Answer :
Answer with explanation:
Let [tex]\mu[/tex] be the population mean.
As per given , we have
[tex]H_0:\mu \leq4\\\\ H_a: \mu >4[/tex]
Since the alternative hypothesis is right-tailed , so the test is a right-tailed test.
Also, population standard deviation is given [tex]\sigma=1[/tex] , so we perform one-tailed z-test.
Test statistic : [tex]z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
, where [tex]\mu[/tex] = Population mean
[tex]\sigma[/tex] = Population standard deviation
n= sample size
[tex]\overline{x}[/tex] = Sample mean
For n= 18 , [tex]\overline{x}=4.50[/tex] , [tex]\sigma=1[/tex] , [tex]\mu =4[/tex], we have
[tex]z=\dfrac{4.5-4}{\dfrac{1}{\sqrt{18}}}\approx2.12[/tex]
P-value (for right tailed test): P(z>2.12) = 1-P(z≤ 2.12) [∵ P(Z>z)=1-P(Z≤z)]\
=1- 0.0340=0.9660
Decision : Since P-value(0.9660) > Significance level (0.01), it means we are failed to reject the null hypothesis.
[We reject null hypothesis if p-value is larger than the significance level . ]
Conclusion : We do not have sufficient evidence to show that the goal is not being met at α = .01 .
The sample of 18 fire calls with a mean response time of 4 minutes 30 seconds doesn't provide sufficient evidence to show that the goal is not being met at α = .01
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.
When to use the z test and when to use the t test?
If the sample taken is of size less than t test, then you can use the t test.
If the sample size is larger or equal to 30, you can use the z test.
It is because as the sample size grows more and more, the z test statistic approximates more and more the normal distribution.
For the considered case, we can use:
- [tex]\mu[/tex] = average response time ( of population)
- Null Hypothesis: [tex]H_0: \mu \leq 4[/tex]
- Alternative hypothesis: [tex]H_1: \mu > 4[/tex]
It is because we want to show that the goal isn't met. Thus, we took alternative hypothesis such that the mean response time is more than 4 minutes.
Now, since the sample size is n = 18 < 30, we will use t-test here.
We evaluate 't' as:
[tex]t = \dfrac{\overline{x} - \mu}{\sigma/\sqrt{n}}[/tex] (we can use sample standard deviation is population standard deviation is not available).
The values are:
- [tex]\overline{x} = 4.5[/tex] (4 minutes and 30 seconds is 4.5 minutes)
- [tex]\sigma = 1[/tex]
- [tex]n = 18[/tex]
- [tex]\mu = 4[/tex]
Thus, we get:
[tex]t = \dfrac{\overline{x} - \mu}{\sigma/\sqrt{n}} = \dfrac{4.5 - 4}{1/\sqrt{18}} = \sqrt{18} \times 0.5 \approx 2.12[/tex]
- The level of significance here is 0.01, and
- The degree of freedom = n-1 = 18-1 = 17
At this degree of freedom and level of significance, the critical value of t-test statistic is [tex]t_{\alpha/2} = 2.567[/tex] (one tailed)
Since [tex]t < t_{\alpha/2}[/tex] we may accept the null hypothesis, and thus, haven't got significant evidence to accept the alternative hypothesis.
(if the obtained value would be bigger than critical value, then we'd reject null hypothesis).
Therefore, the sample of 18 fire calls with a mean response time of 4 minutes 30 seconds doesn't provide sufficient evidence to show that the goal is not being met at α = .01
Learn more about t-test for single mean here:
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