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juin [17]
3 years ago
15

(2) Mele scored 70% on

Mathematics
1 answer:
Katen [24]3 years ago
5 0
She got 7/10 of the question right.
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A survey shows that college students spend on average about $20 per month on Yum Good Ramen. a. Assume that the survey was condu
lawyer [7]

Answer:

a) The 95% confidence interval would be given by (19.168;20.832)

b) The 95% confidence interval would be given by (19.182;20.416)

c) The 95% confidence interval would be given by (19.412;20.588)

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 represent the sample mean

\mu population mean (variable of interest)

\sigma represent the population standard deviation

n represent the sample size  

Part a

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

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

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: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96

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

20-1.96\frac{6}{\sqrt{200}}=19.168    

20+1.96\frac{6}{\sqrt{200}}=20.832

So on this case the 95% confidence interval would be given by (19.168;20.832)

Part b

For this case we just need to change the value for the deviation

20-1.96\frac{3}{\sqrt{200}}=19.584    

20+1.96\frac{3}{\sqrt{200}}=20.416

So on this case the 95% confidence interval would be given by (19.182;20.416)

Part c

For this case we change again the deviation and the sample size.

20-1.96\frac{6}{\sqrt{400}}=19.412    

20+1.96\frac{6}{\sqrt{400}}=20.588

So on this case the 95% confidence interval would be given by (19.412;20.588)

7 0
3 years ago
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