Question

We considered the differences between the temperature readings in January 1 of 1968 and 2008 at 51 locations in the continental US in Exercise 5.19. The mean and standard deviation of the reported differences are 1.1 degrees and 4.9 degrees respectively.Calculate a 90% confidence interval for the average difference between the temperature measurements between 1968 and 2008.lower bound: ______ degrees(please round to two decimal places)upper bound: ______ degrees(please round to two decimal places)

Answer 1

Lower bound: -0.43 degrees

Upper bound: 2.56 degrees

Step-by-step explanation:

To calculate a 90% confidence interval for the average difference in temperature measurements, we follow these steps:

Calculate the standard error:

Standard error = standard deviation / square root of sample size

Standard error = 4.9 degrees / sqrt(51) ≈ 0.33 degrees

Find the critical value:

For a 90% confidence interval, the critical value (z-score) is approximately 1.645.

Calculate the confidence interval:

Lower bound = mean - critical value * standard error

Lower bound = 1.1 degrees - 1.645 * 0.33 degrees ≈ -0.43 degrees

Upper bound = mean + critical value * standard error

Upper bound = 1.1 degrees + 1.645 * 0.33 degrees ≈ 2.56 degrees

Therefore, we can be 90% confident that the true average difference in temperature measurements between 1968 and 2008 falls within the range of -0.43 degrees to 2.56 degrees.

Answer 2

`1.1-2.02(4.9)/(√(50))=-0.30`

`1.1+2.02(4.9)/(√(50))=2.50`

So on this case the 90% confidence interval would be given by (-0.30;2.50)

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=1.1` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s=4.9 represent the sample standard deviation

n=51 represent the sample size

Solution to the problem

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=51-1=50`

Since the Confidence is 0.90 or 90%, the value of
`\alpha=0.1` and
`\alpha/2 =0.05`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,50)".And we see that
`t_(\alpha/2)=2.02`

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

`1.1-2.02(4.9)/(√(50))=-0.30`

`1.1+2.02(4.9)/(√(50))=2.50`

So on this case the 90% confidence interval would be given by (-0.30;2.50)