In a climate survey, it was determined that in a random sample of 15 days, the average temperature in Kalamazoo at 2:00 PM in the months of

Question

In a climate survey, it was determined that in a random sample of 15 days, the average temperature in Kalamazoo at 2:00 PM in the months of July and August is 73.38 degrees with a standard deviation of 3.798 degrees. Using this information, a 95% confidence interval for the average is (71.28, 75.48). Which of the following is the appropriate interpretation of this interval?

1)

We are certain that 95% of the days in the months of August and July will have a temperature at 2:00 PM between 71.28 and 75.48.
2)

We cannot determine the proper interpretation of this interval.
3)

We are 95% confident that the average daily temperature for the months of July and August at 2:00 PM for the days recorded is between 71.28 and 75.48.
4)

We are 95% confident that the proportion of all days’ temperatures will fall between 71.28 and 75.48.
5)

We are 95% confident that the average daily temperature for the months of July and August at 2:00 PM is between 71.28 and 75.48.

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Alaia 2 weeks 2021-10-11T15:51:50+00:00 1 Answer 0

Answers ( )

    0
    2021-10-11T15:53:24+00:00

    Answer:

    73.38-1.76\frac{3.798}{\sqrt{15}}=71.65  

    73.38+1.76\frac{3.798}{\sqrt{15}}=75.11  

    So on this case the 95% confidence interval would be given by (71.65;75.11)

    The interval given for this case by the problem is:

     (71.28, 75.48)

    And the best conclusion is given by:

    4) We are 95% confident that the average daily temperature for the months of July and August at 2:00 PM is between 71.28 and 75.48.

    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=73.78 represent the sample mean  

    \mu population mean (variable of interest)  

    s=3.798 represent the sample standard deviation  

    n=15 represent the sample size  

    95% confidence interval  

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

    \bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

    The degrees of freedom are given:

     df= n-1 = 15-1=14

    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 tabele to find the critical value. The excel command would be: “=-T.INV(0.05,14)”.And we see that t_{\alpha/2}=1.76  

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

    73.38-1.76\frac{3.798}{\sqrt{15}}=71.65  

    73.38+1.76\frac{3.798}{\sqrt{15}}=75.11  

    So on this case the 95% confidence interval would be given by (71.65;75.11)

    The interval given for this case by the problem is:

     (71.28, 75.48)

    And the best conclusion is given by:

    4) We are 95% confident that the average daily temperature for the months of July and August at 2:00 PM is between 71.28 and 75.48.

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