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The time (in number of days) until maturity of a certain variety of tomato plant is normally distributed with mean µ and standard deviation σ = 2.4. If you select a simple random sample of four plants of this variety and measure the time until maturity. The four times, in days are: 63, 69, 62, 66. If I wanted the margin of error for the 99% confidence interval to be ± 1 inch, I should select a simple random sample of size

a. Based on these data, a 99% confidence interval for μ , in days, is______ .
b. If I wanted the margin of error for the 99% confidence interval to be ± 1 inch, I should select a simple random sample of size.

User Jsherk
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1 Answer

2 votes

Answer:

a)
63.5-2.58(2.4)/(√(4))=60.404


63.5+2.58(2.4)/(√(4))=66.596

So on this case the 99% confidence interval would be given by (60.404;66.596)

b)
n=((2.58(2.4))/(1))^2 =38.34 \approx 39

So the answer for this case would be n=39 rounded up to the nearest integer

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 for the sample


\mu population mean (variable of interest)

s represent the sample 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)(\sigma)/(√(n)) (1)

In order to calculate the mean we can use the following formula:


\bar X= \sum_(i=1)^n (x_i)/(n) (2)

The mean calculated for this case is
\bar X=63.5

Since the Confidence is 0.99 or 99%, the value of
\alpha=0.01 and
\alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that
z_(\alpha/2)=2.58

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


63.5-2.58(2.4)/(√(4))=60.404


63.5+2.58(2.4)/(√(4))=66.596

So on this case the 99% confidence interval would be given by (60.404;66.596)

Part b

The margin of error is given by this formula:


ME=z_(\alpha/2)(\sigma)/(√(n)) (a)

And on this case we have that ME =1 and we are interested in order to find the value of n, if we solve n from equation (a) we got:


n=((z_(\alpha/2) \sigma)/(ME))^2 (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.005;0;1)", and we got
z_(\alpha/2)=2.58, replacing into formula (b) we got:


n=((2.58(2.4))/(1))^2 =38.34 \approx 39

So the answer for this case would be n=39 rounded up to the nearest integer

User Johannes Overmann
by
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