Question

The mean weight of 10 randomly selected newborn babies at a local hospital is 7.14 lbs and the standard deviation is 0.87 lbs. Assume the weight of newborn babies has approximately normal distribution. a. Find the margin of error for the 90% confidence interval for the mean weight of all newborn babies at this hospital. b. Use information from part (a) to fill in the blanks in the following sentence: __________ of all samples of size _____ have sample means within _______ of the population mean. c. Find a 90% confidence interval for the mean weight of all newborn babies at this hospital. d. Does the confidence interval, at 90% confidence level, provide sufficient evidence that the mean weight of a newborn at this hospital is above 6.5 lb? Write the appropriate inequality to justify your answer. e. If you increase the confidence level (1-), will the confidence interval estimate be wider or narrower? Explain.

Answer 1

Answer:

a) $ME= 1.83 \frac{0.87}{\sqrt{10}}=0.5035$  

b) 90% of all samples of size 10 have sample means within 0.5035 of the population mean.

c) The 90% confidence interval would be given by (6.636;7.643)    

d) Yes, since the lower limit for the 90% confidence interval is higher than the value of 6.5 we can conclude that the true mean is significantly higher than 6.5 at 10% of significance. (6.636>6.5)

e) If we increase the confidence level that implies increase the margin of error. Since with more confidence level the value for the critical value $t_{\alpha/2}$ increase. So then the new interval would be wider than the original.

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".

$\bar X=7.14$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s=0.87 represent the sample standard deviation

n=10 represent the sample size  

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)

And the margin of error is given by:

$ME= t_{\alpha/2}\frac{s}{\sqrt{n}}$  

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=10-1=9$

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(0.05,9)".And we see that $t_{\alpha/2}=1.83$

$ME= 1.83 \frac{0.87}{\sqrt{10}}=0.5035$  

Part b

90% of all samples of size 10 have sample means within 0.5035 of the population mean.

Part c

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

$7.14-1.83\frac{0.87}{\sqrt{10}}=6.637$    

$7.14+1.83\frac{0.87}{\sqrt{10}}=7.643$    

So on this case the 90% confidence interval would be given by (6.636;7.643)    

Part d

Yes, since the lower limit for the 90% confidence interval is higher than the value of 6.5 we can conclude that the true mean is significantly higher than 6.5 at 10% of significance. (6.636>6.5)

Part e

If we increase the confidence level that implies increase the margin of error. Since with more confidence level the value for the critical value $t_{\alpha/2}$ increase. So then the new interval would be wider than the original.