Question

Suppose a baker claims that the average bread height is more than 15cm. Several of this customers do not believe him. To persuade his customers that he is right, the baker decides to do a hypothesis test. He bakes 10 loaves of bread. The mean height of the sample loaves is 17 cm with a sample standard deviation of 1.9 cm. The heights of all bread loaves are assumed to be normally distributed. The baker is now interested in obtaining a 95% confidence interval for the true mean height of his loaves. What is the lower bound to this confidence interval? 2 cm (round to 2 decimal places) What is the upper bound to this confidence interval? cm (round to 2 decimal places) For the following situations, use RStudio to find the appropriate t-critical values that would be needed to construct a confidence interval. Round all critical values to the second decimal place. 1. n = 15, confidence level is 95%, x= 35 and s = 2.7, t-critical value- 2, n = 37, confidence level is 99%, x= 82 and s = 5.9 t-critical value- 2 3, n 1009, confidence level is 90%, x 0.9 and s-0.04 t- critical value = 2 2

Answer 1

Answer:

a) $17-2.26\frac{1.9}{\sqrt{10}}=15.64$  

$17+2.26\frac{1.9}{\sqrt{10}}=18.36$  

So on this case the 95% confidence interval would be given by (15.64;18.36)

b) 1. n=15, conf =95% $\bar X= 35$ s=2.7

> round(qt(p=1-0.025,df=15-1),2)

[1] 2.14

> round(qt(p=0.025,df=15-1),2)

[1] -2.14

2. n=37, conf =99% $\bar X= 82$ s=5.9

> round(qt(p=1-0.005,df=37-1),2)

[1] 2.72

> round(qt(p=0.005,df=37-1),2)

[1] -2.72

3. n=1009, conf =90% $\bar X= 0.9$ s=0.04

> round(qt(p=1-0.05,df=1009-1),2)

[1] 1.65

> round(qt(p=0.05,df=1009-1),2)

[1] -1.65

Step-by-step explanation:

Part a: What is the lower bound to this confidence interval? 2 cm (round to 2 decimal places) What is the upper bound to this confidence interval? cm (round to 2 decimal places)

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".  

Solution to the problem

We have the following data:

$\bar x= 17$ represent the sample mean

$s = 1.9$ represent the sample deviation

n =10 represent the sample size

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:  

$df=n-1=10-1=9$  

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(0.025,9)".And we see that $t_{\alpha/2}=2.26$  

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

$17-2.26\frac{1.9}{\sqrt{10}}=15.64$  

$17+2.26\frac{1.9}{\sqrt{10}}=18.36$  

So on this case the 95% confidence interval would be given by (15.64;18.36)

Part b

1. n=15, conf =95% $\bar X= 35$ s=2.7

> round(qt(p=1-0.025,df=15-1),2)

[1] 2.14

> round(qt(p=0.025,df=15-1),2)

[1] -2.14

2. n=37, conf =99% $\bar X= 82$ s=5.9

> round(qt(p=1-0.005,df=37-1),2)

[1] 2.72

> round(qt(p=0.005,df=37-1),2)

[1] -2.72

3. n=1009, conf =90% $\bar X= 0.9$ s=0.04

> round(qt(p=1-0.05,df=1009-1),2)

[1] 1.65

> round(qt(p=0.05,df=1009-1),2)

[1] -1.65