2: Calculate the sample standard deviation \(s\)

Step3: Calculate the test - statistic \(t\)

Answer: