Question

A technical magazine stated that the mean retail cost of all models of a particular laptop was $600. A random sample of 10 stores that sell that laptop in Los Angeles found it sold with a mean cost of $586.50 and a standard deviation of $26.77. Does this indicate that the mean cost of that laptop in Los Angeles is less than $600?

Answer 1

We conclude that the mean cost of that laptop in Los Angeles is same as $600 at 0.05 significance level.

Explanation:

We are given the following in the question:

Population mean, μ = $600

Sample mean,
`\bar{x}` = $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

`H_(0): \mu = 600\text{ dollars}\\H_A: \mu < 600\text{ dollars}`

We use One-tailed(left) t test to perform this hypothesis.

Formula:

`t_(stat) = \displaystyle\frac{\bar{x} - \mu}{(s)/(√(n)) }`

Putting all the values, we have

`t_(stat) = \displaystyle(586.50- 600)/((26.77)/(√(10)) ) =-1.59`

Now,

`t_(critical) \text{ at 0.05 level of significance, 9 degree of freedom } =-1.83`

Since,

`t_(stat) > t_(critical)`

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.

Answer 2

No neccesarily

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.