Question

1. A sample of 150 Publix cashiers showed a mean hourly wage of $ 8.00, with a standard deviation of $ 1. Give an interval that is likely to contain about 95% of the sampled cashiers’ hourly wages.

A. (5, 10)
B. (6, 10)
C. (7, 9)
D. (6, 9)

2. Practically interpret the sample mean (x-bar) from the question above.

A. 95% of the sampled cashiers will have an hourly wage that falls at $ 8.00 per hour.
B. Each cashier can earn $ 8.00 per hour.
C. The mean hourly wage of the sampled cashiers is $ 8.00 per hour.
D. The true hourly wage for the Publix cashiers is $ 8.00 per hour.

Answer 1

a)
`8-1.976(1)/(√(150))=7.83`

`8+1.976(1)/(√(150))=8.16`

So on this case the 95% confidence interval would be given by (7.83;8.16)

So then the best option seems to be:

C. (7, 9)

b) The mean calculated comes from the following formula:

`\bar X = (\sum_(i=1)^n X_i)/(n) =8`

So then the best interpretation for this case would be:

C. The mean hourly wage of the sampled cashiers is $ 8.00 per hour.

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X=8` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Part a

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=150-1=149`

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,149)".And we see that
`t_(\alpha/2)=1.976`

Now we have everything in order to replace into formula (1):

`8-1.976(1)/(√(150))=7.83`

`8+1.976(1)/(√(150))=8.16`

So on this case the 95% confidence interval would be given by (7.83;8.16)

So then the best option seems to be:

C. (7, 9)

Part b

The mean calculated comes from the following formula:

`\bar X = (\sum_(i=1)^n X_i)/(n) =8`

So then the best interpretation for this case would be:

C. The mean hourly wage of the sampled cashiers is $ 8.00 per hour.