Mathematics, 20.02.2020 08:31 sarahppalenz9519

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.

Answers: 2

Show answers

Answers

Answer from: hncriciacheichi

a) $8-1.976\frac{1}{\sqrt{150}}=7.83$

$8+1.976\frac{1}{\sqrt{150}}=8.16$

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:

$\bar X = \frac{\sum_{i=1}^n X_i}{n} =8$

So then the best interpretation for this case would be:

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

Step-by-step 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (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):

$8-1.976\frac{1}{\sqrt{150}}=7.83$

$8+1.976\frac{1}{\sqrt{150}}=8.16$

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:

$\bar X = \frac{\sum_{i=1}^n X_i}{n} =8$

So then the best interpretation for this case would be:

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

Answer from: Quest

answer is more than one

step-by-step explanation:

Answer from: Quest

When rates are expressed as a quantity of 1, such as 2 feet per second or 5 miles per hour, they are called unit rates. if you have a multiple-unit rate such as 120 students for every 3 buses, and want to find the single-unit rate, write a ratio equal to the multiple-unit rate with 1 as the second term.

Another question on Mathematics

Mathematics, 21.06.2019 20:00

Apatient is to be given 35 milligrams of demerol every 4 hours. you have demerol 50 milligrams/milliliter in stock. how many milliliters should be given per dose?

Answers: 2

Answer

Mathematics, 21.06.2019 22:40

(10. in the figure, a aabc is drawn tocircumscribe a circle of radius 3 cm ,such thatthe segments bd and dc are respectively oflength 6 cm and 9 cm. find the length ofsides ab and ac.

Answers: 3

Answer

Mathematics, 21.06.2019 23:10

Point s lies between points r and t on . if rt is 10 centimeters long, what is st? 2 centimeters 4 centimeters 6 centimeters 8 centimeters

Answers: 2

Answer

Mathematics, 22.06.2019 00:10

Write the slope intercept equation for line ab.

Answers: 2

Answer

You know the right answer?

1. A sample of 150 Publix cashiers showed a mean hourly wage of $ 8.00, with a standard deviation of...