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.

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".

[tex]\bar X[/tex] represent the sample mean for the sample

[tex]\mu[/tex] 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:

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

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

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)

The mean calculated for this case is [tex]\bar X=63.5[/tex]

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], 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 [tex]z_{\alpha/2}=2.58[/tex]

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

[tex]63.5-2.58\frac{2.4}{\sqrt{4}}=60.404[/tex]

[tex]63.5+2.58\frac{2.4}{\sqrt{4}}=66.596[/tex]

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:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (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:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (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 [tex]z_{\alpha/2}=2.58[/tex], replacing into formula (b) we got:

[tex]n=(\frac{2.58(2.4)}{1})^2 =38.34 \approx 39[/tex]

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