Mathematics, 07.03.2020 00:05, leahglover19

Part 1. An astronomer makes 576 measurements of the diameter of Venus and finds that the mean is 7848 miles. Assume that the standard deviation of all the measurements of the diameter of similar planet is 310 miles. Construct a 95% confidence interval estimate for the mean measurement of the diameter of Venus. Make sure to include a conclusion.

Part 2. What sample size is required so that the margin of error in determining the diameter of Venus is only 20 miles?

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Answers

answered: sophx

1) $7848-1.96\frac{310}{\sqrt{576}}=7822.68$

$7848+1.96\frac{310}{\sqrt{576}}=7873.32$

So on this case the 95% confidence interval would be given by (7822.68;7873.32)

2) $n=(\frac{1.960(310)}{20})^2 =922.94 \approx 923$

So the answer for this case would be n=923 rounded up to the nearest integer

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".

$\bar X=7848$ represent the sample mean

$\mu$ population mean (variable of interest)

s=310 represent the sample standard deviation

n=576 represent the sample size

Part 1

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=576-1=575

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 tabel to find the critical value. The excel command would be: "=-T.INV(0.025,575)".And we see that $t_{\alpha/2}=1.96$