Respuesta :
Answer:
No neccesarily
Step-by-step explanation:
10 is not a big enough number to estimate the mean of the cost of the laptop. In Los Angeles there are hundreads, if not thousands, of laptop stores, and you need a sample with at least 50 or maybe more stores to obtain more precese results. In a sample of ten there might be one or two stores that sell the laptop at $550, which shoudnt be strange due to the standard deviation of $26.77, and those 2 samples might make your sample mean a lot smaller than $600, but that alone doesnt guarantee that the mean cost on Los Angeles is less than $600.
Answer:
We conclude that the mean cost of that laptop in Los Angeles is same as $600 at 0.05 significance level.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = $600
Sample mean, [tex]\bar{x}[/tex] = $586.50
Sample size, n = 10
Alpha, α = 0.05
Sample standard deviation, s = $26.77
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 600\text{ dollars}\\H_A: \mu < 600\text{ dollars}[/tex]
We use One-tailed(left) t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{586.50- 600}{\frac{26.77}{\sqrt{10}} } =-1.59[/tex]
Now,
[tex]t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } =-1.83[/tex]
Since,
[tex]t_{stat} > t_{critical}[/tex]
We fail to reject the null hypothesis and accept the null hypothesis.
We conclude that the mean cost of that laptop in Los Angeles is same as $600 at 0.05 significance level.
