Question

A sample of 100 workers located in Atlanta has an average daily work time of 6.5 hours with a standard deviation of 0.5 hours. A sample of 110 workers located in Chicago has an average daily work time of 6.7 hours with a standard deviation of 0.7 hours. With 95% confidence, can you say that the average hours worked daily in Atlanta is different than Chicago?

Answer 1

No, I can't say precisely that. Because there are common values within both working hours intervals.

Explanation:

1)Let's do it by parts. For a Confidence Interval of 95%, i.e. covering 95% of the area of the Graph of this distribution, with known mean and Standard Deviation we have to plug it in the formula below.

Notice that in the formula the part:

`Z(s)/(√(n))`

2)We can find the margin of error.

Atlanta:

Average Daily Work Time

100 workers

`\bar{x}=6.5`

`s=0.5`

`\bar{x}\pm Z(s)/(√(n))\\6.5\pm 1.96(0.5)/(√(100))\\6.5 \pm 0.098\\6.40\: to\: 6.59`

Atlanta workers may have a an average of 6.4 to 6.59 daily working hours

Chicago

110 workers (observations)

`\bar{x}:6.7\\s=0.7`

`\bar{x}\pm Z(s)/(√(n))\\6.7\pm 1.96(0.7)/(√(110))\\6.7\pm 0.13\\6.57\: to\: 6.83`

Chicago workers may have worked an average of 6.57 to 6.83 daily working hours

Answer 2

`t=\frac{6.5-6.7}{\sqrt{(0.5^2)/(100)+(0.7^2)/(110)}}}=-2.398`

`df = n_1 +n_2 -2= 100+110-2= 208`

Since is a bilateral test the p value would be:

`p_v =2*P(t_(208)<-2.398)=0.01110`

Comparing the p value with the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we can reject the null hypothesis, and we have significant differences between the two groups at 5% of significance.

Explanation:

Data given and notation

`\bar X_(1)=6.5` represent the sample mean for Atlanta

`\bar X_(2)=6.7` represent the sample mean for Chicago

`s_(1)=0.5` represent the sample deviation for Atlanta

`s_(2)=0.7` represent the sample standard deviation for Chicago

`n_(1)=100` sample size for the group Atlanta

`n_(2)=110` sample size for the group Chicago

t would represent the statistic (variable of interest)

`\alpha=0.01` significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the meanfor atlanta is different from the mean of Chicago, the system of hypothesis would be:

Null hypothesis:
`\mu_(1)=\mu_(2)`

Alternative hypothesis:
`\mu_(1) \\eq \mu_(2)`

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

`t=\frac{\bar X_(1)-\bar X_(2)}{\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}}` (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

`t=\frac{6.5-6.7}{\sqrt{(0.5^2)/(100)+(0.7^2)/(110)}}}=-2.398`

What is the p-value for this hypothesis test?

The degrees of freedom are given by:

`df = n_1 +n_2 -2= 100+110-2= 208`

Since is a bilateral test the p value would be:

`p_v =2*P(t_(208)<-2.398)=0.01110`

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we can reject the null hypothesis, and we have significant differences between the two groups at 5% of significance.