With its old software, a robot in an evaporator assembly line applied a mean of $3.2 \mathrm{~mL}$ of lubricant per evaporator, with a standard deviation of 0.70 . The lead engineer thinks that, following a recent software update, the standard deviation, $\sigma$, is now less than 0.70 . To see if this is the case, he tests a random sample of 24 evaporators that were assembled following the update. He finds that, for the evaporators in the sample, the robot applied a mean of $3.1 \mathrm{~mL}$ of lubricant per evaporator, with a standard deviation of 0.55 . Is there enough evidence to support the engineer's claim at the 0.05 level of significance? Assume that the amounts of lubricant applied per evaporator following the software update are approximately normally distributed.
Perform a one-tailed test. Then complete the parts below.
Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.)
(a) State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$.
\[
\begin{array}{l}
H_{0}: \sigma=0.70 \\
H_{1}: \sigma< 0.70
\end{array}
\]
(b) Determine the type of test statistic to use.
Chi-square Degrees of freedom:
(c) Find the value of the test statistic. (Round to three or more decimal places.) 14.199
(d) Find the $p$-value. (Round to three or more decimal places.)
\[
0,05
\]
(e) Can we support the claim that the standard deviation of amount of lubricant applied per evaporator following the software update is less than 0.70 ?
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Answer
Since the p-value is less than the level of significance (0.05), we reject the null hypothesis. Therefore, there is enough evidence to support the claim that the standard deviation of the amount of lubricant applied per evaporator following the software update is less than 0.70. This is represented as \(\boxed{\text{Reject } H_{0}}\)
Step 1 ：State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$ as: $H_{0}: \sigma=0.70$ and $H_{1}: \sigma<0.70$
Step 2 ：Identify the type of test statistic to use, which is a Chi-square test. The degrees of freedom are calculated as n-1 = 24-1 = 23
Step 3 ：Calculate the value of the test statistic using the formula: $\chi^2 = (n-1) * (s^2 / \sigma^2)$, which gives $\chi^2 = (23) * (0.55^2 / 0.70^2)$, resulting in $\chi^2 = 14.199$ (rounded to three decimal places)
Step 4 ：Determine the p-value, which is the probability that a Chi-square statistic having 23 degrees of freedom is more extreme than 14.199. Using a Chi-square distribution table or a statistical calculator, we find that the p-value is less than 0.05
Step 5 ：Since the p-value is less than the level of significance (0.05), we reject the null hypothesis. Therefore, there is enough evidence to support the claim that the standard deviation of the amount of lubricant applied per evaporator following the software update is less than 0.70. This is represented as \(\boxed{\text{Reject } H_{0}}\)
