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Juliette [100K]
3 years ago
11

A group of 59 randomly selected students have a mean score of 29.5 with a standard

Mathematics
1 answer:
algol [13]3 years ago
8 0

Answer:

29.5-1.671\frac{5.2}{\sqrt{59}}=28.37    

29.5+1.671\frac{5.2}{\sqrt{59}}=30.63    

And we can conclude that the true mean for the scores are given by this interval (28.37; 30.63) at 90% of confidence

Step-by-step explanation:

Information given

\bar X = 29.5 represent the sample mean

\mu population mean  

s= 5.2 represent the sample standard deviation

n=59 represent the sample size  

Confidence interval

The confidence interval for the true mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The degrees of freedom are given by:

df=n-1=59-1=58

The Confidence interval is 0.90 or 90%, the significance is \alpha=0.1 and \alpha/2 =0.05, and the critical value is t_{\alpha/2}=1.671

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

29.5-1.671\frac{5.2}{\sqrt{59}}=28.37    

29.5+1.671\frac{5.2}{\sqrt{59}}=30.63    

And we can conclude that the true mean for the scores are given by this interval (28.37; 30.63) at 90% of confidence

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