Respuesta :
Answer:
We accept the alternate hypothesis and conclude that the assembly time using the new method is faster.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 42.3 minutes
Sample mean, [tex]\bar{x}[/tex] = 40.6 minutes
Sample size, n = 24
Alpha, α = 0.10
Sample standard deviation, s = 2.7 minutes
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 42.3\text{ minutes}\\H_A: \mu < 42.3\text{ minutes}[/tex]
We use One-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n-1}} }[/tex] Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{40.6 - 42.3}{\frac{2.7}{\sqrt{23}} } = -3.0195[/tex]
Now, [tex]t_{critical} \text{ at 0.10 level of significance, 23 degree of freedom } = -1.3195[/tex]
Since,
[tex]t_{stat} < t_{critical}[/tex]
We reject the null hypothesis and fail to accept it.
We accept the alternate hypothesis and conclude that the assembly time using the new method is faster.
At 0.10 level of significance, the assembly time using the new method is shown to be faster
How to form the hypotheses?
There are two hypotheses. First one is called null hypothesis and it is chosen such that it predicts nullity or no change in a thing. It is usually the hypothesis against which we do the test. The hypothesis which we put against null hypothesis is alternate hypothesis.
Null hypothesis is the one which researchers try to disprove.
For this case, we want to check if the mean of the population (if would be using the new method) would be smaller than the mean of the population using present method.
Thus, we get:
Null hypothesis: New method doesn't have extra effect, so it is same efficient old method, and therefore, population mean time by current method (42.3 minutes) = population mean time by old method ([tex]\mu[/tex] minutes)
or [tex]H_0: \mu = 42.3[/tex]
Alternate hypothesis: new method reduces average time.
or [tex]H_1: \mu < 42.3[/tex]
(thus, our test is going to be for one tailed as the critical area of the distribution is one sided)
We're provided that:
- Sample size = n = 24
- Sample mean = [tex]\overline{x}[/tex] = 40.6 minutes
- Sample standard deviation = s = 2.7 minutes
Thus, as sample size is < 24, we use t-test.
The value of t test statistic is evaluated as:
[tex]t = \dfrac{\overline{x} - \mu}{s/\sqrt{n}}[/tex]
(we're testing null hypothesis, so we take [tex]\mu = 42.3[/tex] as hypothesized.)
[tex]t = \dfrac{\overline{x} - \mu}{s/\sqrt{n}} = \dfrac{40.6 - 42.3}{2.7/\sqrt{24}} \approx -3.08[/tex]
The degree of freedom is n - 1 = 24 - 1 = 23
The level of significance is 0.10 = 10%
At this level of significance and degree of freedom, we get the critical value of t as: [tex]t_{\alpha/2} = 1.319[/tex] (one tailed)
We see that: [tex]|t| = 3.08 > 1.319 = t_{\alpha/2}[/tex] (that's why we can reject the null hypothesis), else if we'd get test statistic < critical value, then we may accept null hypothesis)
Thus, we accept the alternative hypothesis that the new method is going to reduce the overall average time.
Learn more about t-test for single mean here:
https://brainly.com/question/25316952
