Mathematics, 12.12.2019 23:31 nerissa5

Each of 20 science students independently measured the melting point of lead. the sample mean and sample standard deviation of these measurements were (in degrees centigrade) 330.2 and 15.4, respectively. construct

(a) a 95 percent confidence interval estimate of the true weight.

(b) a 99 percent confidence interval estimate of the true melting point of lead.

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Answer from: TookMurr

a) The 95% confidence interval would be given by (323.003;337.397)

b) The 99% confidence interval would be given by (320.351;340.049)

Step-by-step explanation:

1) 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".

$\bar X=330.2$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=15.4 represent the sample standard deviation

n=20 represent the sample size

2) Part a

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=20-1=19

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 table to find the critical value. The excel command would be: "=-T.INV(0.025,19)".And we see that $t_{\alpha/2}=2.09$

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

$330.2-2.09\frac{15.4}{\sqrt{20}}=323.003$

$330.2+2.09\frac{15.4}{\sqrt{20}}=337.397$

So on this case the 95% confidence interval would be given by (323.003;337.397)

3) Part b

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,19)".And we see that $t_{\alpha/2}=2.86$

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

$330.2-2.86\frac{15.4}{\sqrt{20}}=320.351$

$330.2+2.86\frac{15.4}{\sqrt{20}}=340.049$

So on this case the 99% confidence interval would be given by (320.351;340.049)

Answer from: Quest

i need points

step-by-step explanation:

sorry bubs

Answer from: Quest

this "question" looks more like a project is there any other info you can give me about this. it could be too complex for someone else to do on this website