Question

Consider results of 20 randomly chosen people who have run a marathon. Their times, in minutes, are as follows: 136, 143, 153, 164, 166, 179, 184, 194, 199, 207, 216, 220, 228, 247, 252, 259, 275, 276, 282, 293 Calculate a 95% upper confidence bound on the mean time of the race. Assume distribution to be normal. Round your answer to the nearest integer (e.g. 9876) the absolute tolerance is +-1

Answer 1

To calculate a 95% upper confidence bound on the mean time of the marathon race, use the formula: Upper Confidence Bound = Sample Mean + (Z * Standard Deviation / sqrt(n)). Plug in the given values and calculate to obtain the 95% upper confidence bound on the mean time.

Step-by-step explanation:

To calculate a 95% upper confidence bound on the mean time of the race, we need to use the formula:

Upper Confidence Bound = Sample Mean + (Z * Standard Deviation / sqrt(n))

where Z is the z-score corresponding to the desired confidence level. For a 95% confidence level, the z-score is approximately 1.96. The sample mean is calculated by summing all the times and dividing by the number of observations (20 in this case), and the standard deviation can be obtained using statistical software or calculators.

Using the given values:

Sample Mean = 205.6

Standard Deviation = 53.609

Plugging these values into the formula, we get:

Upper Confidence Bound = 205.6 + (1.96 * 53.609 / sqrt(20)) = 217.075

Rounded to the nearest integer, the 95% upper confidence bound on the mean time of the race is 217 minutes.

Answer 2

`213.65-2.093(49.028)/(√(20))=190.704`

`213.65+2.093(49.028)/(√(20))=236.596`

And the confidence interval would be given by:

`190.704 \leq \mu \leq 236.596`

Step-by-step explanation:

Information given

136, 143, 153, 164, 166, 179, 184, 194, 199, 207, 216, 220, 228, 247, 252, 259, 275, 276, 282, 293

In order to calculate the mean and the sample deviation we can use the following formulas:

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\sum_(i=1)^n (x_i-\bar X))/(n-1)}` (3)

`\bar X= 213.65` represent the sample mean

`\mu` population mean (variable of interest)

s=49.028 represent the sample standard deviation

n=20 represent the sample size

Confidence interval

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=20-1=19`

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and the critical value for this case is
`t_(\alpha/2)=2.093`

And replacing we got:

`213.65-2.093(49.028)/(√(20))=190.704`

`213.65+2.093(49.028)/(√(20))=236.596`

And the confidence interval would be given by:

`190.704 \leq \mu \leq 236.596`