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Alexeev081 [22]
3 years ago
6

Suppose a baker claims that the average bread height is more than 15cm. Several of this customers do not believe him. To persuad

e 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? What is the upper bound to this confidence interval?
Mathematics
1 answer:
laila [671]3 years ago
4 0

Answer:

17-2.262\frac{1.9}{\sqrt{10}}=15.641    

17+2.262\frac{1.9}{\sqrt{10}}=18.359    

So on this case the 95% confidence interval would be given by (15.641;18.359)    

And since the lower limit for the confidence interval is higher than 15 we can conclude that at 5% of significance the true mean is higher than 15 cm

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=17 represent the sample mean

\mu population mean (variable of interest)

s=1.9 represent the sample standard deviation

n=10 represent the sample size  

Confidence interval

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}=2.262

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

17-2.262\frac{1.9}{\sqrt{10}}=15.641    

17+2.262\frac{1.9}{\sqrt{10}}=18.359    

So on this case the 95% confidence interval would be given by (15.641;18.359)    

And since the lower limit for the confidence interval is higher than 15 we can conclude that at 5% of significance the true mean is higher than 15 cm

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Step-by-step explanation:

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Solution:

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Step-by-step explanation:

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